SKATEBOARDING PHYSICS



THE MATHEMATICS OF VERT SKATING




Vert skating is the type of skateboarding where the rider rides on a half-pipe where at the top of each wall, is completely vertical space.
Mathematics can be used to find exactly how high you will go for a given speed, or how much speed you need for a given height.

To go higher, more work must be done to attain higher speeds. Air resistance plays so little of a role in this equation, it doesn't have it's own variable.

For these equations, these variables will be used to substitute the state of the energy of the rider and other variables:

  • A= All potential, no kinetic
  • B= Potential energy becoming kinetic
  • C= All kinetic, no potential
  • D= Kinetic becoming potential
  • E= All potential, no kinetic
  • H= The height of the ramp
  • R= The radius of the ramp
  • A= The height of the air
  • M= The mass of the skateboarder
  • S= The height of the skateboarder

Using these equations, one can find the amount of work needed to reach any height:

  • L = mvD r
  • E3 = E2
  • K3 + U3 = K2 + U2
  • Efinal = Einitial
  • Kfinal + Ufinal = Kinitial2 + Uinitial
  • W=Fd
  • E2 = E1 + W

Here is a problem solved using these numbers:

  • E2 = E1 + W
  • K2 + U2 = K1 + U1 + W
  • ½ mv32 + mgh3 = ½ mv22 + mgh2
  • (.5)(80kg)(7.15 m/s)2 +(80kg)((3.05m)=(.5)(80kg)(v1)2 + (80kg)(9.8 m/s2)(3.05m) + (80kg)[ 7.15 m/s)2 /2.11](.254)
  • 2045 kg m2/s2 + 2391 kg m2/s2 = 40 kg (v1)2 + 492 kg m2/s2
  • 40 kg(v1)2 = 3944 kg m2/s2
  • v1 = 9.93 m/s

Therefore, in this equation, the rider would have to go at a rate of 9.93 meters per second to go about 6 meters high.

This equation was taken from sammilechmann.com