Raikage vs Tsunade

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.

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

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.

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.

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' 🙂

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

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

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.

[Large (volume)] = [Small (volume)] * ([new length]/[previous length])^3 = 1/4 * L^3 / 8 = L^3 / 2^5

Now lets control check:

[length]*[width]*[breadth] = 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)

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.

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?

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.

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

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.

[Large (volume)] = [Small (volume)] * ([new length]/[previous length])^3 = 1/4 * L^3 / 8 [b]= L^3 / 2^5

Now lets control check:

[length]*[width]*[breadth] = L * L/4 * L/8 = L^3 / 2^5

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

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:

([Length of Gamabuta's katana]/[Length of a regular katana])^3 * [Density of a regular katana] = [Mass of Gamabunta's katana]

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"😉. 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
([Length of Gamabuta's katana]/[Length of a regular katana])^3 * [Density of a regular katana] = [Mass of Gamabunta's katana]

(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.) 🙂

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. 😐

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.

Originally posted by dadudemon
Several things:

1. Not a rectangle, but a rectangular cuboid.

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.

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.

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.

Example: Sphere

Radius: r

Volume = c*r^3
Where: c = (4 π / 3)

Determine the volume of a sphere with radius R.

[New body] = [Original body]*( [new radius] / [original radius] ) = 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:

([Length of Gamabuta's katana]/[Length of a regular katana])^3 * [Density of a regular katana] = [Mass of Gamabunta's katana]

The actual calculation would be:

[Mass of Gamabuta's katana] = ([Gamabuta's katana width]/[Original katana width])^3 * [Original katana volume] * [Original katana density] = [Mass of Gamabuta's katana] = ([Gamabuta's katana width]/[Original katana width])^3 * [Original katana mass] = (2/.5)^3[-] * 2[kg] = 250 000[kg] = 250[tons]

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.

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).

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.

facepalm @ everything

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.

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. 😐

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.

[New body] = [Original body]*( [new radius] / [original radius] ) = c*r^3 * (R^3 / r^3) = [b]c*R^3

Works for every body. Period.[/B]

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. 😆 😆 😆 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? 😆
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 [Original katana's mass] I wrote [Original katana's density]. Yeah, let me fix that.

The actual calculation would be:

[Mass of Gamabuta's katana] = ([Gamabuta's katana width]/[Original katana width])^3 * [b][Original katana volume] * [Original katana density] = [Mass of Gamabuta's katana] = ([Gamabuta's katana width]/[Original katana width])^3 * [Original katana mass] = (2/.5)^3[-] * 2[kg] = 250 000[kg] = 250[tons][/B]

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[kg] = 250 000[kg] = 250[tons]

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?

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

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.

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. 😐

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.

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. 😆 😆 😆 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? 😆
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 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.

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.

2000[cm]^3 * 6[g][cm]^-3 = 12000[g] || 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.

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.

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.

Originally posted by dadudemon
One last complaint, Astner: kilograms never convert to tons by dividing by 1000.

Originally posted by dadudemon
This is mostly physics......I think.

Originally posted by Astner
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.

No I'm not. It's quite specific:

Originally posted by dadudemon
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.

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.

So did I. 2, .5 and .25 aren't arbitrary. They're fixed values.

Originally posted by Astner
[list=1][*]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.[/list=1]

1. Oh really? You mean the "calculation" that applies to rectangular cuboids as well? Nice try. Oh, right, it's simple formula.

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.

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.

Full disclosure: I figured you were just an online expert and did not have any real math skills. That should be apparent by your atrocious calculations. 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.

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. Can't help you there, pal. You were supposed to get a laugh out of it and facepalm at what you did: not miss it entirely and contradict yourself. It was a joke, clowning, pulling your chain, etc. Whatever you want to call it. It was an setup that would either make you look foolish or make you laugh. I HOPED for the latter. 🙁
3. But it's not. facepalm Why don't you disprove what I did, if it's erroneous with more botched math, please. 😐

Originally posted by Astner
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 [kg] and [m].

crylaugh

OMG!

crylaugh

Boy, you sure pwned the shit out of me, man.

crylaugh

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.

Originally posted by Astner
Do you even understand what you're doing? I think not, because you're directly implying that a regular katana weights:

2000[cm]^3 * 6[g][cm]^-3 = 12000[g] || [b]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.[/B]

All you can do is fault my numbers for not translating into real world masses for katanas? 😬

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? Imagine that! I must of just...improvised those numbers because they were easier to work with. Damn me and my forward thinking. DAMN ME!

But, let's look at what I ACTUALLY was illustrating to you and that you missed entirely: your math was wrong. 😐 You screwed up on so many levels it's ridiculous.

Originally posted by Astner
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.

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

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.

Originally posted by Astner
I'll give you this, the sword weighs 128 tons.

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, I use SI-units. I never dropped the density. As:

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

hahahaha

So if I post something more credible than that, I win, right?

http://www.wisegeek.com/what-is-a-metric-ton.htm

"The metric ton is often spelled as tonne, and in the United States may also be called a tonneau. A metric ton is not to be confused with the short ton unit, known simply as a ton in the system used in the United States. This ton is equal to 2000 pounds, or roughly 907kg. It should also not be confused with a long ton, a unit no longer in common usage in the United States, which is equal to 2240 pounds.

...

Because of the similarity in pronunciation — and in some circles, spelling — there can often be confusion between a short ton, a long ton, and a metric ton. In general, in the metric-using world the term ton or tonne alone will be used to refer to a metric ton, while the distinction long ton or short ton will be used to refer to the measure of the standard or Imperial system. In the United States, the term ton will be used to refer to a short ton, although in some industries — such as freight — a ton may be assumed to be a long ton. The term metric ton is then used to distinguish the metric unit. As a rule of thumb, it is a good idea to distinguish which unit of measurement you intend, if there is any doubt that your listeners might misconstrue your meaning."

So, not only does your website tell you to start using the correct word, other places say the same exact thing as I do: in America, a "ton" is 2000lbs, a tonne is 1000Kg,

Originally posted by marwash22
lol, that's not quite the point.

I know, I know. But I like physics.

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.

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.

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*

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.

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.

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.

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.[/quite]
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.

[QUOTE=12734202]Originally posted by dadudemon
Full disclosure: I figured you were just an online expert and did not have any real math skills.

Originally posted by dadudemon
That should be apparent by your 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.

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)| &#8800; 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.

[Solution provided by request] (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.

Originally posted by dadudemon
Boy, you sure pwned the shit out of me, man.

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.

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

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?

{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.

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. 😐 You screwed up on so many levels it's ridiculous.

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

The symbol for angles is the angle symbol: &#8736;

And the symbol for the friction coefficient is mu: &#956;

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

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.

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.

Originally posted by dadudemon
You incorrectly simplified out ALL of density when you shouldn't have.

Originally posted by dadudemon
So if I post something more credible than that, I win, right?

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.

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.

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.

😆

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. 😬

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.

😆 😆

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? 😆

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. 🙂

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. 😬

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:
[list][*]Mass isn't measured in m, it's measured in [kg].
[*]Force isn't measured in F, it's measured in [kg][m][s]^-2 (Newton)[/list]

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.

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.