Raikage vs Tsunade

dadudemon

Originally posted by Astner
From my understanding Tsunande is a class-100 warrior. This can be proven with a simple calculation of when she's using Gamabuta's katana.

A regular katana handle is approximately 4 cm, and the length of Gamabuta's katana's handle is a approximately 3/2 of Tsunande's length (or 2 meters).

A reasonable assumption would be that Gamabuta's katana has the same density as a regular katana.

So the formula is basically:

(/)^3 * =

(2/.05)^3 * 2 = 250 000 kg (= 250 tons)

Raikage (Also known as "A" wink

seem to be more deadly than strong. He's as fast as the Yellow Flash, his lightning shield increases his defense and offense dramaticly and he's pretty powerful to begin with.

I'm going with Tsunsande for now (in terms of strength) based on the evidence for now. But I'm going to reserve judgement until I see the upper limits of A's strength.

While I love the fact that there is someone on here actually using their brain to figure shit out, you're numbers are way off, here, and the calculation isn't working for me, at all.

First, find the density of the handle. Then, find the density of the metal. Then, find the volume of the handle and complete blade.

Find the volume similar to how archimedes calculated a more precise version of pi: get more accurate with a geometric figure: in this instance, it would be a regular cuboid, then angle the edges off and recalculate volume, then add two edges to that one edge, and recalculate just the change in volume for the "right triangle" of the new sub-figure, and so forth.

This is similar to how you calculate your own volume in physics for some cool projects you can do in college.

Anywho, we can assume the handle is wood or a firm rock...we can do the calculation for both, actually.

As I don't feel like taking the time to draw this shizer out and do the trig for it, feel free.

Tell us what you come up with. If you need me to better explain what I mean, just send me a PM.

Then you could really find out what the mass is, which I think you were close with.

Demonic Phoenix

Originally posted by Q99
Also hypothetically, Tsunade could use her forehead-chakra to power any jutsu, including a strength-based one.

We've just happened to have only seen her use it for defensive jutsu, but if you think about it that chakra store is a major threat.

That, while interesting to watch, would be severely unnecessary as she's insanely strong already; unless she's up against something like a Tailed Beast, or a hypothetical Juubi-enhanced Madara and wants to punch them to Kingdom Come. ermm

Astner

Originally posted by dadudemon
While I love the fact that there is someone on here actually using their brain to figure shit out, you're numbers are way off, here, and the calculation isn't working for me, at all.
The numbers are completely accurate under the given assumption that Gamabuta's oversized katana has the same properties of a regular katana. In other words the same uniform or nonuniform matter distribution. Which is the best estimation one could make.

Originally posted by dadudemon
First, find the density of the handle. Then, find the density of the metal. Then, find the volume of the handle and complete blade.
Completely unnecessary and would mean that we would have to resort to further estimations that would lower the accuracy of the results.

Originally posted by dadudemon
Anywho, we can assume the handle is wood or a firm rock...we can do the calculation for both, actually.
It's likely to be made out of regular wood, like any ordinary katana. We can further see the swirl shaped pattern at the handle indicating the very same.

You could make a calculation for substituting it with some form of rock. This would in fact be easy if you at home had a similar katana with a detachable blade (a scale and a 500 ml cup of water). But it would only lead to a higher "low estimation" beneath anyone's interest.

dadudemon

Originally posted by Astner
The numbers are completely accurate under the given assumption that Gamabuta's oversized katana has the same properties of a regular katana. In other words the same uniform or nonuniform matter distribution. Which is the best estimation one could make.

Wrong. You cubed the same number...that's not accurate (unless the katana is a perfect cube). You shouldn't have done that. You also cubed a relational value between your estimate on a regular katana and gama's katana: closer to reality, but not accurate as Gama's katana is not proportional to a regular Katana AND your "regular" katana measures are off.

To find the volume of a regular cuboid: length*width*height (x*y*z)

Your calculation should look like this (x*y*z)*dh+(a*b*c)*db

x, y, and z are the dimensions of the hilt. a, b, and c are the dimensions of the blade. dh is the density of the hilt. db is the density of the blade.

That assumes the the hilt and the blade are regular cuboids (each side of the six sides of the cuboid are rectangles.) That's not true, and would give an estimate too far over the actual values. The blade, alone, would be less than half the volume of the cuboid estimate.

Originally posted by Astner
Completely unnecessary and would mean that we would have to resort to further estimations that would lower the accuracy of the results.

Completely NOT unnecessary as your numbers are off AND it would INCREASE accuracy to have a proper density and volume calculation. Please tell me, Astner, how the **** increasing the accuracy of the volume of the blade is going to DECREASE the accuracy of the measurement?

Originally posted by Astner
It's likely to be made out of regular wood, like any ordinary katana. We can further see the swirl shaped pattern at the handle indicating the very same.

But, you see, it's not an ordinary Katana, if it were wood, it'd be treated, probably be more dense than some drift wood, and you'd also have to consider that the hilt may be something other than wood. But, we can almost be sure that the hilt is unwrapped Ho, which was the wood usually used for katanas. It'd be treated and more dense than just some Ho you cut out of the tree, as well. DO you have that density measure?

Originally posted by Astner
You could make a calculation for substituting it with some form of rock. This would in fact be easy if you at home had a similar katana with a detachable blade (a scale and a 500 ml cup of water). But it would only lead to a higher "low estimation" beneath anyone's interest.

No, it would lead to a more accurate measure.

Also, the sword would have to be made of a really strong metal: not simply super steel. It would have bent and flexed under such abrupt movement, from Tsunade. This tells us that the sword is made of materials other than hardened steel and Ho. This also tells us that the sword is just supposed to be really ****in' huge and a measure is not really obtainable, but we can still estimate.

Because you are not interested in actually making the correct calculation, I'll go ahead and do it for you. I'll draw pictures and so forth.

Q99

Originally posted by Demonic Phoenix
That, while interesting to watch, would be severely unnecessary as she's insanely strong already; unless she's up against something like a Tailed Beast, or a hypothetical Juubi-enhanced Madara and wants to punch them to Kingdom Come. ermm

True, I'm just sayin' smile

psycho gundam

that'd be cool if tsunade could use her stored chackra to amp her physical blows

Astner

Originally posted by dadudemon
Wrong. You cubed the same number...that's not accurate (unless the katana is a perfect cube).
Comprehension failure.

Example: Rectangle (i.e. not perfect cube)

Dimensions:

Length: 2 m
Width: .5 m (2/4)
Breadth: .25 m (2/8)

Volume: 1 * 0.5 * 0.25 = 0.25 m^3

We want the volume of the same type of rectangle with an arbitrary length L.

Apply the same calculation.

= * (/)^3 = 1/4 * L^3 / 8 = L^3 / 2^5

Now lets control check:

** = L * L/4 * L/8 = L^3 / 2^5

Of course this works for any given function of x, y, z as explained beneath.

Originally posted by dadudemon
To find the volume of a regular cuboid: length*width*height (x*y*z)
Where x, y and z may be an arbitrary function (in other words, it also covers any other conceivable shape).

For instance, apply spherical coordinates and:

x = r*cos(a)*sin(b)
y = r*sin(a)*sin(b)
z = r*cos(b)

And you'll get a sphere.

Originally posted by dadudemon
Your calculation should look like this (x*y*z)*dh+(a*b*c)*db

x, y, and z are the dimensions of the hilt. a, b, and c are the dimensions of the blade. dh is the density of the hilt. db is the density of the blade.
Once again, completely unnecessary. The formula above is derived from a triple integral that covers an arbitrary non-uniform distribution of mass over a three dimensional body.

Originally posted by dadudemon
Completely NOT unnecessary as your numbers are off AND it would INCREASE accuracy to have a proper density and volume calculation. Please tell me, Astner, how the **** increasing the accuracy of the volume of the blade is going to DECREASE the accuracy of the measurement?
Fact is, you're decreasing the accuracy of the volume by treating it as rectangles, like you do.

Originally posted by dadudemon
Because you are not interested in actually making the correct calculation, I'll go ahead and do it for you. I'll draw pictures and so forth.
Once again, it's accurate. At best your rectangles will come 90% to that of my result.

Astner

Here's the sphere (MATLAB is an numerical program, so it's not 100% smooth but down to the pixel it should be).

http://img163.imageshack.us/img163/6012/bild7tx.th.png

dadudemon

Originally posted by Astner
Comprehension failure.

Example: Rectangle (i.e. not perfect cube)

Dimensions:

Length: 2 m
Width: .5 m (2/4)
Breadth: .25 m (2/8)

Volume: 1 * 0.5 * 0.25 = 0.25 m^3

We want the volume of the same type of rectangle with an arbitrary length L.

Apply the same calculation.

= * (/)^3 = 1/4 * L^3 / 8 = L^3 / 2^5

Now lets control check:

** = L * L/4 * L/8 = L^3 / 2^5

Of course this works for any given function of x, y, z as explained beneath.

Several things:

1. Not a rectangle, but a rectangular cuboid.
2. Where are you getting your numbers from? Those seem arbitrary. Also, if those numbers are supposed to be used for Gama's sword, they are way off.
3. 1*.5*.25 does not equal .25*.25*.25. I don't even have to do that math on that to know that you're way off base, there. I have no idea what you were trying to accomplish, there. No rearranging of decimals or adding in a factor corrects for your .25 error.
4. You're volume measure is correct. If you run across a congruent sword that has a depth measure (because that would be the quickest and easiest way to measure it) of 100cm, it is 100^3 times voluminous (see, very easy to do). Simply multiply 1000cm (the small sword's volume) by 1,000,000 (the cube of the factor) to get the right answer of 1,000,000,000cm^3. If you don't believe me, here everything is, worked out, the long way: length 10000cm (100*100cm), width 1000cm (100*10cm), depth 100 (100*1cm). Volume = 1,000,000,000cm^3. That's just assuming a rectangular cuboid, though. It's not as accurate as other forms of measure, but it will be close than other things you could do. It's just faster that way.
5. Reading over your first post, I see that there is not even congruency with what you just posted, above. You originally posted this:

(/)^3 * =

Then calculated this:

(2/.05)^3 * 2 = 250 000 kg (= 250 tons)

That has so many problems with it, it's ridiculous.

Here are all the problems:

a.)The Density of a regular katana is significantly greater than 2. It's closer to 7. Density of high quality Japanese Katana steel is probably above 8, or right below 8. Hilt will decrease the density, significantly, of course, but probably not much below 7.
b.) You incorrectly estimated the length of the sword as 2 meters. The length is significantly greater than that and any degree of miss-measure is very significant to the final outcome because the factor is cubed.

This scan here shows us a different story:
http://www.mangafox.com/manga/naruto/v19/c170/16.html

The top right panel shows a Tsunade, at an angle to the width of the blade (not a straight on perpendicular shot), has Tsunade almost standing all the way up. The next shot shows the sword's smallest length on (the depth of the hilt). It's even "longer" than Tsunade standing straight up (even account for a bit of her legs being cut off, due to the "upwards" angle of the "shot" wink

. Tsunade is 163cm in height. I used a straight edge and measured, and came up the hilt being about 3 times as "wide" as Tsunade is tall. (I accounted for the angle and Tsuande's slightly bent knee. That's 163*3 = 489cm.
c.) Let's check out the triple integral.

Take rectangular cuboid "A" whose dimensions are (x, y, z): 4, 2, 3cm. Volume of A is 24cm^3.

