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Topic, The Formula for Frictional Dynamics... | ||||||
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You must register or log in to post a message.if V=10 and F=2 then it goes 10 8 6 4 2 0 0 0 0 0 that isnt frictional deceleration thats just taking 2 off each time should be V/(1+F)^n [edit] in other words if you have 2 balls with the same frictional resistance and a(ball 1) is moving twice as fast as b(ball 2), then a initially experiences double the resistance to b in yours they both would experience the same What is 3 divided by 0? friction=repetitive division thats all you need to know divide the speed by about 1.01 every frame at 30fps I dont think we need excessively precise stuff :P so with this formula, if the frictional deceleration is larger then the velocity, then you don't go forwards? The number of pixels moved from the start, D, is defined by the formula: where V is the initial velocity and F is the amount of frictional deceleration. | GeneralFirst post of the topicThe number of pixels moved from the start, D, is defined by the formula: where V is the initial velocity and F is the amount of frictional deceleration. |
it just subtracts F from V over and over, or if V<0 (the V>0 requirement is a lie, should be V=/=0) it adds F to V over and over
the formula itself is a lie, and cars show this
two identical cars are rolling along a road, no pedals are pressed, no brakes are on, nothing stopping them except friction and nothing pushing them along
one is going at 100mph and the other at 20mph, according to that formula the speeds both decrease linearly and thus the cars after whatever period of time would be at 80 and 0 simultaniously. this is wrong, the fast car would infact be going less than that and the slow car will never have completely stopped(it will in real life due to other factors).
[edit] another flaw is that the formula doesnt stop itself by checking if nF is larger than V, meaning that the speed will become negative and there are 3 possible outcomes(0 and +-Inf)