KillerMovies - Movies That Matter!

REGISTER HERE TO JOIN IN! - It's easy and it's free!
Home » Misc » Computer / Video Games Discussion » Game Math.

Game Math.
Started by: dadudemon

Forum Jump:
Post New Thread    Post A Reply
Pages (7): « 1 2 [3] 4 5 » ... Last »   Last Thread   Next Thread
Author
Thread
Morridini
Morridini

Gender: Male
Location: Norway

There's way too much speculation going around for you to need to concern yourself with such details, do simplifications. The 300 ton has got to be an estimate, I set g=10 instead of 9.81, the height he raises it and the time spent are both very unsure, all in all making the margin for error huge. Just go by a decent guess, like 0.02 m^2.


__________________
GoldenEye for XBLA!!

Thanks to Raoul for the sig!

Old Post Jul 26th, 2011 01:59 PM
Morridini is currently offline Click here to Send Morridini a Private Message Find more posts by Morridini Edit/Delete Message Reply w/Quote Quick Quote
Burning thought
Restricted

Gender: Male
Location: The end of eternity

Account Restricted

300 tons was calculated here;

quote: (post)
Originally posted by BloodRain

Obelisks tip;
The square based pyramid tips height is equal to its lengths which are equal to Raziel's height.

V= 1/3 b2*h
V= 1/3 175^2*175
V= 1/3 30,625*175
V= 1/3 5,359,375

V= 1,786,458.33 cm3
Density = 2.6 g/cm3 [granite type density]

1,786,458.33*2.6 = 4,644,791.66 g = 5.12tons.

Obelisks frustum body;
The base of the obelisk is 2x Raziel's height at 350cm. The height is 4x the base for 1,400cm

Pyramidal frustum volume = ⅓h(a1+a2+(√a1*a2))
h is height, a1 is area of base square and a2 is area of top square.
= (1400cm/3)*(350cm^2+175cm^2+√(350cm^2*175cm^2))
= 466.666*(122500+30625+√3751562500)
= 466.666*(122500+30625+61250)
= 466.666*214375
= 100,041,524cm3

100,041,524*2.6 = 260,107,962 g = 286.719949 tons

286.719949 ton body + 5.12 ton tip = 291.839949 tons.

But because a friend has another way of calcing it that had slightly higher results I'll use the middle ground of 300 tons.


Using the density as solid granite, the average building material for obelisks, granite seems the closest we can get but yes, its an estimation. in a fictional world its nigh impossible to get facts without a specific canon statement which most games vs games do not have.

Using that calculation of yours we get; 3 900 000 /0.02 m^2

195000000 pascals (195 megapascals)- nearly 2000 bar, about 400 tons of force over his hands, which makes sense if he was lifting the whole 300 tons at that speed but this is not the case.

Seems a fair enough number, although I guess to get the final figuire I will have to wait until a kind soul with some time on their hands can calculate how the lever would work.

Taking into account I will want to be mathing how much pressure he can create in a punch, moving far faster than he does in the video we are mathing now. How would I use his strength found in this current calculation in conjunction with new acceleration/movements?


__________________

Last edited by Burning thought on Jul 26th, 2011 at 02:27 PM

Old Post Jul 26th, 2011 02:13 PM
Burning thought is currently offline Click here to Send Burning thought a Private Message Find more posts by Burning thought Edit/Delete Message Reply w/Quote Quick Quote
BloodRain
Knight of Retribution

Gender: Male
Location: Midcyru

Soo.. F = M(a+g) where M= 272155.422 kg, a=2.8m/s^2 [1.4m in 1 sec] and g=9.81m/s.

3431879.87142 kgm/s^2 = 3.431x10^6 N = 3.431 Mega Newton. [To my understanding]


Can you explain the 'x = x0 + v0*t + 0.5a*t^2' formula and why the 'x0' wasnt used? I'm only used to seeing the 'a = 2(distance in meters - (initial velocity*time))/t^2' that sorta matches your 'a=2x/t^2', and I take it the 'v0*t' is the initial velocity which is a stationary 0m/s. So whats that x0?


__________________

"Gonna need more chloroform..."



"If you look down on us, judge us, condemn us... then you shall fear us!"

Old Post Jul 26th, 2011 02:34 PM
BloodRain is currently offline Click here to Send BloodRain a Private Message Find more posts by BloodRain Edit/Delete Message Reply w/Quote Quick Quote
Burning thought
Restricted

Gender: Male
Location: The end of eternity

Account Restricted

I am wondering if its even possible to find the pressure used based on a lever, wouldnt the starting weight of the object be near its max? when its only like, an inch or two off the ground, then gradually the force required would be less as weight gets transfered to the axis.