Take congruent figure "B". We only know one side of B and that is z. It is 6. (simply by that, we know all of the other sides, as well: 8 and 4.) Using your math, that comes to:

24cm^3 (old volume) * (6(new length)/3(previous length)^3) =

8*24cm = 192

Does that check?

6*8*4 = 192

Yes.

But, wait...is that the same thing you were talking about earlier?

Let's check:

You said this:

Originally posted by Astner
(/)^3 * =

(2/.05)^3 * 2 = 250 000 kg (= 250 tons)

No, that's not what you said earlier. You have the density of regular katana in there. Well, later, you did clarify that it was very simply the z dimension length, no doubt. So, I assume that that was just a mistake on your part and you actually meant the volume of the original object. In which case, you'd be correct (I"m a nice guy, giving you the benefit of the doubt.) However, there's still the problem with you concluding "mass" in there, totally throwing off your calculation, big time. I apologize if this comes off as rude, but, your math is totally off, there. I could understand if you did a stupid mistake, like I have, by reducing like terms in a force calculation, only ending up with weight as your final measure (trust me, I did that..forgot that my answer would be in 32.17lbs*ft/sec/sec). But this is not anything like that. It's so far out there that I have no clue what you were doing there.

However, I do have some dimensions of a bamboo "hilted" sword:
http://www.nihonzashi.com/shinken_bamboo_stick_katana.aspx

The nakago width is 2.8448 cm. Let's go with that as our measure for Gama's sword. However.....how the hell are you going to calculate the volume of your bamboo stick Katana? Oh, that's right, you'll have to do it my way. (unless you have one sitting around that you'd like to try a fluid (water) displacement measure with.) smile

Density? Easy. No problem. The metal is easy to figure out as we already have stuff on that. The bamboo is another problem because it's a treated nakago. Bamboo is about .97g/cm^3, untreated. Just guessing, but I'd put the treated blade at 3g/cm^3. Now we just got to figure out the volume of the hilt and then the entire blade. facepalm BTW, just looking on the internet...it looks like the average nakago "depth" is 17mm. That's the dimension you were trying to use, btw.

We can then use my way of gradually making things more accurate until we come up with a number that is close...without going too in depth. Or, dunk a blade (not preferable...unless you have one?)

Originally posted by Astner
Where x, y and z may be an arbitrary function (in other words, it also covers any other conceivable shape).

For instance, apply spherical coordinates and:

x = r*cos(a)*sin(b)
y = r*sin(a)*sin(b)
z = r*cos(b)

And you'll get a sphere.

No, where x, y, and z are the Cartesian Coordinate system (to be read "x axis, y axis, and z axis.) I apologize for assuming too much as I thought everyone would know what I was talking about, especially since I clarified that just a few short sentences later.

You're way would have been this way: f(x), f(y), and f(z) (or simply f(x,y,z).

Originally posted by Astner
Once again, completely unnecessary. The formula above is derived from a triple integral that covers an arbitrary non-uniform distribution of mass over a three dimensional body.

No-no, that works. As I stated several sentences above, the measure works just fine for volume....but not mass like you tried to do.

Originally posted by Astner
Fact is, you're decreasing the accuracy of the volume by treating it as rectangles, like you do.

That's not what I was doing, btw. I was saying to use that as a starting point and then work your way down to something more accurate until you're satisfied with the volumetric results (in other words, you would get to some arbitrary point to where the volume decrease caused by adding another "edge" to the "curve", for the nakago, is too small to matter. That could be 10cm^3...it could be 1cm^3. (This is what computer animators do: the curves on their animations are actually polygons...with lots and lots of edges.)

Originally posted by Astner
Once again, it's accurate. At best your rectangles will come 90% to that of my result.

Yeah, cause a rectangular cuboid is exactly what I said you should use for the final volumetric estimate. no expression

I'll still draw pictures for you and do the measure my way....next weekend. I am out of time.

Just had an epiphany: for the hilt, it's as easy as find the radius for the curve on each side of the hilt. Then, calculate the volume of the newly formed "rectangular cuboid" left after calculating the volume of the "cylinder" and add the two volumes together. You'd only have to measure from the center of the hilt to the top edge, from the part the the hilt starts to curve. Then subject the double of that measure (cause that would be the diameter) from the top to bottom measure of the nakago/hilt to end up with the y coordinate for a the new rectangular cuboid. Again, just simply add the volumes of the two together to come up the measure. In fact, there should be a relational property between the top to bottom measure, and the radius measure: it should vary directly.

Astner

Originally posted by dadudemon
Several things:

1. Not a rectangle, but a rectangular cuboid.
Or a rectangular parallelepiped. How does this apply? It doesn't. The body is obviously clear to anyone reading it.

Originally posted by dadudemon
2. Where are you getting your numbers from? Those seem arbitrary. Also, if those numbers are supposed to be used for Gama's sword, they are way off.
I used an arbitrary L for the scalar, and fixed values for the original body.

Originally posted by dadudemon
3. 1*.5*.25 does not equal .25*.25*.25. I don't even have to do that math on that to know that you're way off base, there. I have no idea what you were trying to accomplish, there. No rearranging of decimals or adding in a factor corrects for your .25 error.
Typo, I originally had different dimensions which I then changed, but point taken.

Originally posted by dadudemon
4. That's just assuming a rectangular cuboid, though. It's not as accurate as other forms of measure, but it will be close than other things you could do. It's just faster that way.
Inaccurate:

Example: Sphere

Radius: r

Volume = c*r^3
Where: c = (4 π / 3)

Determine the volume of a sphere with radius R.

= *( / ) = c*r^3 * (R^3 / r^3) = c*R^3

Works for every body. Period.

Originally posted by dadudemon
5. Reading over your first post, I see that there is not even congruency with what you just posted, above. You originally posted this:

(/)^3 * =
Heh, I actually noticed that instead of writing I wrote . Yeah, let me fix that.

The actual calculation would be:

= (/)^3 * * = = (/)^3 * = (2/.5)^3 * 2 = 250 000 = 250

Originally posted by dadudemon
b.) You incorrectly estimated the length of the sword as 2 meters. The length is significantly greater than that and any degree of miss-measure is very significant to the final outcome because the factor is cubed.
Once again, I meant the width quit playing with semantics. As the outcome is identical.

Originally posted by dadudemon
No, where x, y, and z are the Cartesian Coordinate system (to be read "x axis, y axis, and z axis.) I apologize for assuming too much as I thought everyone would know what I was talking about, especially since I clarified that just a few short sentences later.

You're way would have been this way: f(x), f(y), and f(z) (or simply f(x,y,z).
Cartesian coordinates can be transformed to spheric coordinates.

Spheres exists in a Cartesian coordinate systems too it's simply written x^2 + y^2 + z^2 = R, where R is the radius.

Originally posted by dadudemon
No-no, that works. As I stated several sentences above, the measure works just fine for volume....but not mass like you tried to do.
Assuming the density is spread out in a similar fashion in both bodies (the original and the resized one) it's identical.

psycho gundam

facepalm @ everything

Demonic Phoenix

Originally posted by psycho gundam
facepalm @ everything

I second this.

Could you guys debate this in PMs plox? Pretty soon, the posts will be dedicated to Math, rather than Tsunade pwning Raikage in the Arm-wrestling contest.

dadudemon

Originally posted by Astner
Or a rectangular parallelepiped. How does this apply? It doesn't. The body is obviously clear to anyone reading it.

No amount of google searching will make your previous statement correct. You called a 3D object a 2D name. And, no, it isn't obvious, which is why I made the clarification. No one cares but you and I. no expression

Originally posted by Astner
I used an arbitrary L for the scalar, and fixed values for the original body.

I was referring to 2, .5, and .25, not the L.

Originally posted by Astner
ITypo, I originally had different dimensions which I then changed, but point taken.

Fair enough...but....that's far from a typo, man.

Originally posted by Astner
IInaccurate:

Example: Sphere

Radius: r

Volume = c*r^3
Where: c = (4 π / 3)

Determine the volume of a sphere with radius R.

= *( / ) = c*r^3 * (R^3 / r^3) = c*R^3

Works for every body. Period.

You misunderstood what I was saying, most definitely.

Here's why.

1. I just got done showing you your own work in a different way.
2. If you had actually understood what I did, you would have realized I just did the same exact calculation you did. laughing

Reread what I did, and see if you can figure out what I did there. I'll give you a hint: the 100 times factor is found with...division. You cube the factor...then multiply that by the old volume. ZOMG! What did I just do? laughing

3. Why didn't you recognize it? I don't want to speculate because I might offend you. You tell me why.

Originally posted by Astner
Heh, I actually noticed that instead of writing I wrote . Yeah, let me fix that.

The actual calculation would be:

= (/)^3 * * = = (/)^3 * = (2/.5)^3 * 2 = 250 000 = 250

LOL!

I see what you this time.

You eliminated like terms in the conversion factor multiplication:

Specifically, you did this:

((1cm^3)/1)*(1g/(cm^3))

Multiply the two fractions straight across:

(1gcm^3)/(1cm^3)

Eliminate like terms:

(1gcm^3 )/(1cm^3)

What are you left with? 1 gram...aka, mass.

That makes MUCH more sense. However, your numbers are still off...because, like I was saying earlier, I eliminated like terms in a force calculation and ended up with just mass, which was the incorrect result. You've reduced too far, man. I'll run an example to show you where you went wrong:

Running easy to use, arbitrary numbers:

Pretend we have a small sword that has a side that is 1cm in length and weighs 300g. Let's say that it's volume is 2000cm^3 and a density of 6.

New, bigger sword, has the same side used above, with a measure of 40cm.

Check:

(40/1)^3 * 2000cm^3 * 6g/cm^3 = 128000000cm^3 * 6g^cm^3 = 768000000gcm^3/cm^3.

Simplify:

768000000g. So its mass is 768000Kg OR, its mass is 768 tonnes.

Let's try it your way:

Using my same numbers (which I chose for an obvious reason, as you'll see):

(40/1)^3 * 300g= 19200000g.

Simplify to kgs then tonnes: 19200Kg and 19.2 tonnes.

Check work:

Does 19.2 tonnes = 768 tonnes?

No.

Let's look at your math to see if and where you made a mistake:

Originally posted by Astner
(2/.5)^3 * 2 = 250 000 = 250

hmmm.

?

What does that mean? That's not any standard mathematical notation that I'm aware of...but, cool. I'll pretend it's not there.

2/.5 = 4.

4^3 = 64

64 * 2 = 128Kg

Check: does 128g = 250000Kg?

No.

Let's review:

What did you do?

You made a typo. 2/.05 is what you had originally put.

That only changes the result by a factor of 1000 (factor of the component itself is 10, but since it's cubed, the final factor is 1000, so you only need to multiply the final result by 1000.)

That results in a new number of 128000Kg.

Recheck for your typo:

Does 128000Kg = 250000kg?

No.

Let's factor in an a value that I got from simply dividing 250000 by 128000:

That's a factor of 1.953125.

Let's take some liberty and assume that you were using nice round numbers and you forgot to put in there that you also multiplied it by a density of 2, then rounded down to 250000.

I have you a HUGE benefit of the doubt.

However, there are several problems with what you've done:

1. You made a typo: it's .05. This is understandable.
2. You dropped density when you shouldn't have. Density is part of the calculation, despite the fact that cm^3 is reduced out of the results, leaving you with just mass.
3. You have some odd notation in your equation.

