1. Attraction
- Posted by simulat <simulat at INTERGATE.BC.CA> Jun 09, 1999
- 420 views
Hi Mark said: >I have an object with a position {x,y} and velocity {x,y} >i also have other objects with position {x,y} these other objects attract or >repell the first object, how do i caluclate a value to add to the first >objects velocity so that it will be pushed directly away from a repelling >object, or pulled toward an attracting object? The force should be less >further away from the object. I wrote a program last year called "linofo" that deals with attraction. Unfortunately, it got lost in a disk crash a few months ago, so I don't have the code anymore. I think I may have uploaded a demo to the list, so if anyone happens to have a copy I'd sure like it if they could send one my way. Anyway - the way it worked was like this. The formulae are: 1--f=m*a 2--s=1/2*a*t^2 3--s=v*t 4--f=G*(m1*m2)/r^2 (4) gives the force of attraction between 2 masses, where the gravitational constant is G and the distance between the objects is r. From (1) we get f=m*a -> a=f/m Substituting that a in (2) we get s=1/2*(f/m)*t^2 --- (5) since f=G*(m1*m2)/r^2 (from (4)) (5) becomes: s1=1/2*((G*(m1*m2)/r^2)/m1)*t^2 for the distance that m1 moves in time t and s2=1/2*((G*(m1*m2)/r^2)/m2)*t^2 for the distance that m2 moves in time t. The only real mystery here is what to do with t? What is the nature of time in a computer simulation? What I do is to divide the continuous stream of time into discrete "ticks". A "tick" is a minumum unit of time. So, when we move our objects, we want to know how far they move in a "tick". Movement then becomes a series of steps, each step being the distance moved in a tick. Since we are dealing with ticks, we can set t in the equations above to be 1 which gives: s1=1/2*((G*(m1*m2)/r^2)/m1) for the distance that m1 moves in a tick and s2=1/2*((G*(m1*m2)/r^2)/m2) for the distance that m2 moves in a tick. So now we have a distance of movement. What is the direction? The direction is the direction of the line joining the two points. So now we have a magitude and a direction for our step. This makes the step into a vector. If you have many masses, then the step of any one mass is just the vector sum of the step caused by all of the other masses. This is done with a nested loop: ie For each mass step = 0 for each other mass step = step + step caused by mass and other mass end for move by step amount end for This method gets very slow if there are a lot of masses, because the calculation time rises with the square of the number of masses. I'm not sure how useful it would be in a game situation. But maybe we have a math whiz among us who could make it run faster by eliminating the need for the nested loop. Bye Martin
2. Re: Attraction
- Posted by Liquid-Nitrogen Software <nitrogen_069 at HOTMAIL.COM> Jun 09, 1999
- 408 views
Thanks for those who suggested some ideas for my problem. I think I've got something working now. What I'm doing is making a small action-puzzle game to get in the mood to make my RPG. Hopefuly I'll have something playable soon. -Mark.