__________________

Old Post Jul 26th, 2011 02:41 PM
Burning thought is currently offline Click here to Send Burning thought a Private Message Find more posts by Burning thought Edit/Delete Message Reply w/Quote Quick Quote
BloodRain
Knight of Retribution

Gender: Male
Location: Midcyru

Wouldn't using a torque calculation be used for this?


__________________

"Gonna need more chloroform..."



"If you look down on us, judge us, condemn us... then you shall fear us!"

Old Post Jul 26th, 2011 03:20 PM
BloodRain is currently offline Click here to Send BloodRain a Private Message Find more posts by BloodRain Edit/Delete Message Reply w/Quote Quick Quote
Burning thought
Restricted

Gender: Male
Location: The end of eternity

Account Restricted

This;

http://www.bondhus.com/tech-library/body-1b.htm

Sort of thing? so if i get it right, the applied force being 3.9MN multiplied by about 2 meters (distance from Raziels claws to the axis?).

So 3.9MN x 2= 7800000 (7.8MN) newtons? Not sure if that can be right, I probably did something wrong, but I am confused.


__________________

Old Post Jul 26th, 2011 03:36 PM
Burning thought is currently offline Click here to Send Burning thought a Private Message Find more posts by Burning thought Edit/Delete Message Reply w/Quote Quick Quote
Morridini
Morridini

Gender: Male
Location: Norway

quote: (post)
Originally posted by BloodRain
Can you explain the 'x = x0 + v0*t + 0.5a*t^2' formula and why the 'x0' wasnt used? I'm only used to seeing the 'a = 2(distance in meters - (initial velocity*time))/t^2' that sorta matches your 'a=2x/t^2', and I take it the 'v0*t' is the initial velocity which is a stationary 0m/s. So whats that x0?


You're right, it was entirely pointless of me to include it here (I just typed it down as I remembered it), x0 stands for the initial position, say he picked something up that was buried 10 cm into the ground, then x0=-0.1. But in our case it was touching the ground, making x0=0.


__________________
GoldenEye for XBLA!!

Thanks to Raoul for the sig!

Old Post Jul 26th, 2011 03:42 PM
Morridini is currently offline Click here to Send Morridini a Private Message Find more posts by Morridini Edit/Delete Message Reply w/Quote Quick Quote
Burning thought
Restricted

Gender: Male
Location: The end of eternity

Account Restricted

Morridini can you shed some light on the Torque subject? can this be used to find out the force? if it was a square cube perfect in shape I assume this would be easier but because its a large obelisk and is taller than it is wide by a fair margin it must have some gravity that works against Raziel even when its on its edge if that makes sense.


__________________

Old Post Jul 26th, 2011 03:50 PM
Burning thought is currently offline Click here to Send Burning thought a Private Message Find more posts by Burning thought Edit/Delete Message Reply w/Quote Quick Quote
BloodRain
Knight of Retribution

Gender: Male
Location: Midcyru

#I found a formula for torque being mgl/2d. Mass*gravity*axis length of 3.5m / distance from the axis to the applied force of also 3.5m*2. Soo...:
(272155.422kg*9.81*3.5m)/ (2*3.5m) = 1334922.34 kg/meter. ##Then I timed by the 1.4m/s lift for 1868891.28 kg m/s = 1.868 mega newtons.

Its probably not torque, #the wrong formula and ##wrong interpretations... But dammit it looks good from where I'm standing compared to what a full lift is :V


__________________

"Gonna need more chloroform..."



"If you look down on us, judge us, condemn us... then you shall fear us!"

Old Post Jul 26th, 2011 04:31 PM
BloodRain is currently offline Click here to Send BloodRain a Private Message Find more posts by BloodRain Edit/Delete Message Reply w/Quote Quick Quote
Burning thought
Restricted

Gender: Male
Location: The end of eternity

Account Restricted

If thats true, then Raziel is excerting 209.971562 tons of force across the area of his hand which using Morridinis estimate 0.02 squared is fair. So it makes some sense that 200 tons can move a portion of the 300 tons. Whether or not its the right figuire I dont know.


__________________

Old Post Jul 26th, 2011 05:12 PM
Burning thought is currently offline Click here to Send Burning thought a Private Message Find more posts by Burning thought Edit/Delete Message Reply w/Quote Quick Quote
dadudemon
Senior Member

Gender: Male
Location: Bacta Tank.

The calculation is much more complicated because the block is so tall. Gravity will act on the y axis of the block, also helping it to be picked up on one side.

Picture two force vectors: one force vector lifting straight up on one side on the bottom of the block (your video game character) and one force vector always pulling down on the very top of the block (gravity). The system is "dynamic" meaning, the more you lift it, the easier it is to lift.


For the sake of the calculation, finding how much force is required to lift it just a tiny tiny bit would be enough to find the maximum force applied.

Why?

