CIM: theoretical time delay from water

Last few posts I documented that, for whatever reason, average times were slower in the 2012 CIM then in the previous 11 years. The year prior, 2001, times were about the same as 2012. Your response might be "Well duh! The weather was appalling both in 2001 and 2012: it rained." But 2012 certainly wasn't bad enough that hypothermia was a concern, or that the rain was so strong that the mechanical action of running was a challenge. It should be possible to estimate why a runner, running at the same metabolic rate, would go slower in the conditions.

There are two candiates for slower speeds: the wind and the rain. I'll start with the rain. The rain presents two mechanisms for increased resistance: hitting drops in the air and displacing water on the ground. I'll consider each of these.

First, displacing drops in the air. If it is raining at a rate R (a rate of precipitation per unit time), then if air droplets are falling at an average speed v_r, and if water has a density d, then the mass density of water in the air = d R / v_r. In a unit time dt, the net volume of water hitting the ground is v_r dt, and since the total water hitting the ground is R dt, the volume fraction of water must be R / v_r, yielding this result.

When I run at speed v water which hits my body on average is accelerated from a zero forward velocity to my forward velocity. I thus add forward momentum to the water. This is a drag term. Falling water can hit me from the top or from the front. Looking down, I present a certain cross section A_top. From the front, I present a different cross-section A_front.

The rate at which water hits me from the top, in terms of mass per unit time, is A_top R d.

The rate at which water hits my front, assuming water is falling vertically, is the volume density of water multiplied by my forward velocity multiplied by my area. This is thus A_front d R v / v_r.

The retarding force provided by this rain equals the rate of change of the rain's momentum. This is the rate of water mass impact multplied by my speed:

F_rain = (A_top + A_front v / v_r) R d v

This is interesting: the retarding force from rain hitting my front is similar to wind resistance, proportional to v^{2, while the retarding force from wind hitting the top of my head is only proportional to v. With wind resistance it is assumed the wind is stationary. However, if wind was blowing towards the ground there would be a similar term.}

Okay: time for some numbers. The typical speed of a falling raindrop (from "The Physics Factbook" , edited by Glenn Elert) is 9 meters/second. An eyeball-estimate of average rainfall rate during the race (measured in North Auburn, KCAAUBUR13) was 0.25 inches / hour, or 1.76 µm/sec. My shoulders are around 42 cm wide so if I estimate my cross-section from above as a square (need to include fact arms and legs are not hanging vertical when I run) that's 0.18 meters squared. My frontal area is around 1.7 meters high by 40 cm wide totalling 0.68 meters squared. My running speed during the race averaged 3.44 meters/second. The density of water is 1 gram / cm³ which equals 1000 kg / m³.

Using these numbers, the total retarding force from raindrops hitting my head is thus 1.09 mN. The retarding force from running into rain drops is 1.57 mN. The sum is then 2.66 mN.

Note the rain drops also hit the top of my head and push down. I need to resist this force, but am not doing any work in doing so, since work is force times distance and in the vertical direction I'm not moving. So it might make my legs tired but is not a force which directly retards my forward motion.

2.66 mN doesn't seem like much and indeed it is not: over the course of a full marathon that's just 112 joules. To see how much this slowed me down I consider an internal metabolic efficiency of on order 25% requires a metabolic expenditure of approximately 450 joules, which is around 8 joules/kg. The per-distance cost of running I have already described, measured by Minetti:

On the flat, it's around 3.4 J/m/kg. So 8 joules/kg slows me by around 2.5 meters over the course of the full marathon. Since I ran at 3.44 meters/second that's around 0.7 seconds.

So now I consider the effect of displacing water when I step. I consider water on he road to be an infinite area of water of a certain height. There was some sections of deep water, but in general the water wasn't enough to be really obvious, since the roads were well crowned for drainage (maybe not to the benefit of my legs: running on a tilted surface is non-ideal). I'll guess the typical water depth was 2 mm. So when my foot lands, it displaces these two mm of water to the height of the infinite sea of water. Since energy is proportional to height difference, the average energy of water thus displaced is the same as the energy raising the water 1 mm. In reality water would be displaced between 0 and 2 mm. But assuming all water raised 1 mm makes the calculation simpler and yields the same result. Of course, there is splashing, so water rises more than 1 mm peak, but I will first check to see if the simple assumption yields a significant result before trying to be more sophisticated.

My running shoes have a surface area on order 30 cm by 5 cm = 150 cm². I do not know how much water I displaced each step, but 2 mm seems like a reasonable approximation. I took approximately 1.5 steps per second with each foot, approximately 3 steps total. So each second, I displaced approximately 90 cm³ of water, raising it approximately 1 mm on average, which with a water density of 1 gram/cm³ and gravity of 9.8 m/sec², multiplying by 3:25 of running (I was probably stationary for 34 seconds) yields 10.9 joules of work done. That's less than a tenth of what I got for for hitting rain drops, and that was only 0.7 seconds delay. So even if I am off here, my net delay is still not much more than one second.

One thing this misses is the kinetic energy imparted on the water. I considered only the potential energy. I'll guess my shoe falls from a height of 10 cm when I run. I run at 1.5 steps each foot per second, and if my foot spends 70% of the time in the air, that's a falling time of 35%, or 0.233 seconds. Assuming uniform acceleration during this interval, the final velocity is twice the average, or 2 × 10 cm / 0.233 seconds = 0.86 meters/seconds. The water depth is 2 mm, so it takes my foot only 2.33 msec to displace the water. From the center of my shoe, the water must move around 2.5 cm. So the average water velocity must be approximately 2.5 cm / 2.33 msec = 10.7 meters/second. The amount of water displaced is approximately 2 mm × 150 cm² × (1 gram / cm³) = 30 grams. The kinetic energy imparted on the water displaced is thus approximately ½ m v² = 1.72 joules. I am doing this three times per second (1.5 each with left and right foot) so the total is 5.17 watts. Since I am running at 3.44 meters/second this is 1.506 joules/m, or 0.026 joules/meter/kg. At a 25% metabolic efficiency this is 0.105 joules/kg/m metabolic cost. The metabolic cost of running is again around 3.4 joules/m/kg. So this is 3%.

This is an upper bound, because of course I assumed my foot was in free-fall, but as soon as it strikes the water it is no longer in free-fall. The kinetic energy imparted on the water was 1.72 Joules, which is equivalent of my center of mass dropping 3 mm, and this is a small fraction of my center of mass drop, so I conclude this approximation is not wildly incorrect. However, it does represent an upper bound, since my body is not rigid, and my foot velocity could be retarded without my center of mass being equally retarded.

Note I concluded times were 3.73% slower this year than in 2011. So this already does a nice job of accounting for that. When the foot lands on standing water, it displaces it, increasing its potential energy, but that is a very small effect. On the other hand, it displaces it extremely rapidly, which imparts a high kinetic energy on the water, and that kinetic energy is significant. A lot depends on the height of the water which I just crudely estimated. But if I assume this energy is wasted then I get the runner is slowed by 3%. It could be argued the energy, if it didn't go to accelerating the water, would have been wasted compressing the shoe: the shoe compresses less because the water has already absorbed some of the runner's kinetic energy. I'll talk more about this next post.

That leaves wind.