blockooJoin Date: 2007-11-08
Post Count: 17202 |
What's the formula I could use to balance an object's weight with an upward BodyForce? |
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AgentFirefoxTop 100 PosterJoin Date: 2008-06-20
Post Count: 22404 |
There is no EXACT formula, but you can use this to get it close:
Force = V * 196.4 |
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blockooJoin Date: 2007-11-08
Post Count: 17202 |
V = the mass of the part, right? |
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zeke505Join Date: 2008-05-26
Post Count: 15765 |
BODYFORCE.force = Vector3.new(0,PART:GetMass() * 196.22, 0) |
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blockooJoin Date: 2007-11-08
Post Count: 17202 |
Wait, do I use 196.22 or 196.4? |
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MrNicNacJoin Date: 2008-08-29
Post Count: 28554 |
Object:GetMass() * 196.2 |
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AgentFirefoxTop 100 PosterJoin Date: 2008-06-20
Post Count: 22404 |
Try using 196.22.
When I was the first person to figure it out, I got 196.2. Now all my bricks are falling a bit. I did not test the 196.4. |
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blockooJoin Date: 2007-11-08
Post Count: 17202 |
Thanks guys |
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pwnedu46Join Date: 2009-05-23
Post Count: 7534 |
"There is no EXACT formula, but you can use this to get it close"
While there isn't a universal formula, you can use vector physics to find the x-component, y-component, and/or z-component force(s) and find the force being exerted on an object by multiplying the mass by the appropriate trig. ratio of theta. (paraphrasing my Pre AP Physics book)
(**IMPORTANT: WHEN USING TRIGONOMETRIC FUNCTIONS TO CALCULATE VECTORS, YOU MUST CONVERT TO DEGREES _NOT RADIANS_**) sry for caps, but I don't want trolls saying "hey that doesn't work1!1!1!!!!!1!1one!," when it does work.
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pwnedu46Join Date: 2009-05-23
Post Count: 7534 |
floodcheck:Remove()
...this works because force is a vector. With a little math, it's very simple.
~pwnedu46~ |
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Uh... No. In ROBLOX, gravity behaves in a much simplified way. It always pulls straight down, and its magnitude and direction do not change at all with respect to your location. No trig or vector math required.
In real life, spherical objects behave like point masses. Basically, they act as if all of their mass were concentrated at a single point, for purposes of gravitation (at least, while you are outside their radius). For irregular, non-spherical objects, or systems with many objects, the calculations become incredibly complex. Not something you would ever be able to do without powerful supercomputers, if at all. |
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pwnedu46Join Date: 2009-05-23
Post Count: 7534 |
@fungal: hmm...you're probably right. Irregular objects would require calculus. No supercomputer would be required, you can use a 3D CAD (more specifically, Autodesk Inventor) program to simulate as close as possible to real-world conditions. On the other hand, when I downloaded that $500+ program (i got a free copy for school :D) it took about 2 hours to download with a 100 Mb/sec connection, and another 3 hours to install it. The total file size was about 8-10 Gb. |
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Pocok5Join Date: 2009-03-13
Post Count: 70 |
I tested 196.4. It worked. |
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ss1122Join Date: 2008-12-16
Post Count: 1601 |
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