Even accounting for the odd typo mistake, you still didn't get it right: you have to keep density in there, or at least the value of density.

Originally posted by Astner
Once again, I meant the width quit playing with semantics. As the outcome is identical.

No, it is not semantics.

You used the length and then proceeded to use the thickness of the hilt to do the division: waaaaay off, dude. .05m = 5cm. Have you EVER seen a regular katana with a nakago width of 5 cm? Have you ever seen a regular nakago with a height of 5cm? It should be no to both.

The outcome is NOT identical in either case because you're doing your math wrong and using absurd numbers. In any case, it is not semantics, at all.

Originally posted by Astner
Cartesian coordinates can be transformed to spheric coordinates.

Spheres exists in a Cartesian coordinate systems too it's simply written x^2 + y^2 + z^2 = R, where R is the radius.

See below:

Originally posted by dadudemon
No, where x, y, and z are the Cartesian Coordinate system (to be read "x axis, y axis, and z axis.) I apologize for assuming too much as I thought everyone would know what I was talking about, especially since I clarified that just a few short sentences later.

You're way would have been this way: f(x), f(y), and f(z) (or simply f(x,y,z).

That reply applies to your post, yet again.

Originally posted by dadudemon
Assuming the density is spread out in a similar fashion in both bodies (the original and the resized one) it's identical.

It doesn't work because you need to apply density to the problem. That's on top of the other mistakes I pointed out.

One last complaint, Astner: kilograms never convert to tons by dividing by 1000. They always convert to tonnes, or if you're in America, metric tons. Now, this item I brought up was truly word semantics. You can disregard. Using the correct terms avoids confusion and it also makes you look smarter (which is needed to be taken seriously when you have a good point).

And, no, this will be argued in here because it determines Tsunade's strength, properly. Also, these same types of discussions NEED to be had much more often. The level of discussion needs to be raised beyond speculation and to actually land up on some more solid math. I wish everyone would start considering math as a viable argument in these debates as it would be much harder to pull a fast one, as long as everyone paid attention.

We can then move on to Raikage.

Also, we need to decide which is better/more powerful: Raikage's Raiton Armor or Medical Ninjutsu's ability to "penetrate". Wouldn't Tsunade's hand bounce right off of Raikage's armor.....or would Tsunade's absurdly awesome medical ninjutsu null and penetrate right through the armor and flesh?

marwash22

WTF! I'm on this forum because I'm trying to ignore my Calc prof...

dadudemon

Originally posted by marwash22
WTF! I'm on this forum because I'm trying to ignore my Calc prof...

This is mostly physics......I think.

Astner

Originally posted by dadudemon
No amount of google searching will make your previous statement correct. You called a 3D object a 2D name. And, no, it isn't obvious, which is why I made the clarification. No one cares but you and I. no expression

Once again you're playing with semantics, that what I was trying to denote. A rectangle in three dimensions is a rectangular parallelepiped.

As for google searching, the calculation of the volume of a parallelepiped is used to prove the function of the determinant, in linear algebra.

Originally posted by dadudemon
I was referring to 2, .5, and .25, not the L.
So did I. 2, .5 and .25 aren't arbitrary. They're fixed values.

Originally posted by dadudemon
You misunderstood what I was saying, most definitely.

Here's why.

1. I just got done showing you your own work in a different way.
2. If you had actually understood what I did, you would have realized I just did the same exact calculation you did. laughing

3. Why didn't you recognize it? I don't want to speculate because I might offend you. You tell me why.
No, after I disproved your claim of that the calculation only applied to perfect cubes, you moved the goalpost (logic fallacy!) to that it only applied to cuboid bodies. Which I further disproved, by showing you that it even work on spheres.
It's not that I don't understood what you did, it's just that it's wrong. Since you go more in-depth with this error further down in this post, I'll address it then.
It's errorous, plain and simple.

Originally posted by dadudemon
LOL!

I see what you this time.

You eliminated like terms in the conversion factor multiplication:

Specifically, you did this:

((1cm^3)/1)*(1g/(cm^3))

Multiply the two fractions straight across:

(1gcm^3)/(1cm^3)

Eliminate like terms:

(1gcm^3 )/(1cm^3)

What are you left with? 1 gram...aka, mass.
Then I'm sorry to inform you of that you have to be blind. In each one of my calculations I used SI-units, in this case and .

Originally posted by dadudemon
Pretend we have a small sword that has a side that is 1cm in length and weighs 300g. Let's say that it's volume is 2000cm^3 and a density of 6.

New, bigger sword, has the same side used above, with a measure of 40cm.

Check:

(40/1)^3 * 2000cm^3 * 6g/cm^3 = 128000000cm^3 * 6g^cm^3 = 768000000gcm^3/cm^3.

Simplify:

768000000g. So its mass is 768000Kg OR, its mass is 768 tonnes.
Do you even understand what you're doing? I think not, because you're directly implying that a regular katana weights:

2000^3 * 6^-3 = 12000 || 12 kg

Considering the fact that katanas are light weapons intended for swift strikes and rarely weights more than 2 kg you're way off.

Originally posted by dadudemon
hmmm.

?

What does that mean? That's not any standard mathematical notation that I'm aware of...but, cool. I'll pretend it's not there.
It's a common notation for a void unit such as angels, the friction coefficient. Pretty much junior high basics, what's sad is that you couldn't even identify it with dimension analysis.

Originally posted by dadudemon
2/.5 = 4.

4^3 = 64

64 * 2 = 128Kg

Check: does 128g = 250000Kg?

Let's review:

What did you do?

You made a typo. 2/.05 is what you had originally put.

That only changes the result by a factor of 1000 (factor of the component itself is 10, but since it's cubed, the final factor is 1000, so you only need to multiply the final result by 1000.)

That results in a new number of 128000Kg.

Recheck for your typo:

Does 128000Kg = 250000kg?

No.
I'll give you this, the sword weighs 128 tons.

Originally posted by dadudemon
2. You dropped density when you shouldn't have. Density is part of the calculation, despite the fact that cm^3 is reduced out of the results, leaving you with just mass.
Once again, I use SI-units. I never dropped the density. As:

http://img297.imageshack.us/img297/480/ton.gif

Originally posted by dadudemon
One last complaint, Astner: kilograms never convert to tons by dividing by 1000.
http://img42.imageshack.us/img42/5143/tons.jpg

marwash22

Originally posted by dadudemon
This is mostly physics......I think. lol, that's not quite the point.

King Kandy

This is too much for me to nit pick, and I don't car enough to try. But I will toss in on one issue: I've never seen used as notation in any math or physics i've taken.

psycho gundam

http://img6.imageshack.us/img6/1407/generalshermansequoia.jpg

general sherman (largest single living thing on earth) weighs approximately 6200 tons, and it's 275 feet tall.

http://en.wikipedia.org/wiki/General_Sherman_%28tree%29#Specifications

pic has human size comparison so i guess it can be used for something *shrugs*

Astner

Originally posted by dadudemon
No I'm not. It's quite specific:

You can call semantics all you want: a rectangular cuboid is not a rectangle. You effed up, back-tracked with some google searching to save face, and now you're pretending it's semantics.
Bullshit. What's you're getting at is semantics. Rectangle or rectangular cuboid, it doesn't matter. It was the title of the example, not the example itself.

Does that work with your college prof-- teacher too? "How should I've known it wasn't a 2 dimensional body, the title confused me, even though no one else was confused by it."

Nevertheless you were disproved, when you suggested that the given formula only applied to perfect cubes. This fact severely wounded you ego and in response to that you got on the fact that I named the example after a 2 dimensional object.

Originally posted by dadudemon
1. Oh really? You mean the "calculation" that applies to rectangular cuboids as well? Nice try. Oh, right, it's simple formula.
Except that I proved that it applied to spheres as well. See example sphere.

Originally posted by dadudemon
You fell into the trap I set up for you, didn't see it, tried to argue against it, and thereby argued against yourself, but you didn't see it, I had a laugh, you didn't see that, and now I'm face palming. You're so out of it, it's ridiculous. It's not funny, anymore, man. It's pathetic. Sorry to be a jerk...but you've taken this too seriously, and are being so blind to what you're doing that this discussion has lost its flavor.
It's painfully obvious that you have no idea of what you're talking about. Unless you planned to make a fool out of yourself, in which case you succeed.

Originally posted by dadudemon
And, no, what you did was add functions to an x,y,z coordinate system that I quite specifically clarified as just coordinates, not functions.
Because functions applies as well you cretin. The formula I devised is general and not a special case, I explained this to you already. But even after a page of posts your ignorance remains.

Originally posted by dadudemon
Full disclosure: I figured you were just an online expert and did not have any real math skills.
Engineering physics, the most prestigious maths and physics related university programme available. You might want to watch what you type before I throw complex curve integrals at you.

Originally posted by dadudemon
That should be apparent by your atrocious calculations.
Numerical errors isn't equivalent to atrocious calculations.

Originally posted by dadudemon
So, I figured on setting up a little test to see if you would pass. You didn't. You failed miserably and on top of that, you had the brazen audacity to tell me what I was doing was wrong, despite it being the very same work you were doing, just worded differently. I wouldn't have clowned you had you kept it civil and just admitted fault. Instead, you had to get combative and insulting.
Let me put you up with a test then. I'm even going to be nice and tell you that it's a test. I further am going to formulate this test directly as a well defined problem.

The functions f and g are analytic in the limited area D, continuous in D<conjugate> (= D <union> <part. diff>D) and have the same root in D, calculating with multiplicity. Show that |f(z)| = |g(z)| ≠ 0 for all z <in> <part. diff>D then there exists a complex number L, |L| = 1 do that f(z) = L*g(z) for all z <in> D<conjugate>.

D<conjugate> = the conjugate of D
<union> = union symbol
<part. diff> = partial differential symbol (dev)
<in> = element symbol

There is no way with your restricted knowledge that you'll solve this problem, which I solved at an exam 6 months ago.

(for you it's the give up button and the only way you'll ever find out).

Originally posted by dadudemon
2. Oh really? You still don't see it? Man, that's very annoying. I even gave you the most absurdly obvious hint ever and you still don't see it.
Yeah, you're so witty playing these "obvious" mind games that only you understand. Especially when 50% of the content of your post is nothing but a long reminisce.

Originally posted by dadudemon
Boy, you sure pwned the shit out of me, man.
I'm aware, sadly you're not.

Originally posted by dadudemon
Okay, dude, seriously.

WTF?

You do realize that I just used smaller units of SI units, right? You know...the whole entire point of the SI system? If you've had any real lab (or book) time, you'd know that 1g/cm^3 is the same thing as 1000kg/m^3.
In the same sense, an inch is a smaller version of the SI-unit meter. It still doesn't make it an SI-unit you dolt. This is basic stuff brought up in junior high.

Originally posted by dadudemon
All you can do is fault my numbers for not translating into real world masses for katanas? erm

No, I was just trying to be polite, not pointing out all your logical shortcomings. But thinking a katana weights as much as an modern drawer takes rare stupidity.

Originally posted by dadudemon
Wow! OH MY GOSH! You mean arbitrary numbers I pulled out of my ass for the sake of a mathematical demonstration didn't work out to be perfect numbers for a real world object?
You seem to struggle with the concept of arbitrary numbers.

