So I have a little game which works with small steps, however those steps vary in time, so for example I sometimes have 10 Steps/second and then I have 20 Steps/second. This changes automatically depending on how many steps the user's computer can take. To avoid inaccurate positioning of the game's player object I use y=v0*dt+g*dt^2/2 to determine my objects y-position, where dt is the time since the last step, v0 is the velocity of my object in the beginning of my step and g is the gravity. To calculate the velocity in the end of a step I use v=v0+g*dt what also gives me correct results, independent of whether I use 2 steps with a dt of for example 20ms or one step with a dt of 40ms.
Now I would like to introduce air drag. For simplicity's sake I use a=k*v^2 where a is the air drag's acceleration (I am aware that it would usually result in a force, but since I assume 1kg for my object's mass the force is the same as the resulting acceleration), k is a constant (in this case I'm using 0.001) and v is the speed.
Now in an infinitely small time interval a is k multiplied by the velocity in this small time interval powered by 2. The problem is that v in the next time interval would depend on the drag of the last which again depends on the v of the last interval and so on... In other words: If I use a=k*v^2 I get different results for my position/velocity when I use 2 steps of 20ms than when I use one step of 40ms.
I used to have this problem for my position too, but adding +g*dt^2/2 to the formula for my position fixed the problem since it takes into account that the position depends on the velocity which changes slightly in every infinitely small time interval.
Does something like that exist for air drag too?
And no, I dont mean anything like Adding air drag to a golf ball trajectory equation or similar, for that kind of method only gives correct results when all my steps are the same.
(I hope you can understand my intermediate english, it's not my main language so I would like to say sorry for all the silly mistakes I might have made in my question)
