> “…If carving is more efficient than skidding, shape skis ought to be less
> tiring. A carved turn is more efficient in terms of maintaining speed, but
> is it more efficient in terms of energy expended by the skier? …”
Hi again, Rusty –
As implied by your question, they are two different types of energy efficiency being discussed in the 1st sentence of yours that I quoted above. Unfortunately, few people distinguish between the two. The first type of energy efficiency in the above statement pertains to the fraction of the kinetic energy of the skier lost per second (or per meter, or whatever…) to the snow. Using this definition, any skidding will cause more energy to be lost per second than when carving.
The second type of energy efficiency in the above statement pertains to the amount of energy the skier has to expend per second in order to control the ski to make it behave in one of two ways, carve or skid/skarve. The two definitions of energy efficiency are almost completely unrelated, so the numbers that come out of them are similarly unrelated.
Here’s an analogy. Consider two cars. The first is a giant old Cadillac with underinflated tires but with so much power assist on the steering that you can steer the thing with one finger. The second car is a small, gas-efficient commuter car without power steering. The Caddy has horrible energy efficiency (using definition #1), but the driver can maneuver it using almost no energy of his own. The commuter car has super energy efficiency (ie, gas mileage) but it takes two hands on the wheel for the driver to get it into a parking place. The two forms of “energy efficiency” in driving are almost completely unrelated, just as they are in skiing.
You are absolutely correct that one shouldn’t mindlessly use the freshman physics formula, W = force * distance, to calculate the metabolic energy a muscle will consume. That’s precisely why you didn’t see me quote that formula (in spite of my screen name and day job), and is also why I mentioned that “…the force the muscles have to overcome is only 1 G...” (ie, the metabolic energy required to support a static load).
However … if you are interested, there are some approximations that one can use to calculate the energy consumption of muscles that are only slightly more complicated than W = f * d, and which work quite well. For more info, look here, http://asb-biomech.org/onlineabs/abs...01/pdf/146.pdf
, and in similar articles. Basically you first realize that the rate of energy consumption is always positive and the rate required to shorten a muscle (under some load) is very closely proportional (~ 5x – 7x ) to the rate of energy consumption required to lengthen that muscle (under the same load). You add the above two positive numbers together and multiply by the length of time the muscle is shortening and lengthening. To this number, you add the metabolic rate for a static load of specified size (times the duration in seconds that the load is indeed static), and finally in add a mechanical power term proportional to the rate of mechanical work being performed.
Tom / PM