{a + b = c}, is a system of arbitrary numbers as each of the parameters a, b and c can be set (technically you have 2-degrees of freedom, but that's beside the point).

{1 + 2 = 3}, isn't a system of arbitrary numbers.

Originally posted by dadudemon
Imagine that! I must of just...improvised those numbers because they were easier to work with.
But they aren't arbitrary.

Originally posted by dadudemon
But, let's look at what I ACTUALLY was illustrating to you and that you missed entirely: your math was wrong. no expression

You screwed up on so many levels it's ridiculous.
The hypocrisy residing within the fact that you haven't been able to point out any of them.

Originally posted by dadudemon
I have never seen that "notation" used in trig/calc/geom, ever.

The symbol for angles is the angle symbol: ∠

And the symbol for the friction coefficient is mu: μ
Which has nothing whatsoever to do with units.

In the SI system:
Mass isn't measured in m, it's measured in .
Force isn't measured in F, it's measured in ^-2 (Newton)

Originally posted by dadudemon
What's sad is that it plays not part in the actual calculation, you continue to throw random words out there as if they have any meaning in context, and you pretend that "" was ever used in any math course I have taken, from middle school to college.
If you haven't come across units and dimension analysis you haven't taken any courses in physics whatsoever (not even junior high). And I have trouble believing that you attend college, but that's neither here or there.

Fact is, it is vital to the calculations made, since it indicates that they're made correctly.

Originally posted by dadudemon
No, you won't give this to me because I showed you where it was wrong and told you it was wrong in my first reply to you. 128 tonnes is still wrong.
Actually I do give you that because it's correct.

Originally posted by dadudemon
You incorrectly simplified out ALL of density when you shouldn't have.
Once again, no I didn't. Instead of committing your stupid mistake of--for no reason--separate the volume from the density when neither is known. I kept them the same. Since the density of the regular katana and Gamabuta's katana is assumed to be identical.

http://img576.imageshack.us/img576/954/mimetex.gif

Originally posted by dadudemon
So if I post something more credible than that, I win, right?
No, it was an attempt to prove to you that you were playing with semantics. "It's tonnes not tons" while true it holds no relevance to the calculation at hand. Sadly this is were your focus is at. But OK I'll promise you that I will stay with kilograms since it apparently makes you feel better.

Originally posted by dadudemon
Feel free to chime in with some sort of help with this guy. It's like...he's trying to use college math but botching it, horribly, and failing to make simple calculations.
Fun fact, all of this is high school level maths (aside from the "test" I gave you, have fun with that).

Either way, failing with calculations and failing to hit the right buttons on the calculator are two different things. I'm used to MATLAB and Mathematica, not the MAC's simple calculator.

dadudemon

Originally posted by Astner
Bullshit. What's you're getting at is semantics. Rectangle or rectangular cuboid, it doesn't matter. It was the title of the example, not the example itself.

Does that work with your college prof-- teacher too? "How should I've known it wasn't a 2 dimensional body, the title confused me, even though no one else was confused by it."

Nevertheless you were disproved, when you suggested that the given formula only applied to perfect cubes. This fact severely wounded you ego and in response to that you got on the fact that I named the example after a 2 dimensional object.

So, basically, you bring in a bunch of strawman arguments because you simply can't say, "I was wrong." My professors or teachers have nothing to do with you posting something wrong.

Also, you need to quote me where I said that the given formula only applied to perfect cubes. Please, I beg you, show me where I said that.

laughing

Originally posted by Astner
Except that I proved that it applied to spheres as well. See example sphere.

facepalm

It's called context and you don't understand it. You can't even follow a simple internet conversation. L*W*H is a very simple formula. That was the context of the convo, you could not follow it.

Originally posted by Astner
It's painfully obvious that you have no idea of what you're talking about. Unless you planned to make a fool out of yourself, in which case you succeed.

Yes, I have no idea what I'm talking about because I make massive mistakes in all of my math, eliminate essential "variables" from my calculations, and then throw a fit when someone points out where I went wrong. Oh, wait, that's you. erm

Originally posted by Astner
Engineering physics, the most prestigious maths and physics related university programme available. You might want to watch what you type before I throw complex curve integrals at you.

laughing

Oh man. It's all too clear, now.

I'm sorry I took you seriously for even a little bit. My bad.

Originally posted by Astner
Numerical errors isn't equivalent to atrocious calculations.

And you've done both, in spades. On top of that, you've simply done your math wrong on almost every occasion math was involved. "Prestigious" huh? laughing

Originally posted by Astner
Let me put you up with a test then. I'm even going to be nice and tell you that it's a test. I further am going to formulate this test directly as a well defined problem

Bla bla bal

So, you want to test me because I clowned you with some very simple math?

Sorry, ain't happening. smile

Originally posted by Astner
Yeah, you're so witty playing these "obvious" mind games that only you understand. Especially when 50% of the content of your post is nothing but a long reminisce.

So this is how you admit you're wrong?

I bet it pisses you off to no end that I reversed your own work against you to see if you'd argue against yourself.

Originally posted by Astner
I'm aware, sadly you're not.

You're kidding, right?

You just replied with a, "I know you are but what am I?"

Originally posted by Astner
In the same sense, an inch is a smaller version of the SI-unit meter. It still doesn't make it an SI-unit you dolt. This is basic stuff brought up in junior high.

Now-now, watch the insults.

And, I will not humor your massive load of fail when it comes to SI. You've gone too far into fail for me to be willing to humor you.

Originally posted by Astner
No, I was just trying to be polite, not pointing out all your logical shortcomings. But thinking a katana weights as much as an modern drawer takes rare stupidity.

Ugh. Man, you've gone too far. Why does the mass matter at all to the actual point? Tell me why these arbitrary numbers even matter a little bit above and beyond you spazing? Do you even know what the actual point was? (No, you don't. If you actually addressed the point, you'd have to admit fault, yet a gain, which you don't want to do, obviously.)

Originally posted by Astner
You seem to struggle with the concept of arbitrary numbers.

{a + b = c}, is a system of arbitrary numbers as each of the parameters a, b and c can be set (technically you have 2-degrees of freedom, but that's beside the point).

{1 + 2 = 3}, isn't a system of arbitrary numbers.

You seem to struggle with the concept of just about everything we've discussed. Dead serious.

If you had any sort of ability to follow conversation context, you'd know that "arbitrary numbers" does not mean what you think it means. It's highly possible that you're pretending to be ignorant of what we are actually talking about to further avoid actually discussion what you've done wrong.

Originally posted by Astner
But they aren't arbitrary.

Yeah, they were. You just don't get it. You think this is a "math definition" of "arbitrary numbers", but it clearly is not. Look up the definition of the word arbitrary, and then get back to me. "Improvised" should have been your first clue. erm

Originally posted by Astner
The hypocrisy residing within the fact that you haven't been able to point out any of them.

Not only did I point out all of them, you continue to miss the points I've made to you. If you want to see where I've pointed them out, just reread my posts.

Originally posted by Astner
Which has nothing whatsoever to do with units.

In the SI system:
Mass isn't measured in m, it's measured in .
Force isn't measured in F, it's measured in ^-2 (Newton)

Are you familiar with "the strawman" argument?

Originally posted by Astner
If you haven't come across units and dimension analysis you haven't taken any courses in physics whatsoever (not even junior high). And I have trouble believing that you attend college, but that's neither here or there.

Fact is, it is vital to the calculations made, since it indicates that they're made correctly.

You have trouble believing I attend college because you have very poor comprehension and math skills. How does that even make sense?

And, no, it is not necessary, even in the slightest, and, like I said, I have never seen it. Please, kind sir, link me to something on the internet that shows when "" is supposed to be used. Please, I beg you.

Originally posted by Astner
Actually I do give you that because it's correct.

Originally posted by dadudemon
No, you won't give this to me because I showed you where it was wrong and told you it was wrong in my first reply to you. 128 tonnes is still wrong. You incorrectly simplified out ALL of density when you shouldn't have. If you actually knew what I showed you one post to you, back, you would have seen where I clearly demonstrated that to you.

Originally posted by Astner
Once again, no I didn't. Instead of committing your stupid mistake of--for no reason--separate the volume from the density when neither is known. I kept them the same. Since the density of the regular katana and Gamabuta's katana is assumed to be identical.

http://img576.imageshack.us/img576/954/mimetex.gif

Originally posted by Astner
No, it was an attempt to prove to you that you were playing with semantics. "It's tonnes not tons" while true it holds no relevance to the calculation at hand. Sadly this is were your focus is at. But OK I'll promise you that I will stay with kilograms since it apparently makes you feel better.

I can't read that image crap you are posting.

Regardless, you still don't get it:

Originally posted by dadudemon
You really did drop density. I reversed your work to show you that you multiplied by an arbitrary value of 1.9.... But you still don't see it.
Yet, you get onto me for using arbitrary numbers? Why is it that you don't even see a very simple demonstration, using math, where you went wrong? You've crossed over into the absurd, now, man. If you won't see where you've clearly messed up, even after I very explicitly and simply showed you where you went wrong, you'll never get it right.

Originally posted by Astner
You Fun fact, all of this is high school level maths (aside from the "test" I gave you, have fun with that).

Either way, failing with calculations and failing to hit the right buttons on the calculator are two different things. I'm used to MATLAB and Mathematica, not the MAC's simple calculator.

I guess you need to retake highschool because you can't even make simple calculations, much less understand the relationship between volume and density to mass.

dadudemon

Ran out of time to edit that post as I left a few lines of response blank:

Basically, Astner, you don't know what you're doing. And, if you do, you're so high when you reply that you don't know left from right.

You have yet to correct the obvious mistakes that I pointed out and are, instead, dancing around the subject. You make simple mistakes that you shouldn't such as calling them "tons" and "rectangles", and failing to understand has SI works completely.

Stick to the topic at hand: correct your mistakes on your calculation of Gama's sword. I'll work on it, properly, on this end. smile

wakkawakkawakka

Umm....really? Was a physics debate really that necessary for this forum? no expression

Why can't people just go for the person with better feats; strength wise that would be Tsunade.

However, on the battlefield, Raikage would murder her unless she used Genesis Rebirth.

Demonic Phoenix

What good would Genesis Rebirth do? A would just murder her again.

Unless she does what Q99 stated, and uses her stored up chakra to punch the Raikage to Kingdom Come. Then again, that's assuming she can even land a hit on A.

dadudemon

Originally posted by Demonic Phoenix
What good would Genesis Rebirth do? A would just murder her again.

Unless she does what Q99 stated, and uses her stored up chakra to punch the Raikage to Kingdom Come. Then again, that's assuming she can even land a hit on A.

That's right, the Raikage is super fast.

hmm

Astner

You'll have to excuse my absence I've basically been studying at full time for my upcoming exams in my second course in quantum mechanics and mathematical modeling. Though I have to say I'm disappointed in your response. But I'll address that point by point.

Originally posted by dadudemon
So, basically, you bring in a bunch of strawman arguments because you simply can't say, "I was wrong." My professors or teachers have nothing to do with you posting something wrong.
It wasn't an argument, thus not a fallacy. The title of the example has no relevance whatsoever. It doesn't make the example or solution wrong, deal with it.

Originally posted by dadudemon
Also, you need to quote me where I said that the given formula only applied to perfect cubes. Please, I beg you, show me where I said that.
Third page, second post:

"Wrong. You cubed the same number...that's not accurate (unless the katana is a perfect cube)."

