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Topic, The Formula for Frictional Dynamics...

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Hexicube 13 years ago
  the formula is designed to look scarier than it is
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)
chris3000 13 years ago
  that is one crazy formula
Ahroo 13 years ago
  ':_

For some reason, I have a feeling that this will do nothing in Flash. |:p
Hexicube 13 years ago
  oh...just remembered
Hexicube 13 years ago
  I just worked it out...that sucks
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
rocky 13 years ago
  That was the joke.
SuperDog 13 years ago
  Nothing... Nothing can be divided by 0
gamelover101 13 years ago
  Garygoh, you're a math expert. Can you solve this equation for me? My calculator won't do it for some reason.

What is 3 divided by 0?
SuperDog 13 years ago
  same here rocky

i know how you feel
rocky 13 years ago
  D:

I'm very good at math but this...
Hexicube 13 years ago
  I like math and I think thats just overdoing it...
friction=repetitive division
gamelover101 13 years ago
  Math? Holy sh- *head asplodes*
Hexicube 13 years ago
  the way friction works is that it divides the speed by a number over and over
thats all you need to know
birjolaxew 13 years ago
  I have no idea what-so-ever that is, but it looks freaking awesome.
Garygoh884 13 years ago
  I think so. If F≥V, then it moves only V pixels. And if it doesn't move all the time (kept adding 0 infinitively), it stopped by friction.
Hexicube 13 years ago
  I just use a simpler formula
divide the speed by about 1.01 every frame at 30fps
I dont think we need excessively precise stuff :P
SimonM 13 years ago
  ah, k, now I get it. :p but if F is larger then V then always (|V-F|+V-F)/2= 0/2 =0
so with this formula, if the frictional deceleration is larger then the velocity, then you don't go forwards?
Garygoh884 13 years ago
  V is expressed as pixels per frame. Also, we use n to be term for this summation.
SimonM 13 years ago
  The formula is fine. What's n? is V expressed in pixels per seconds?
allyally 13 years ago
  I get what its for, but a lot of your equation is a mystery to me. Any I dont really know about putting formulae into games, :P. You've just got to make a game now,
Garygoh884 13 years ago
  This is useful if you want a launch speed that satisfies the equation given the amount of friction and desired length.
allyally 13 years ago
  My head-
Garygoh884 13 years ago
  After hours of experimenting the frictional dynamics in Game Maker 8, here, I made this formula by myself!

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.

[1]

General

First post of the topic

Garygoh884 13 years ago
  After hours of experimenting the frictional dynamics in Game Maker 8, here, I made this formula by myself!

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.
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