- #1
krsbuilt
- 15
- 0
I'm new to this forum so please don't mind my being a newbie. I found this equation at school, .5((90-theta)/90)*G=Ag where theta is the angle on the base (not the right angle)
|\
| \ <ramp
|_ \<Theta
^
right angle
and G is the acceleration due and Ag is acceleration in the direction of the ramp. i found this bu knowing if it's a 90 (horizontal/flat) degree angle then there is no acceleration and if it's 0 (vertical) degrees then there's full acceleration. so every degree in between 0 and 90 it adds (1/2)1/90*G to the acceleration to the acceleration in the direction of the ramp.
|\
| \ <ramp
|_ \<Theta
^
right angle
and G is the acceleration due and Ag is acceleration in the direction of the ramp. i found this bu knowing if it's a 90 (horizontal/flat) degree angle then there is no acceleration and if it's 0 (vertical) degrees then there's full acceleration. so every degree in between 0 and 90 it adds (1/2)1/90*G to the acceleration to the acceleration in the direction of the ramp.