Further ironically not what I did in that formula.

Originally posted by dadudemon
It's called context and you don't understand it. You can't even follow a simple internet conversation. L*W*H is a very simple formula. That was the context of the convo, you could not follow it.
So I couldn't follow it because I was able to point out that it didn't only work for fixed values but also for any given function? After pointing that out and your failure to give a proper response I'd say that you were the one that couldn't follow.

Originally posted by dadudemon
Yes, I have no idea what I'm talking about because I make massive mistakes in all of my math, eliminate essential "variables" from my calculations, and then throw a fit when someone points out where I went wrong. Oh, wait, that's you. erm

You have to understand the difference between a mathematical and a numerical problem, which again is basic algebra. I'm not denying the fact that I made a numerical error. However, I formulated the mathematical problem correctly and you fail to understand it.

Originally posted by dadudemon
Oh man. It's all too clear, now.

I'm sorry I took you seriously for even a little bit. My bad.
Ironic. Especially since you were unable to comply with the mathematical problem I gave you, one which I solved several years back.

Originally posted by dadudemon
So this is how you admit you're wrong?
That was sarcasm, though I can't say I expected more of you.

I bet it pisses you off to no end that I reversed your own work against you to see if you'd argue against yourself.

Originally posted by dadudemon
And, I will not humor your massive load of fail when it comes to SI. You've gone too far into fail for me to be willing to humor you.
It's saddening to see that you fail to admit that you're wrong when you said that grams were an SI-unit, it's not more of an SI-unit than a stone or pound.

Originally posted by dadudemon
Ugh. Man, you've gone too far. Why does the mass matter at all to the actual point?
Because that's what we're trying to calculate, if it didn't matter. We wouldn't do the calculation. Either--if not both--of your estimations (density and volume of an actual katana) are ridiculous.

Originally posted by dadudemon
Yeah, they were. You just don't get it. You think this is a "math definition" of "arbitrary numbers", but it clearly is not.
Ironically, mathematical definitions or axioms are regious and injective (one-to-one) if they weren't problems wouldn't be well defined to begin with, rendering maths useless.

Originally posted by dadudemon
And, no, it is not necessary, even in the slightest, and, like I said, I have never seen it. Please, kind sir, link me to something on the internet that shows when "" is supposed to be used. Please, I beg you.
More often than not it's simply let out, however regious solution always contain the units in brackets the dimensionless unit being wrote out as either: , or .

I couldn't find any notation of the dimensionless (or even with dimension) unit in brackets. Even though it's scientifically stipulated.

As for references: is Kusse, Bruce R.'s notation in "Mathematical Physics: Applied Mathematics for Scientists and Engineers"

But then again, with junior high dimension analysis you should had figured it out. * = . Naturally can't have any unit.

Originally posted by dadudemon
I can't read that image crap you are posting.

Regardless, you still don't get it
And that's an ad hominem. I like how you not once in this entire post do any attempt to do any actual refutations. Rather you go on and on with you logical fallacies "it's wrong because it stupid", sorry life isn't that simple.

Originally posted by dadudemon
I guess you need to retake highschool because you can't even make simple calculations, much less understand the relationship between volume and density to mass.
Is that how it is? I'm not the one failing to understand the relation between volume and mass, you are. As I said, and you dismissed.

Gamabuta's katana length =
Original katana length =
Gamabuta's katana volume =
Original katana volume =
Gamabuta's katana desnity =
Original katana density =
Gamabuta's katana mass =
Original katana mass =

= (/)^3 * - The volume formula applies to a arbitrary body (cubes, cuboids, spheres, etc.).

Now the and have the same density if they're made of the same materials (unaffected by scale). So = .

If we now decide to multiply the LHS and the RHS of the equation with the density we would get.

* = (/)^3 * *

or

= (/)^3 * Q.E.D.

Now if you make an actual response and manage to refute any of these points, then I will address these points.

However if your following post, like your previous one is void of any legit argument I'll ignore it. No, "your wrong, this is stupid" isn't an argument against anything, it's an ad homiem.

Neither is addressing my previous numerical error (like you did 4-5 times in this post) which I conceded in was incorrect several posts back.

dadudemon

Originally posted by Astner
You'll have to excuse my absence I've basically been studying at full time for my upcoming exams in my second course in quantum mechanics and mathematical modeling. Though I have to say I'm disappointed in your response. But I'll address that point by point.

It wasn't an argument, thus not a fallacy. The title of the example has no relevance whatsoever. It doesn't make the example or solution wrong, deal with it.

Third page, second post:

"Wrong. You cubed the same number...that's not accurate (unless the katana is a perfect cube)."

Further ironically not what I did in that formula.

So I couldn't follow it because I was able to point out that it didn't only work for fixed values but also for any given function? After pointing that out and your failure to give a proper response I'd say that you were the one that couldn't follow.

You have to understand the difference between a mathematical and a numerical problem, which again is basic algebra. I'm not denying the fact that I made a numerical error. However, I formulated the mathematical problem correctly and you fail to understand it.

Ironic. Especially since you were unable to comply with the mathematical problem I gave you, one which I solved several years back.

That was sarcasm, though I can't say I expected more of you.

I bet it pisses you off to no end that I reversed your own work against you to see if you'd argue against yourself.

It's saddening to see that you fail to admit that you're wrong when you said that grams were an SI-unit, it's not more of an SI-unit than a stone or pound.

Because that's what we're trying to calculate, if it didn't matter. We wouldn't do the calculation. Either--if not both--of your estimations (density and volume of an actual katana) are ridiculous.

Ironically, mathematical definitions or axioms are regious and injective (one-to-one) if they weren't problems wouldn't be well defined to begin with, rendering maths useless.

More often than not it's simply let out, however regious solution always contain the units in brackets the dimensionless unit being wrote out as either: , or .

I couldn't find any notation of the dimensionless (or even with dimension) unit in brackets. Even though it's scientifically stipulated.

As for references: is Kusse, Bruce R.'s notation in "Mathematical Physics: Applied Mathematics for Scientists and Engineers"

But then again, with junior high dimension analysis you should had figured it out. * = . Naturally can't have any unit.

And that's an ad hominem. I like how you not once in this entire post do any attempt to do any actual refutations. Rather you go on and on with you logical fallacies "it's wrong because it stupid", sorry life isn't that simple.

Is that how it is? I'm not the one failing to understand the relation between volume and mass, you are. As I said, and you dismissed.

Gamabuta's katana length =
Original katana length =
Gamabuta's katana volume =
Original katana volume =
Gamabuta's katana desnity =
Original katana density =
Gamabuta's katana mass =
Original katana mass =

= (/)^3 * - The volume formula applies to a arbitrary body (cubes, cuboids, spheres, etc.).

Now the and have the same density if they're made of the same materials (unaffected by scale). So = .

If we now decide to multiply the LHS and the RHS of the equation with the density we would get.

* = (/)^3 * *

or

= (/)^3 * Q.E.D.

Now if you make an actual response and manage to refute any of these points, then I will address these points.

However if your following post, like your previous one is void of any legit argument I'll ignore it. No, "your wrong, this is stupid" isn't an argument against anything, it's an ad homiem.

Neither is addressing my previous numerical error (like you did 4-5 times in this post) which I conceded in was incorrect several posts back.

Hi, Astner.

I didn't read any of that.

I will not respond to your posts on this subject any longer.

smile

I'll work out the correct calculation when I feel up to it. My homie has some 3D physics software that I can model this shizer from that will help me calculate a correct number.

Like I said, earlier, though, the sword has to be made from a material other than steel....cause it's too big to actually use without it bending/breaking. teehee

Astner

Originally posted by dadudemon
Hi, Astner.

I didn't read any of that.

I will not respond to your posts on this subject any longer.

smile

Concession accepted.

Originally posted by dadudemon
I'll work out the correct calculation when I feel up to it. My homie has some 3D physics software that I can model this shizer from that will help me calculate a correct number.
Some 3D physics software, you didn't research that one through did you?

Even if that was the case and you just had no idea of what you're talking about. You should get about 128 metric tons if you're doing it correctly.

Originally posted by dadudemon
Like I said, earlier, though, the sword has to be made from a material other than steel....cause it's too big to actually use without it bending/breaking. teehee
So would it if it was made of any other material too, even graphene wouldn't keep it together.

King Kandy

Originally posted by Astner
So would it if it was made of any other material too, even graphene wouldn't keep it together.
Yes, fortunately we're using a fictional world where materials can have as much structural integrity as we want.

dadudemon

Originally posted by Astner
Concession accepted.

But I didn't concede. Only thing I did was show you where your math was wrong while you pretended it was right while making painful mistake after mistake.

I just don't like to entertain huge wastes of my time.

Originally posted by Astner
Some 3D physics software, you didn't research that one through did you?

Even if that was the case and you just had no idea of what you're talking about. You should get about 128 metric tons if you're doing it correctly.

Yeah, you know, software that allows people to render...you know.....3d graphics? durr

He's an animator.

You didn't think before responding, did you? Of course not, you haven't very much on all of this.

Also, there are tons of software programs out there that allow you to model complex physics based scenarios. You didn't get to use any of those in "college" did you?

Anywho, most of those software programs give you TONS of information on the objects you draw including volume. You can set things such as the gravity constant, volumetirc info, etc. It's quite nice. smile

Originally posted by Astner
So would it if it was made of any other material too, even graphene wouldn't keep it together.

English, please. smile

Astner

Originally posted by dadudemon
But I didn't concede. Only thing I did was show you where your math was wrong while you pretended it was right while making painful mistake after mistake.
No, what you did was acting as if I made mathematical mistakes.

Originally posted by dadudemon
Yeah, you know, software that allows people to render...you know.....3d graphics? durr

He's an animator.

You didn't think before responding, did you? Of course not, you haven't very much on all of this.

Also, there are tons of software programs out there that allow you to model complex physics based scenarios. You didn't get to use any of those in "college" did you?
An animator, really? And no, I don't physic based programs that allows you to model complex physic related scenarios because frankly they don't exist. What we learn, however, is programing using mathematical software such as MATLAB, Python, GNU Octave and Mathematica in which we can solve virtually any scenario.

In fact, here is a recent program I made studying the relation between the structure of black holes and their rotation. Enjoy.

http://img84.imageshack.us/img84/6900/blhz.jpg

Originally posted by dadudemon
Anywho, most of those software programs give you TONS of information on the objects you draw including volume. You can set things such as the gravity constant, volumetirc info, etc. It's quite nice. smile

Funny, you've yet to mention one. Fact is there is no point in creating a visual based physic program because it would be needlessly abstract and take up far too much space.

Originally posted by dadudemon
English, please. smile

That's right you don't understand scientific terminology. Graphene is the prime carbon structure used in nano technology.

Originally posted by King Kandy
Yes, fortunately we're using a fictional world where materials can have as much structural integrity as we want.
You would still have to define its properies.

dadudemon

Originally posted by Astner
No, what you did was acting as if I made mathematical mistakes.

Really? I mean...really, dude? You forgot density, made tons of math mistakes on almost every last calculation, and still don't see it. Yet, I acted?

Originally posted by Astner
An animator, really? And no, I don't physic based programs that allows you to model complex physic related scenarios because frankly they don't exist. What we learn, however, is programing using mathematical software such as MATLAB, Python, GNU Octave and Mathematica in which we can solve virtually any scenario.

