2009-10-18, 01:01 PM
Shidoshi Wrote:Take the derivative of the inclination, dy/dx, then take the arctg(dy/dx) for the angle of inclination, then multiply g = 9,81m/s² by the sine(arctg(dy/dx)) and you have your acceleration.
by 'arctg', do you mean arctan, tan^-1, etc?
g*sin[arctan(dy/dx)] is the accelaration at any given point... so you would integrate from the initial point to the current point to find the velocity at any given distance x? and again to find the distance it traveled in a certain time...?
I think I just lost myself....