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## Never ask a woman her weight…but her kinetic energy is fineAugust 2, 2014

Posted by mareserinitatis in math, physics, running, science.
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Today, I had one of the most awesome runs I’ve ever had.  In particular, I sustained a much faster pace than I have over a 3 mile distance.

I couldn’t help but wonder, however, about the factor weight plays in one’s speed.  As much as I try not to worry about weight and focus on being healthy, there’s this part of me that thinks it would be cool to lose a bit of weight because then I would go SO MUCH FASTER.  Or at least that’s what I tell myself.  However, I wondered if maybe I was exaggerating a bit, so I decided to check it out.

While it’s a bit of an oversimplification (that doesn’t take into account muscle tone, lung capacity, hydration, electrolyte levels, altitude adjustment, and the 18 bazillion other things that can affect a runner, even as stupid as that kink that’s still in your neck from last Thursday’s swim (okay, that only affects the triathletes here)), a quick check is to use the kinetic energy equation.

First, of course, we have to assume a perfectly spherical runner.  Or a Blerch:

(As an aside, if you don’t know what the Blerch is, you must check out the Oatmeal’s wonderful cartoon on running.  We all have a Blerch deep inside of us.)  Either way, perfectly spherical things are happy for physicists because of all the lovely simplifications we can use in learning about them.  So, if you’re a perfectly spherical runner, remember that physicists will love you.

Anyway, our hypothetical runner will have a mass (m), which is, of course, directly proportional to weight.  (Weight, of course, is also referred to as gravitational attraction, so the more you have of it, the more attractive you are, at least from the perspective of the planetary body you’re closest to.  Also, it may start to be more attracted to you if your velocity starts to approach the speed of light.  Maybe this is why many humans also find runners attractive?  Not sure.)  The unit of mass is the kilogram.  The runner will also have to maintain an average speed velocity (v), and of course your pace is inversely proportional to your velocity.  Your velocity is probably measured in miles per hour by your local race, but since we’re being scientific, we could also use SI units of meters/second.  That being said, if you double your speed in one unit, it will also double in the other.  There’s nothing fancy that happens because you’re using one unit or the other.

The kinetic energy of our runner, assuming an average velocity, can be written as

(1) KE=½ mv2

If we have the kinetic energy and mass, but want to find out the velocity, we first divide both sides of the equation by the mass and then take the square root of both sides.  This leaves us with the following result:

(2) v=√(2 KE/m)

Let’s take an example.  If we have a runner who has a velocity of 5 mph (or 2.2352 m/s) and a weight of 140 lbs. (or 63.5 kg).  If we use SI units to compute this runner’s velocity, it turns out her initial kinetic energy (KEi) is 158.63 J.

On the other hand, we don’t really need to know how much initial kinetic energy the runner has, in terms of numbers.  We can just define it as the quantity KEi. It turns out that physicists are kind of lazy about using numbers, so we’ll try to go without them because, in my opinion, it sort of confuses things. (You’ll see why later.)

How this this help us?  Well, if you want to take a drastic example, let’s assume a runner loses half of her body weight.

First, let’s establish that her initial kinetic energy is defined also by an initial mass mi and velocity vi.  (These would be the same as the 5 mph and 140 lbs. above.)  This means her initial kinetic energy can be written as

(3) KEi=½ mivi2

and her initial velocity would therefore be

(4) vi=√(2 KEi/mi).

If her weight drops by half, we can write this as her initial weight divided by 2:

(5) m=mi/2

If we put (5) into our velocity equation (2) as our new mass and keep the same initial kinetic energy, we get

(6) vnew=√(2 KEi/m)=√(2 KEi/(mi/2))=√2*(2 KEi/(mi))=√2(2 KEi/(mi))

You can see that the last part in six is basically the square root of two times our initial velocity from (3).  That means that by losing half her weight, our runner would run about 1.4 times as fast, or 40% faster.

Now what if she only loses 10% of her weight?  It turns out that (5) would become

(7) m=mi/1.1

so our new velocity would be the initial velocity times the square root of 1.1, which is about 1.05.  Losing 10% of her weight only makes her 5% faster.

After spending time looking at this, I decided that going on a massive diet definitely isn’t going to help me speed up significantly.  (In fact, if I manage to go from my current weight to my ideal, I would maybe get a gain of a bit over 1/2 mph.)  It’s the fact that the mass doesn’t play as strong a role as velocity does because velocity gets squared and mass doesn’t.  If you want to go faster, you are better off practicing running faster.

So please pass the ice cream!  I need it for my fartlek recovery.

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