They don't exist, eh?

That's funny.

Cause like, you have no clue what you're talking about.

I love it.

http://lmgtfy.com/?q=physics+engine

hehehehehe

Originally posted by Astner
Funny, you've yet to mention one. Fact is there is no point in creating a visual based physic program because it would be needlessly abstract and take up far too much space.

http://lmgtfy.com/?q=physics+engine

Originally posted by Astner
That's right you don't understand scientific terminology. Graphene is the prime carbon structure used in nano technology.

laughing

That's not even close to what I was talking about.

But, Astner, if you want to pretend you're right to save your ego, go ahead.

Just make sure you pass your on-topic math by me before you post it in a vs. thread. I actually don't mind talking about this stuff and I wish more people would make an effort to use math in vs. debates, when applicable. Originally, I was happy that you tried it. Sure, you did a horrible job of actually doing things correctly, I was still proud of you.

marwash22

^
lol, that was soooooo condescending.

King Kandy

Originally posted by Astner
You would still have to define its properies.
No, because we have no information by which to base that on, except that it retains structural integrity at that size. There you go, there's the one known property about it.

draxx_tOfU

Kandy is right...

dadudemon

Originally posted by marwash22
^
lol, that was soooooo condescending.

He started it! mad

(No, really, he did. In his first reply to me, after I had complimented him, he replied with meanie stuff. sad

)

Originally posted by draxx_tOfU
Kandy is right...

He usually is.

King Kandy

Originally posted by dadudemon
He usually is.
Yeah, and you'd best remember that when you start going off on the McDonalds lawsuit again. mad

dadudemon

Originally posted by King Kandy
Yeah, and you'd best remember that when you start going off on the McDonalds lawsuit again. mad

*Thinks about that McDonald's lawsuit, again* 313

yungz22

genisis makes sure u dont die no matter what while its activated

King Kandy

Originally posted by yungz22
genisis makes sure u dont die no matter what while its activated
No matter what? What if she fought, like, Galactus? How about God, could God kill Tsunade while she was using genesis?

Astner

Originally posted by dadudemon
Really? I mean...really, dude? You forgot density, made tons of math mistakes on almost every last calculation, and still don't see it. Yet, I acted?
I didn't forget the density, but rather than separating density from volume which is completely unnecessary for a calculation when the density is a scalar. I kept it as mass.

Tell me, what's wrong with this formula.

http://img714.imageshack.us/img714/2787/mimetexa.gif

a represents the linear dimension of any given parameter (i.e. circumference, length, width, breadth, curve, etc.) prior the volume transformation.

b represents the linear dimension of any given parameter (i.e. circumference, length, width, breadth, curve, etc.) after the volume transformation.

m represents the mass before the volume transformation.

M represents the mass after the volume transformation.

I proved to you that this works on, cubes, cuboid and spheres. You might test any other fix body, but as long as they have the same density and shape it will be accurate.

Originally posted by dadudemon
They don't exist, eh?

That's funny.

Cause like, you have no clue what you're talking about.

I love it.

http://lmgtfy.com/?q=physics+engine
While standing corrected on the matter of that they do exist, you were wrong regarding their complexity and versatility.

I quote:

"A physics engine is computer software that provides an approximate simulation of certain simple physical systems, such as rigid body dynamics (including collision detection), soft body dynamics, and fluid dynamics, of use in the domains of computer graphics, video games and film."

It's a simple program used for simple problems. You wouldn't rely on the same program you use to enhance the realism of computer generated effects that you would creating an airplane.

So there's been a week now, did you make an attempt on the visual calculations you promised or was it all bullshit as I predicted?

Originally posted by dadudemon
That's not even close to what I was talking about.
That was what you were talking about, why else would you've responded. In the manner you did?

Originally posted by Astner
So would it if it was made of any other material too, even graphene wouldn't keep it together.
Originally posted by dadudemon
English, please. smile

There are no gramamtical errors, no spelling errors or any other writing errors one could think of.

Originally posted by dadudemon
But, Astner, if you want to pretend you're right to save your ego, go ahead.
Saving my ego from empty assaults? I've told you, once you find a mathematical error in my calculations I'll address them accurately.

Originally posted by dadudemon
Just make sure you pass your on-topic math by me before you post it in a vs. thread. I actually don't mind talking about this stuff and I wish more people would make an effort to use math in vs. debates, when applicable. Originally, I was happy that you tried it. Sure, you did a horrible job of actually doing things correctly, I was still proud of you.
You're wrong, plain and simple. You haven't mentioned one flaw in the formula above, you even choose to ignore one of my responses.

For all the miserable attempts you've made I've addressed all of them with no exceptions. Because unlike you I'm more interested in the truth than being right. That's also why I've made 2 concessions, and you have yet to make one. Even though you were directly disproved that the formula above only applies to perfect cubes.

Astner

I'd further like to add that the same formula puts the Rhino Naruto threw, assuming it has the shape and density of a white rhino (close approximation) and the length of 22 m it should weigh 500 tons.

dadudemon

Originally posted by Astner
I didn't forget the density, but rather than separating density from volume which is completely unnecessary for a calculation when the density is a scalar. I kept it as mass.

Tell me, what's wrong with this formula.

http://img714.imageshack.us/img714/2787/mimetexa.gif

a represents the linear dimension of any given parameter (i.e. circumference, length, width, breadth, curve, etc.) prior the volume transformation.

b represents the linear dimension of any given parameter (i.e. circumference, length, width, breadth, curve, etc.) after the volume transformation.

m represents the mass before the volume transformation.

M represents the mass after the volume transformation.

I proved to you that this works on, cubes, cuboid and spheres. You might test any other fix body, but as long as they have the same density and shape it will be accurate.

Bla bla bla.

We've gotten past this sooooo very long ago.

Originally posted by Astner
While standing corrected on the matter of that they do exist, you were wrong regarding their complexity and versatility.

I quote:

"A physics engine is computer software that provides an approximate simulation of certain simple physical systems, such as rigid body dynamics (including collision detection), soft body dynamics, and fluid dynamics, of use in the domains of computer graphics, video games and film."

It's a simple program used for simple problems. You wouldn't rely on the same program you use to enhance the realism of computer generated effects that you would creating an airplane.

So there's been a week now, did you make an attempt on the visual calculations you promised or was it all bullshit as I predicted?

Wait, so you were wrong, but now you are claiming that they are SIMPLER than what I referred to? You know, the exact OPPOSITE of what you should have concluded? They are far more complex than the very simple "description" I gave them.

Cut it out. That's very annoying.

You're crossing into the territory of trolling, now.

Originally posted by Astner
That was what you were talking about, why else would you've responded. In the manner you did?

There are no gramamtical errors, no spelling errors or any other writing errors one could think of.

facepalm

I said:

"Like I said, earlier, though, the sword has to be made from a material other than steel....cause it's too big to actually use without it bending/breaking. teehee "

You said in reply:

"So would it if it was made of any other material too, even graphene wouldn't keep it together."

To which I replied:

"English, please. smile

" Instead, I figured I'd make a joke about your poor use of pronouns as that sh*t gets annoying as hell and I was tired of arguing about why you're wrong and miss everything.

Originally posted by Astner
Saving my ego from empty assaults? I've told you, once you find a mathematical error in my calculations I'll address them accurately.

You're wrong, plain and simple. You haven't mentioned one flaw in the formula above, you even choose to ignore one of my responses.

For all the miserable attempts you've made I've addressed all of them with no exceptions. Because unlike you I'm more interested in the truth than being right. That's also why I've made 2 concessions, and you have yet to make one. Even though you were directly disproved that the formula above only applies to perfect cubes.

1. You know you love me.

2. I've pointed out some errors already.

3. I was trying to be cool about stuff but you got all butthurt over a very minor passing comment I made about your atrocious math.

4. ???

5. Profit.

Astner

Originally posted by dadudemon
Bla bla bla.

We've gotten past this sooooo very long ago.
Thank you for your insightful and elaborate response.

Originally posted by dadudemon
Wait, so you were wrong, but now you are claiming that they are SIMPLER than what I referred to? You know, the exact OPPOSITE of what you should have concluded? They are far more complex than the very simple "description" I gave them.
Ironically that's not what the link you provided me with said. The programs you refer to are mainly used to induce a level of realism in games and movies. They are not used as tools for optimizing the shape of aircrafts, whereas MATLAB--among other things--is.

Originally posted by dadudemon
Cut it out. That's very annoying
You promised us calculation results, results you haven't and most likely won't provide.

Mainly because the programs you refer to aren't used for calculating the volume of objects of "unknown" size.

Originally posted by dadudemon
I said:

"Like I said, earlier, though, the sword has to be made from a material other than steel....cause it's too big to actually use without it bending/breaking. teehee "

You said in reply:

"So would it if it was made of any other material too, even graphene wouldn't keep it together."

To which I replied:

"English, please. smile

"

Now, stop right there...why do I think I included a smart*ss smiley face?

Could it have been the atrocious amgibugous pronoun references?

Or is it what you thought it was: me not understanding how graphene even made sense in that sentence when, in context, it would make perfect sense even to an idiot.
You think graphene is a ambiguous reference? It's the most basic material in the field of nanotechnology, to which even highschool students are introduced to.

It have relevance due to the fact that it's currently the most durable material.

Originally posted by dadudemon
You assumed the latter so you would have an excuse to pontificate, yet again, because you're strill trying to save "face" to a bunch of teenage boys in an anime/manga thread.
If I were to "save my face" why didn't I respond earlier? Why would I lets your posts ridiculing me taint my reputation for a week? Perhaps because I'm not that interested in my Internet reputation, Endless Mike even postulated that I sought to be infamous.

Originally posted by dadudemon
Also, if I REALLY wanted to get pedantic instead of making a joke about your overuse of "it", I would have said something like: "Right, cause sheets of carbon atoms can be made into gigantic swords. no expression

" Instead, I figured I'd make a joke about your poor use of pronouns as that sh*t gets annoying as hell and I was tired of arguing about why you're wrong and miss everything.
I wouldn't call paraphrasing Wikipedia pedantic. Mathematically being able to explain the molecular structure's behavior on a subatomic level would be though.

Originally posted by dadudemon
1. You know you love me.
Yeah embarrasment

Originally posted by dadudemon
2. I've pointed out some errors already.
No you didn't, the fact that this entire reply doesn't contain one shred of mathematical rebuttal proves your capabilities.

Please point out one mathematical body to which this formula doesn't apply. As mentioned before, there is none.

Originally posted by dadudemon
3. I was trying to be cool about stuff but you got all butthurt over a very minor passing comment I made about your atrocious math.
Yes, of course.

Originally posted by dadudemon
5. Profit.
Nonsensical to the given response.

King Kandy

Ugh, people shut up already. I already showed why trying to calculate it is pointless, this is just argument for argument's sake.

Demonic Phoenix

Originally posted by dadudemon
1. You know you love me.

Everybody loves you 313

Originally posted by King Kandy
Ugh, people shut up already. I already showed why trying to calculate it is pointless, this is just argument for argument's sake.

This.

dadudemon

Originally posted by Astner
Thank you for your insightful and elaborate response.

