Tweet
Why not a thread on the basic forces of skiing, complete with some concise formulae? It has always seemed to me that terms in this regard (ski forces)seem to get used a lot but perhaps without a useful level of precision.
I found this short little bit, which seems to offer, perhaps, a decent place to start.
I'll start and hopefully others will follow. There's nothing here difficult to comprehend that requires any math mastery beyond a little Algebra , Trig and a few concepts relevant to Calculus.
Ready, set:
1) F=MA ..The overall force acting on a skier equals the product of the skiers mass and the rate of acceleration (ideally gravitational acceleration-- in freefall) ..It's key to understand that acceleration is a matter of the change in velocity and not simple "speed"..It's also key to understand that mass is not "weight"..
Formulae to consider initially:
And
I found this short little bit, which seems to offer, perhaps, a decent place to start.
I'll start and hopefully others will follow. There's nothing here difficult to comprehend that requires any math mastery beyond a little Algebra , Trig and a few concepts relevant to Calculus.
Ready, set:
1) F=MA ..The overall force acting on a skier equals the product of the skiers mass and the rate of acceleration (ideally gravitational acceleration-- in freefall) ..It's key to understand that acceleration is a matter of the change in velocity and not simple "speed"..It's also key to understand that mass is not "weight"..
Formulae to consider initially:
- Now we can look at the individual terms:
- m*A = inertial (body) forces
- Ffriction = mu*m*g*cos(theta)
- Fdrag = (Cd*Ap*rho*V^2)/2 - drag force, opposes gravity
- Fgravity = m*g*sin(theta) - gravitational force
And
- The force Fgravity is the gravitational force. The inertial forces are zero if the skier is stopped or is moving at constant velocity. Ffriction and Fdrag are the frictional and drag forces. Both of these oppose gravity. The remaining terms are defined as
- A - acceleration
- Cd*Ap*rho - drag coefficient times frontal area times air density
- g - gravitational acceleration
- mu - dynamic friction coefficient
- V - velocity
- Balancing these forces gives the equation of motion:
m*A = m*g*sin(theta) - mu*m*g*cos(theta) - (Cd*Ap*rho*V^2)/2and if we divide by mass, we get
A = g*sin(theta) - mu*g*cos(theta) - (Cd*Ap*rho*V^2)/(2*m)The only term depending on mass is the last term, and mass appears in the denominator. That term opposes gravity because (Cd*Ap*rho*V^2/m) is not negative. - It should be pretty clear that the benefits (or lack thereof) of changing mass depend upon what happens to (Cd*Ap/m) as mass is scaled upwards. It is expected that the frontal area will increase for a more massive skier, but it seems likely that Cd*Ap increases more slowly than m. Therefore, a bigger skier is likely to go faster.
A few important things in this analysis have been ignored:- A heavier skier will probably have more trouble turning, especially if the added weight doesn't take the form of muscle in the right places.
- The friction at the ski/snow interface was treated as coulombic. The friction is generated in a film of meltwater in almost pure shear. So maybe the friction model is wrong, but this does not impact on the importance of mass.
"or even ventured to Notre Dame". How far to Marquette U, brother ?
Comment