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  Calculating acceleration and decelleration to reach a certain point in time  (Read 9535 times)
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Offline Eli Delventhal

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« Reply #30 - Posted 2009-04-20 22:42:39 »

A=(2*(d-V0*t))/t²


Yeah, that's as far as I got but I don't want to use time as an input, instead I want to replace that input with V(f). But naturally to find how long it takes to get from V(0) to V(f) I need to know A in the first place.  Tongue

So I don't think I can use this equation.

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Offline DzzD
« Reply #31 - Posted 2009-04-20 22:43:44 »

Quote
Hm I don't think V = V0 + t*A is correct. It should be V = V0 + A + V0 + 2A + V0 + 3A ... v0 + tA which simplifies to V = t*V0 + ((t * (t+1))/2) * A .
V=V0 + t*A

this is ok, why not ? this is just derivated by t => instant velocity


EDIT :

this one may work  Huh ....

A=-0.5*(V0-V)*(V+V0)/D

Offline Eli Delventhal

JGO Kernel


Medals: 42
Projects: 11
Exp: 10 years


Game Engineer


« Reply #32 - Posted 2009-04-20 23:55:01 »

Thinking about it, I guess I do need t, because you could decelerate very very slowly to get to a certain point, or very very quickly. Like the above example I have you could increase the time if you wanted to just decelerate half as quickly. So I guess once I know that all that's really missing is the final velocity I want, which I suppose I can probably figure out the place for. The equation above really calculates the value to bring the from 0 to the passed amount, no?

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Offline DzzD
« Reply #33 - Posted 2009-04-21 01:17:47 »

I think yes, if you want V=50 unit/s and D=215 unit and V0=10 unit/s :

A=-0.5*(V0-V)*(V+V0)/D

A=5.58


t is :

A=(V-V0)/t
t=(V-V0)/A
t=7.16


the solution is at t=7.16


Verification :


D at t=7.16 :

(D0 is removed because it is a constant/offset)

D=V0*t+0.5*A*t²
D=10*7.16+0.5*5.58*7.16²
D= 72+143
D=215


and V at t=7.16 :

V=V0+t*A
V=10*+7.16*5.68
V=50


good that's all right !

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