04-26-2007, 07:37 AM
...and an addendum to the addendum to point 1 above! I figured I'd explain how that calculation can be performed. ;D
First, a little physics review. The position s of an object moving under constant acceleration is s = x0 + (v0)t + (1/2)at^2, where x0 is the initial position, v0 the initial velocity, a the acceleration, and t the time.
Adapting this formula to our MM falling objects gives d = (1/2)gt^2, where d is the distance of the fall and g is the MM gravity (= 1.3 pixels / frame^2). Solving for t gives t = sqrt (2d/g). This is the time required for any object to fall a given distance in MM.
You can then find how far that object travels horizontally while falling as vt, or the object's speed times the time required for the fall.
There are many potential uses for this sort of calculation. For example, you could use this to make sure that a scorpion that falls from above will land on a moving platform below, without any trial-and-error effort at all.
Ah, the joys of nerdhood.
First, a little physics review. The position s of an object moving under constant acceleration is s = x0 + (v0)t + (1/2)at^2, where x0 is the initial position, v0 the initial velocity, a the acceleration, and t the time.
Adapting this formula to our MM falling objects gives d = (1/2)gt^2, where d is the distance of the fall and g is the MM gravity (= 1.3 pixels / frame^2). Solving for t gives t = sqrt (2d/g). This is the time required for any object to fall a given distance in MM.
You can then find how far that object travels horizontally while falling as vt, or the object's speed times the time required for the fall.
There are many potential uses for this sort of calculation. For example, you could use this to make sure that a scorpion that falls from above will land on a moving platform below, without any trial-and-error effort at all.
Ah, the joys of nerdhood.