Because that would give us the upper limit without having to involve gravity too much. Remember, only do enough to come up with your answer. This is a "simple" second class lever system, if you do it that way.

Fe = l dl / de

FE = force effort

l = load

dl = distance of load from fulcrum

de = distance of effort.





IIRC, from my first year physics class...you pretend the load is at the center and divide by 2 to find the "load" of a lever that is actually the load.



Meaning...pretend you're picking up a support beam on one side. The Fulcrum is the other side where it touches the ground. The load is the support beam itself, meaning, the load is distributed evenly across the whole lever.


One way to test my memory to see if I'm remember this right is to split the load into 4-8 even parts across an imaginary lever arm. Calculate for each point, and add the system up/together. Compare that to my idea of putting all the load in the center and dividing it by two.



If the two are very close, then I'm remembering this correctly.


Then we just find the area of his hands. That should not be too hard. His hands look to be, if you consider that he only has 2 fingers and one thumb, about the same area as a large man's hand.


__________________

Old Post Jul 26th, 2011 05:41 PM
dadudemon is currently offline Click here to Send dadudemon a Private Message Find more posts by dadudemon Edit/Delete Message Reply w/Quote Quick Quote
Burning thought
Restricted

Gender: Male
Location: The end of eternity

Account Restricted

Raziel seems to lift it up to about his chest, then as he turns his hands around the block comes down a little on him and then he pushes it back up and over.

Do you not have to calculate the energy of the push to get it over the "axis" as well? or just the pulling up of the thing?

I will probably wait for someone who understands it more like Morridini to do this, as I dont fully understand it.

The load is 300 tons? But whats the fulcrum? the axis that touches the ground as Raziel lifts?


__________________

Old Post Jul 26th, 2011 06:00 PM
Burning thought is currently offline Click here to Send Burning thought a Private Message Find more posts by Burning thought Edit/Delete Message Reply w/Quote Quick Quote
dadudemon
Senior Member

Gender: Male
Location: Bacta Tank.

quote: (post)
Originally posted by Burning thought
Raziel seems to lift it up to about his chest, then as he turns his hands around the block comes down a little on him and then he pushes it back up and over.

Do you not have to calculate the energy of the push to get it over the "axis" as well? or just the pulling up of the thing?

I will probably wait for someone who understands it more like Morridini to do this, as I dont fully understand it.

The load is 300 tons? But whats the fulcrum? the axis that touches the ground as Raziel lifts?


Oh.


I thought he picked up one side and tipped it over.

If that's the case then BloodRain's answer will do. Just figure out the surface area of your hands if you wear and XL glove. Then you have your answer.


__________________

Old Post Jul 26th, 2011 06:12 PM
dadudemon is currently offline Click here to Send dadudemon a Private Message Find more posts by dadudemon Edit/Delete Message Reply w/Quote Quick Quote
Nephthys
The Gr8est!!!!!!!!

Gender: Male
Location: The End

A video would likely make this easier.


__________________

Old Post Jul 26th, 2011 06:19 PM
Nephthys is currently offline Click here to Send Nephthys a Private Message Find more posts by Nephthys Edit/Delete Message Reply w/Quote Quick Quote
Burning thought
Restricted

Gender: Male
Location: The end of eternity

Account Restricted

quote: (post)
Originally posted by dadudemon
Oh.


I thought he picked up one side and tipped it over.

If that's the case then BloodRain's answer will do. Just figure out the surface area of your hands if you wear and XL glove. Then you have your answer.


He lifted it up to about his chest with his hands underneath, then moved them round to push it with his palms over;

(please log in to view the image)

Uploaded with ImageShack.us

(please log in to view the image)

Uploaded with ImageShack.us

(please log in to view the image)

Uploaded with ImageShack.us

Morridini said 0.02m^2 was a fair size for Raziels hands? I dont wear gloves so I could not tell, I guess I could ask someone who does if thats the best way.


__________________

Old Post Jul 26th, 2011 06:33 PM
Burning thought is currently offline Click here to Send Burning thought a Private Message Find more posts by Burning thought Edit/Delete Message Reply w/Quote Quick Quote
Morridini
Morridini

Gender: Male
Location: Norway

quote: (post)
Originally posted by Burning thought
I will probably wait for someone who understands it more like Morridini to do this, as I dont fully understand it.


I haven't done any "torque" kinda calculations since Mechanics at the Uni several years ago, so I'm actually not that good on this part. I still think the clue lies in making a free body diagram and looking at how the force gets spread, so I might take a look at it tomorrow.

As for the hand size, I simply went to the Wikipedia page for hand, looked at average width and length of a male hand, and multiplied them together (which would be ok if our hands were square-shaped), and rounded up to 0.2 as Raziel seems to have slightly larger hands.