Thank you for wasting time, yet again. *eats some food*

Originally posted by Astner
Ironically that's not what the link you provided me with said. The programs you refer to are mainly used to induce a level of realism in games and movies. They are not used as tools for optimizing the shape of aircrafts, whereas MATLAB--among other things--is.

IS that ironic? IS it? dur

So, tell me, dude, how does optimizing the shapes of aircrafts relate to topic? (You know, making objects in 3 dimensions and getting a volumetric measure for that object...assigning it density and the environment a gravity constant. Wait, what? You missed the point again and now are arguing against stuff that doesn't even exist? PLLEEEEEEEEEEEEEEEEEEEEEEASE stop wasting my time! sad

)

Originally posted by Astner
You promised us calculation results, results you haven't and most likely won't provide.

I already calculated it. I'll give you a hint: it starts with a 9.

But, like I said, the calculation is useless due to the object NOT being made out of hardened steel.

Originally posted by Astner
Mainly because the programs you refer to aren't used for calculating the volume of objects of "unknown" size.

You're right, it gives you a read out on the right, based on this little "units" of area, (or wherever you want to throw you statistics window while building 3D objects. Imagine that: back to Archimedes, like I effin' said. WEEEEEE!)

Originally posted by Astner
You think graphene is a ambiguous reference? It's the most basic material in the field of nanotechnology, to which even highschool students are introduced to.

It have relevance due to the fact that it's currently the most durable material.

Good GOD man! I very very explicitly told you what the damn problem was.

"...ambiguous pronoun reference" refers to what?

"...overuse of 'it'" refers to what?

Tell me the truth, I'll be cool about it, I swear: are you autistic? 100% honest, I will be waaaaaaay nicer to you if you are autistic.

Originally posted by Astner
If I were to "save my face" why didn't I respond earlier? Why would I lets your posts ridiculing me taint my reputation for a week? Perhaps because I'm not that interested in my Internet reputation, Endless Mike even postulated that I sought to be infamous.

The length of time between posts means nothing, only that you continue to post replies. I thought this thread was done with you wasting my time. *farts*

Originally posted by Astner
I wouldn't call paraphrasing Wikipedia pedantic. Mathematically being able to explain the molecular structure's behavior on a subatomic level would be though.

PAH! (That's a really funny pun, if you can get it.) I'm the last person you should accuse of being an internet tiger...

I was actually considering carbon-nano-tubes as a viable option, but making one into a blade is really silly: I have no idea how that would work, but you could make a rope out of dat shit, fo sho. I still don't see how graphene would be even remotely usable because it's a damn sheet.

Originally posted by Astner
Yeah embarrasment

Teehee

Originally posted by Astner
No you didn't, the fact that this entire reply doesn't contain one shred of mathematical rebuttal proves your capabilities.

Please point out one mathematical body to which this formula doesn't apply. As mentioned before, there is none.

You see, that's where the problem lies: you DID make mathematical errors...and we have LOOOOOOOONG been over the formula (remember me throwing it back at you in an expanded form to see if you would argue against it, which you did?)

Originally posted by Astner
Yes, of course.

Hey, this wouldn't have gotten this far had you just corrected your math mistakes and your path to find mass.

Originally posted by Astner
Nonsensical to the given response.

It's an internet meme, dude.

Astner

Originally posted by dadudemon
Thank you for wasting time, yet again. *eats some food*
Are you compelled reply, regardless whether or not the content is credible or not?

You're not attempting to make an argument, the post I'm replying to is nothing short of an attempt to end this topic granting you the final reply.

Originally posted by dadudemon
So, tell me, dude, how does optimizing the shapes of aircrafts relate to topic?
It refutes your claim of that those physics engines are complex programs applied in the frontiers of science.

Originally posted by dadudemon
I already calculated it. I'll give you a hint: it starts with a 9.
No you didn't calculate it. If that was the case you've would have provided the answer you promised as soon as you made the calculation rather than behave in this childish manner.

Originally posted by dadudemon
But, like I said, the calculation is useless due to the object NOT being made out of hardened steel.
So you calculated the mass of an object without without any density related factors? Do I have to point out to you that this is impossible?

Originally posted by dadudemon
You're right, it gives you a read out on the right, based on this little "units" of area, (or wherever you want to throw you statistics window while building 3D objects. Imagine that: back to Archimedes, like I effin' said. WEEEEEE!)
So you pretend has you've build a perfect katana by manually adding volume elements. Do I need to approach this argument? Were you even trying before posting this blatant lie? If you manually added rectangular volume elements you would have gotten the exact volume by multiplying their volume by the amount volume elements you added, easily accomplished with any basic calculator. Even if you didn't know it you would've realized while playing with the program you supposedly have.

Originally posted by dadudemon
Tell me the truth, I'll be cool about it, I swear: are you autistic? 100% honest, I will be waaaaaaay nicer to you if you are autistic.
No, I'm not diagnosed with autism. But then again, this is just another attempt of an Ad Hominem attack to avoid reaching the core of your hollow argument.

Originally posted by dadudemon
The length of time between posts means nothing, only that you continue to post replies.
Apparently you suggest that I'm attempting to "save my ego", implying that I'm taking your position of being right regardless of the facts, even though I've made two concessions in this thread, and you haven't made one even though it's blatantly obvious that you're wrong on a number of accounts and were even directly proven to be regarding the fact that the formula above only applies to perfect cubes.

Aside from being a fallacious argument, it by itself isn't true. If this topic had such an impact on me, then I would've responded earlier.

Originally posted by dadudemon
I was actually considering carbon-nano-tubes as a viable option, but making one into a blade is really silly: I have no idea how that would work, but you could make a rope out of dat shit, fo sho.
Yes, I'm sure you were considering the options of nanotubes without the slightest hint of what graphene was.

Originally posted by dadudemon
I still don't see how graphene would be even remotely usable because it's a damn sheet.
Orthogonally bounded.

Originally posted by dadudemon
You see, that's where the problem lies: you DID make mathematical errors...and we have LOOOOOOOONG been over the formula
No you haven't.

Originally posted by dadudemon
(remember me throwing it back at you in an expanded form to see if you would argue against it, which you did?)
No, I don't recall you doing that. Furthermore that's not a mathematical rebuttal.

Originally posted by dadudemon
Hey, this wouldn't have gotten this far had you just corrected your math mistakes and your path to find mass.
It wouldn't have gotten that far if you, from scratch, realized that I applied the same density to both of the katana allowing me to make use of the distributivity and treat mass as volume times density.

Originally posted by dadudemon
It's an internet meme, dude.
I actually expected a mature response.

King Kandy

Originally posted by Astner
So you calculated the mass of an object without without any density related factors? Do I have to point out to you that this is impossible?
And the irony of this statement... goes right over your head.

Astner

Originally posted by King Kandy
And the irony of this statement... goes right over your head.
One would think it'd be common sense to read through a reply before taking a segment of it out of context.

It's not that I don't know what I'm talking about, it's that you don't know what I'm talking about.

Originally posted by Astner
It wouldn't have gotten that far if you, from scratch, realized that I applied the same density to both of the katana allowing me to make use of the distributivity and treat mass as volume times density.
Here I explicitly address a density related factor, i.e. mass.

And if you follow the debate in detail you'd also take notice of my previous post.

Originally posted by Astner
I didn't forget the density, but rather than separating density from volume which is completely unnecessary for a calculation when the density is a scalar. I kept it as mass.
Once again, addressing the same logical fallacy thrown at me.

I've been parroting this matter since the debate started at page 3, in this thread.

Originally posted by dadudemon
First, find the density of the handle. Then, find the density of the metal. Then, find the volume of the handle and complete blade.
Originally posted by Astner
Completely unnecessary and would mean that we would have to resort to further estimations that would lower the accuracy of the results.
As can be seen, I never made any calculations without any density related factors.

King Kandy

Originally posted by Astner
One would think it'd be common sense to read through a reply before taking a segment of it out of context.

It's not that I don't know what I'm talking about, it's that you don't know what I'm talking about.

Here I explicitly address a density related factor, i.e. mass.

And if you follow the debate in detail you'd also take notice of my previous post.

Once again, addressing the same logical fallacy thrown at me.

I've been parroting this matter since the debate started at page 3, in this thread.

As can be seen, I never made any calculations without any density related factors.
laughing

And the part of that line I thought was amusing... also went right over your head.

Astner

Originally posted by King Kandy
And the part of that line I thought was amusing... also went right over your head.
No rebuttal? Concession accepted.

King Kandy

There's no concession... I bet everyone else visiting the thread will understand, because if you have to explain a joke it's not funny anymore.

Astner

You didn't expect me to take your Appeal to Ridicule fallacy seriously did you?

The fact that you argue using logical fallacies proves your inability to debate properly.

King Kandy

If you thought my point was that you dropped the density, then you didn't get what I was trying to say.

Astner

Originally posted by King Kandy
If you thought my point was that you dropped the density, then you didn't get what I was trying to say.
Did you attempt to read the post you were ridiculing? Density and density related factors aren't necessarily the same. Newton, mass, Joule, etc. are all density related factors.

King Kandy

Originally posted by Astner
Did you attempt to read the post you were ridiculing? Density and density related factors aren't necessarily the same. Newton, mass, Joule, etc. are all density related factors.
Still not my point.

Astner

Originally posted by King Kandy
Still not my point.
Then once again, Appeal to Ridicule.

Allow me to recite the, to you, enigmatic description of the fallacy.

"X is amusing, absurd. Therefore it is false.

Mock the other person's claim and argument. Make fun of it. Get people to laugh at it.

Alternatively, mock the alternatives that they might choose, giving them only one option that you have not mocked."

King Kandy

Well given that you don't understand what my point was, I can see why you also wouldn't see why it actually makes sense.

Astner

Originally posted by King Kandy
Well given that you don't understand what my point was, I can see why you also wouldn't see why it actually makes sense.
Tell me. Why would you argue using logical fallacies if you had a coherent argument?

King Kandy

It's not a fallacy, because I am not making an argument based on mockery. I simply presented a concise, logical argument that you didn't understand and in error declared a fallacy.

Astner

Originally posted by King Kandy
It's not a fallacy, because I am not making an argument based on mockery. I simply presented a concise, logical argument that you didn't understand and in error declared a fallacy.
Are you oblivious to the information provided?

"X is amusing, absurd. Therefore it is false"

That's the category "The argument went way over your head, I win again" fall under.

Suggesting you've provided me with sufficient information for some vaguely defined position and then act as I don't understand that position is not how you settle a debate, any high school social studies class should have provided you with that much knowledge at least.

If you truly had a valid point you would've clearly defined it. But your inability to debate have led you to think that the only way you'll "win" is by committing logical fallacies.

draxx_tOfU

Holy...

no offense Astner but you are sounding like a soulless automaton...

can't we all just get along...

dadudemon

Originally posted by Astner
Are you compelled reply, regardless whether or not the content is credible or not?

You're not attempting to make an argument, the post I'm replying to is nothing short of an attempt to end this topic granting you the final reply.

Tsunade grabbed a huge sword and swung it with super strength.

Originally posted by Astner
It refutes your claim of that those physics engines are complex programs applied in the frontiers of science.

It does not. It, with absurd strawman, only "proves" that there are drawing applications that are really complex. (Guess what? You can assign properties to the components of those objects...like volume and density. So, you really just further proved my point that you had no clue what you were talking about, tried to back pedal, tried use a strawman to throw us off topic just to argue about something else, and ended up bringing us back to square (teehee) one which is what I was talking about in the first place. In other words, you were wrong, didn't know what you were talking about, and indirectly admitted I was right while trying to use a logical fallacy. In other words, PWNED!)

