**Pump up off the bump?**

More on using a notion of a virtual bump to understand how a skier can pump energy into the system. This is what I've been searching for from the start. The VB was incidental.

Ref: "The Physics of Skiing" (Lind and Sanders) Page 231 Technote 10 "Pumping to increase velocity"

**Child's Swing**

A pendulum idea as in a child's swing can be used to imagine skiing down the fall-line over real bumps, anticipating the bumps.

The child can pump positive feedback energy in by extending at the lowest point. There are centripetal forces being resisted by the child's legs.

If the child pumps negative feedback by compressing with (rather than extending against) the centripetal force and moving the c.o.m towards the seat of the swing as the swing passes the lowest point, the energy in the system is reduced.

**Looking at how a real bump works**

Similar to the negative feedback of the child's swing above - The skier in a bump-field taking the high-line/low-line can pump the legs up into a retraction on the peak (moving c.o.m. relatively downwards) and pump the legs down in the troughs (Moving the c.o.m. upwards as the extended legs raise the skier up vertically). This controls the displacement of the centre of mass to a manageable average by taking kinetic energy out of the system.

**Looking at the same forces, but on a smooth (no real bumps) slope**

By translating this into the race course, but using positive feedback the expert skier can displace the centre of mass towards the instantaneous centre of the turn and generate energy. (Lind and Sanders explain that this is a gain in kinetic energy of h times the Centripetal force, where h is the distance moved in).

The gain is achieved by doing work against the centripetal forces generated in the turn.

Here's a worked example: A skier of mass= 70kg negotiates a GS gate at a velocity v=10m/s with a tightest part of the turn r-4m. The skier moves his com inside the turn by a value h= 0.5m, and therefore pumps energy into the turn.

What is

1) The ratio of in kinetic energy KE to the Change in KE after the move inside?

And

2) The change in the skier's weight?

**Here's the working principle:**

The amount of work done against the Centripetal force Fc is (by principle of conservation of energy) exactly equal to the change in kinetic energy (KE).

The centripetal force increases with velocity and so the work done against Fc, and hence the increase in KE, increases with velocity.

Fc = Mv(squared)/r

The work done against Fc is exactly equal to the change in Kinetic energy which is worked out by (h/r ) * m.v(squared)

The Ratio of (Change in KE caused by the pumping) to the (KE of the skier)

is:

2h/r, because KE is 1/2 mv(squared)

That works out an answer for question 1) as The (Change in KE) divided by the original KE is 1/4 = 0.25

**Moving on to the weight force:**

The acceleration a = v squared/r

Putting in the values for v=10, r=4

a= 100/4 = 25 metres per second squared

Centripetal force F=ma so for a mass of 70kg times this acceleration; F works out at 70*25 = 1750 Newtons.

Compare this to the skier's normal weight:

The skier's rest weight would be mg where g is the acceleration due to gravity which we'll approximate roughly to 10 m/(second squared) i.e. 70*10 =700Newtons.

Depending on the slope and trajectory, there will be added on a component F= mgSin(alpha)*Sin(beta) where alpha is the slope angle and beta is the instantaneous traverse angle.

**Is there more?**

We have worked out the percentage change of KE as having a value of 0.25 * 100 = 25%

The skier would need to be skilled enough to direct this extra energy in the correct direction. Realistically, to get a 0.5m movement inside would be difficult at the start of the turn.

This, though is the way the GS skiers appear to accelerate out of the turn and past the gate-post.

**Summary, and Vertical/ Horizontal equivalence:-**

That's for a lateral movement calculation of a skier on a flat inclined plane.

The same calculation works vertically over a 4m trough, but the skier's weight is even more, as the acceleration into the trough a is 25 which is approx 2.5 times gravity, but the skier's own full weight of mg (not a vector component of it) needs added on as it is in the vertical in the middle of the trough.

This is a discussion - not to be quoted as "Gospel". If anyone can improve it - please do so.

Concluding today:

Here's a better link to the virtual Voltaire:-

http://www.planetperplex.com/en/item/slave-market-with-the-disappearing-bust-of-voltaire/