__________________
GoldenEye for XBLA!!

Thanks to Raoul for the sig!

Old Post Jul 26th, 2011 06:49 PM
Morridini is currently offline Click here to Send Morridini a Private Message Find more posts by Morridini Edit/Delete Message Reply w/Quote Quick Quote
BloodRain
Knight of Retribution

Gender: Male
Location: Midcyru

Fe = l dl / de

l = 277155.422 kg
dl= 7.85m [center of the mass? Heck if I know what the dl is]
de= 1.4m/s [Not sure what other one to use, or at what minimal height]

Fe = (277155.422*7.85) / 1.4
Fe = 1558999.25 kgm/s [1.558 MegaNewtons]
1 N = 1 kgm/s

^ Admittedly rough and.. partially skim read and didnt understand the half of the length of the center of something...

The lift in video form.

And the area of what part of the hand? The fingers, palm and fingers, the tips and palm[like in a thrust], the tips alone?

Btw the surface area for an open palm+fingers is about 0.015m2 [where as a rectangle itd 10cmx20cm = 0.02m2] for a guy, Raziel has a 30cmx12cm hand. Given that its roughly the same form as a persons [0.02 block/ 0.015 area = 1.333~] his palm+fingers surface area would be 0.027m2. [0.036 block / 1.333~]


quote:
If that's the case then BloodRain's answer will do


Well thats surprising.


__________________

"Gonna need more chloroform..."



"If you look down on us, judge us, condemn us... then you shall fear us!"

Old Post Jul 26th, 2011 07:28 PM
BloodRain is currently offline Click here to Send BloodRain a Private Message Find more posts by BloodRain Edit/Delete Message Reply w/Quote Quick Quote
Burning thought
Restricted

Gender: Male
Location: The end of eternity

Account Restricted

ok, if we take in mind Raziel can create 1.868 meganewtons of force. Or 200 tons across his hands to push manip this thing.

How do I calculate what with accleration and such being different, this;

http://www.youtube.com/watch?v=Jws63E9dNN8#t=2m5s

Now granted he is not using one of his arms, but taking into account earlier strength, the distance he jumps and the arc+distance his clawed hand travels overall how would I calculate how much force he is using? Can I use previous strength of 1.868MN (find out how much one less arm makes in difference) and then use the acceleration shown in this video?

Just the first strike, the further kick and such is probably something different.


__________________

Last edited by Burning thought on Jul 26th, 2011 at 08:00 PM

Old Post Jul 26th, 2011 07:52 PM
Burning thought is currently offline Click here to Send Burning thought a Private Message Find more posts by Burning thought Edit/Delete Message Reply w/Quote Quick Quote
Morridini
Morridini

Gender: Male
Location: Norway

You can't really use this info for that situation, we don't know if Razienl put all his effort into lifting the monolith or not, nor do we know if he want all out in that fight either.

quote: (post)
Originally posted by BloodRain
l = 277155.422 kg

Why did you set the moment of inertia I equal to the mass? Surely it must be something else, especially since you rotate around one edge and not the axis.


__________________
GoldenEye for XBLA!!

Thanks to Raoul for the sig!

Last edited by Morridini on Jul 26th, 2011 at 08:00 PM

Old Post Jul 26th, 2011 07:56 PM
Morridini is currently offline Click here to Send Morridini a Private Message Find more posts by Morridini Edit/Delete Message Reply w/Quote Quick Quote
Burning thought
Restricted

Gender: Male
Location: The end of eternity

Account Restricted

Well what if we assume he put all his effort into the monolith AND in this fight? In games vs we usually take the best strength feat they have and claim that as their upper limit, I dont know how to find an upper limit if it was deemed Raziel hardly tried (he does not seem to try much more here than he does moving around 50 ton blocks earlier in the game but I am not sure it works).

So assume both are max power? How would the calculation look to find out force in this example, assuming hes using max power of 1.8MN?


__________________

Old Post Jul 26th, 2011 08:02 PM
Burning thought is currently offline Click here to Send Burning thought a Private Message Find more posts by Burning thought Edit/Delete Message Reply w/Quote Quick Quote
All times are UTC. The time now is 08:24 PM.
Pages (7): « 1 2 [3] 4 5 » ... Last »   Last Thread   Next Thread

Home » Misc » Computer / Video Games Discussion » Game Math.

Email this Page
Subscribe to this Thread
   Post New Thread  Post A Reply

Forum Jump:
Search by user:
 

Forum Rules:
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts
HTML code is OFF
vB code is ON
Smilies are ON
[IMG] code is ON

Text-only version
 

< - KillerMovies.com - Forum Archive - Forum Rules >


© Copyright 2000-2006, KillerMovies.com. All Rights Reserved.
Powered by: vBulletin, copyright ©2000-2006, Jelsoft Enterprises Limited.