So, I need you to quote me where I said that 'those physics engines are complex programs applied in the frontiers of science."

If you're referring to video games physics engines (because that would be the context of the conversation), wouldn't you sh*t yourself if a video game engine was adapted and used in the "frontiers of science?" laughing

http://www.gamepro.com/article/news/213103/unity-game-engine-used-to-solve-crime/

That's just one example that I found in less than 2 seconds of google searching (first hit).

laughing

So even your botched and out of context attempt at an argument failed.

So, what have we learned? You didn't know about physics engines, 3D modeling, and using video game physics engines in scientific ways. But, now you do.

Originally posted by Astner
No you didn't calculate it. If that was the case you've would have provided the answer you promised as soon as you made the calculation rather than behave in this childish manner.

I did calculate it. Again, it starts with a 9. I'll give you a second hint: it ends in a number greater than 0.

Do you want another hint?

Also, why do you think I dropped that whole debate to begin with? (Again, this is not a trick question as I have directly indicated why, multiple times.)

Originally posted by Astner
So you calculated the mass of an object without without any density related factors? Do I have to point out to you that this is impossible?

If you actually paid attention, you'd know that we DO have the mass of a bamboo katana as I posted a link to one.

On top of this, all we need to know is average density. Modeling the object can net us a volume and since we have the mass, we can find average density.

Keep in mind, this whole debate started with your horrendous math and formula.

Remember posting this ugly post:

Originally posted by Astner
A regular katana handle is approximately 4 cm, and the length of Gamabuta's katana's handle is a approximately 3/2 of Tsunande's length (or 2 meters).

A reasonable assumption would be that Gamabuta's katana has the same density as a regular katana.

So the formula is basically:

(/)^3 * =

(2/.05)^3 * 2 = 250 000 kg (= 250 tons)

So, if you can even muster the ability to admit fault, please do. Just say, "Man, I was horribly wrong. I apologize for dragging this out for so long."

I actually showed you why you were wrong and you STILL didn't admit it.

But, here's another example, because I have time.

Assume rectangular cuboid A with dimensions 2x4x3 cm

Assume density is uniform and density is 2g/cm^3.

Its mass is found as follows:

Find volume:

2*4*3 = 24

Mutiply density by volume to find the mass:

(2g/cm^3)*(24cm^3/1) = 48g

Easy, right?

Now, let's assume rectangular cuboid B, which is congruent to rectangular cuboid A.

It has dimensions 6x12x9

Volume = 648

Mass = 1296 g

Can we figure out the mass of that object from JUST the measure of ONE side only?

Check:

Pretend we only know one side: 6.

Pretend that this is a far more complex geometric figure than a rectangular cuboid and we can't guess the other sides (despite the fact that we really could, quite easily...just humor me, here.)

Let's try your way:

Pretend we only know that Object A masses 48g and has a side measuring 2cm

Pretend we only know that Object B is congruent to Object A and has the same congruent side measuring 6cm.

Let's do it your way:

(6cm/2cm)^3*48g = 432g

Does 432 = 1296?

No.

Does 432g*3 = 1296?

Yup.

So what did we learn? You did NOT get it correct, even after you backpedaled and changed everything up, again.

Do we need to know the density to find it's mass? Nope. Not at all. smile

But we had to multiply the result by 3 to get the proper mass...something I am not seeing in your work, anywhere.

What is 3? shifty

Do you want me to show you the proper formula for doing this calculation?

Now, I already showed where your math was wrong on using "density", but instead of admitting where you were wrong, you got angry about the numbers not being realistic for a Katana. I see no need to rehash that as the point still stands.

Is that what you said the first time? Nope. You said this:

Originally posted by Astner
(/)^3 * =

(2/.05)^3 * 2 = 250 000 kg (= 250 tons)

Did I criticize your math, very politely, the first time? Yup, I sure did. did you respond with anything other than admitting fault? Yup, you sure did. Should you have admitted fault? Yup. Would our conversation have extended for as long as it did if you would have admitted fault? No, not at all. In fact, I would have been super happy to help you works towards a solution: I would have called you, if necessary, cause I like doing these things.

Originally posted by Astner
So you pretend has you've build a perfect katana by manually adding volume elements...bla bla bla... have.

I didn't pretend anything.

Quote me where I said I used a software to do a calculation on this topic.

And, lol at you pretending that you need to "manually add volume elements". HA! You just draw the figure and look at the volume measure for which it occupies on the right. I clearly stated that in the portion you quoted, already.

But, I did find out from my homie that it is not super accurate as it rounds up based on the "cubes" of space it takes up. To make it more accurate, you'd just have to make the object bigger and use the volume measurement inside the program multiplied by factor each cube represents your unit of measure (such as cm or meter.)

Like you said, that would be less accurate than doing it with a straight relational calculation.

Originally posted by Astner
No, I'm not diagnosed with autism. But then again, this is just another attempt of an Ad Hominem attack to avoid reaching the core of your hollow argument.

No, it is not. It is a way to explain why you seem to miss points, context, and use strawman arguments. Following conversation context is a problem with some forms of autism. It would have made me feel better about our convo and I would probably have felt bad for some of the tone in my posts.

Like I said, I would be completely understanding and nice about it, but it looks like you don't have it, so what now? You're just a poopypants! HA! laughing

Originally posted by Astner
Apparently you suggest that I'm attempting to "save my ego", implying ...bla bla bla...cubes.

You've admitted fault twice, but still fail to admit fault on the real topic: the original reason I responded to you.

Also, it's not hard to admit fault when you've made simple math errors and accidentally called yourself wrong.

Jugglenaut

Hurr durr, Pencil can't grasp the concept of fiction again.

Astner

I would like to thank dadudemon for at least attempting this time.

Originally posted by dadudemon
It does not. It, with absurd strawman, only "proves" that there are drawing applications that are really complex. (Guess what? You can assign properties to the components of those objects...like volume and density. So, you really just further proved my point that you had no clue what you were talking about, tried to back pedal, tried use a strawman to throw us off topic just to argue about something else, and ended up bringing us back to square (teehee) one which is what I was talking about in the first place. In other words, you were wrong, didn't know what you were talking about, and indirectly admitted I was right while trying to use a logical fallacy. In other words, PWNED!)

So, I need you to quote me where I said that 'those physics engines are complex programs applied in the frontiers of science."

If you're referring to video games physics engines (because that would be the context of the conversation), wouldn't you sh*t yourself if a video game engine was adapted and used in the "frontiers of science?" laughing

http://www.gamepro.com/article/news/213103/unity-game-engine-used-to-solve-crime/

That's just one example that I found in less than 2 seconds of google searching (first hit).

laughing

So even your botched and out of context attempt at an argument failed.

So, what have we learned? You didn't know about physics engines, 3D modeling, and using video game physics engines in scientific ways. But, now you do.
I thought it was rather obvious that when I mentioned the application of science, I meant natural sciences, as in physics, chemistry and biology. Mainly physics as it's the current subject of interest, and not some "off the charts so I can win" argument.

But granted, physics engines can be applied to crime scenes as well.

Originally posted by dadudemon
I did calculate it. Again, it starts with a 9. I'll give you a second hint: it ends in a number greater than 0.

Do you want another hint?
Thank you for once again proving my point.

Originally posted by dadudemon
Also, why do you think I dropped that whole debate to begin with? (Again, this is not a trick question as I have directly indicated why, multiple times.)
Because you didn't solve it at all.

Originally posted by dadudemon
If you actually paid attention, you'd know that we DO have the mass of a bamboo katana as I posted a link to one.

On top of this, all we need to know is average density. Modeling the object can net us a volume and since we have the mass, we can find average density.
Even if the density of a regular katana isn't the same, it's a somewhat plausible estimation. The density of a material has three main varying factors, the mass of the molecules (or atoms), the molecular structure and material irregularities.

Originally posted by dadudemon
So, if you can even muster the ability to admit fault, please do. Just say, "Man, I was horribly wrong. I apologize for dragging this out for so long."

I actually showed you why you were wrong and you STILL didn't admit it.

But, here's another example, because I have time.

Assume rectangular cuboid A with dimensions 2x4x3 cm

Assume density is uniform and density is 2g/cm^3.

Its mass is found as follows:

Find volume:

2*4*3 = 24

Mutiply density by volume to find the mass:

(2g/cm^3)*(24cm^3/1) = 48g

Easy, right?

Now, let's assume rectangular cuboid B, which is congruent to rectangular cuboid A.

It has dimensions 6x12x9

Volume = 648

Mass = 1296 g

Can we figure out the mass of that object from JUST the measure of ONE side only?

Check:

Pretend we only know one side: 6.

Pretend that this is a far more complex geometric figure than a rectangular cuboid and we can't guess the other sides (despite the fact that we really could, quite easily...just humor me, here.)

Let's try your way:

Pretend we only know that Object A masses 48g and has a side measuring 2cm

Pretend we only know that Object B is congruent to Object A and has the same congruent side measuring 6cm.

Let's do it your way:

(6cm/2cm)^3*48g = 432g

Does 432 = 1296?

No.

Does 432g*3 = 1296?

Yup.

So what did we learn? You did NOT get it correct, even after you backpedaled and changed everything up, again.
Did you blame me for making a numerical error?

(6/2)^3 * 48 = 1296

(3^3 * 48 = 27 * 48 = 1296)

Originally posted by dadudemon
Quote me where I said I used a software to do a calculation on this topic.
Originally posted by dadudemon
You're right, it gives you a read out on the right, based on this little "units" of area, (or wherever you want to throw you statistics window while building 3D objects. Imagine that: back to Archimedes, like I effin' said. WEEEEEE!)

Originally posted by dadudemon
But, I did find out from my homie that it is not super accurate as it rounds up based on the "cubes" of space it takes up. To make it more accurate, you'd just have to make the object bigger and use the volume measurement inside the program multiplied by factor each cube represents your unit of measure (such as cm or meter.)
I could care less what supposedly your underqualified friend have told you. Even if he was a physics professor from the institute of fundamental physics proof would still be required

In other words, provide the source and explanation for the calculations.

Originally posted by dadudemon
Aside from what I said not being a fallacious "argument" at all, you fail to realize that it doesn't require you to be rabidly obsessed to have to save face.

Responding to defend, period, is the only requirement from my statement. You could be 103 years old, on your deathbed, and if you responded to defend your ego, I would still be correct: You're trying to save your ego.
But my intent is interest. Not that it's of any value, as the intent of ones arguments doesn't affect ones arguments at all.

Originally posted by dadudemon
You are partially correct.

I just got done writing a paper on the future of computing. It covered mostly memristors and graphene transistors. Would you like to read that paper? smile

I'm sure. One week you have no clue of what it is, the next you're writing a paper of it.

Originally posted by dadudemon
- Asnter, if you respond to me, only respond to the sword stuff, but none of the stuff that does not directly apply to Tsunade's strength. That other stuff, like graphene, ego, etc., just send me a PM.
The article you posted, read through IV. STATIC DEFLECTION MEASUREMENTS because you must have missed that.

Originally posted by Jugglenaut
Hurr durr, Pencil can't grasp the concept of fiction again.
Thank you for your insightful comment.