Saturday, December 1, 2012

Physics of Ski Jumping

By Paul Donohue

Ski jumping performance is a result of not only the aerodynamic equipment used but also the motor skills and size of the jumper. Anorexia Nervosa and Bulimia were common disorders among elite jumpers up until recent regulations. The forces involved from takeoff to landing are numerous including gravity, air resistance, torque, centrifugal, and even lift which will be the focus of my investigation. I’d like to estimate just how much lift force is generated by those large skis.

On the ramp the jumper must try to minimize the frictional force between his/her legs and the snow (i.e. being light) and also the aerodynamic drag force to reach a maximum v0 Because the ramp is curved at its end the jumper experiences a good deal of centrifugal force for which he/or she must counteract to obtain a maximum momentum perpendicular to the ramp’s takeoff. At takeoff, the jumper must try to anticipate the torque force exerted by the air stream that threatens to tip him/her backwards into a bad flight position. This is done by directing angular momentum forwards….though not too far or tumbling can occur! Once in flight gravity, lift, and drag affect the jumper

Lift can be solved for the equation FL=(1/2)LcL ρv2 where L refers to surface area and cL is the lift coefficient with a value of .4 (typical values range from .4 - 1)
• First we’ll set v to 25 m/s: This is the sum of v0 and vp which is the
velocity perpendicular to the ramp.
• The surface area of one ski is .32 m2 and the density of air is 1.225 kg/m3
(1/2)(.32*2)(.4)(252)=80 N of Lift!

The reason ski jumping isn’t as dangerous as it may look is that the landings are engineered to only be roughly 3o below the axis of the skier’s trajectory If the launch angle is -9.5o above the horizontal (ski jumps actually don’t slope upwards) and the launch velocity is 25 m/s then…25sin(-9.5)= -4.13 m/s in the y-direction and 25cos(-9.5)=24.66 m/s in the x-direction If the skier travels 55.0 m horizontally to the landing zone then 55.0/24.66=2.23 s

During this time, gravity will increase the rate of to descent to 4.13+(-9.8)(2.23)= -17.7 m/s
Tan-1(-17.72/24.66)= -35.7: The angle of the skier’s trajectory is 35.7o below the horizontal which minus 3o would mean 32.7o below the horizontal for the landing slope!

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