The heat’s toll on the human body and how you pay the price
In this post, we looked at the four primary ways your body gets rid of heat as it converts food calories into energy and heat in the form of waste energy from exercise. Now we’re going to take a closer look at how much heat you naturally produce and expel during exercise. That will serve as our basis of comparison when determining if particular types of cooling garments actually live up to expectations.
When running, we can calculate your heat production based on the following formula:
Heat (in watts) = 4 x [your body mass (kg)] x [running speed (meters/second)]
Here’s a helpful little link to make the conversion from your mile split to that pesky metric speed. Just remember that time is decimals, so a 6:30/mile gets entered as “6.5.” Just to demonstrate, let’s get a hypothetical triathlete—yoink!
Say he’s a 145-pound (66kg) cyclist who will run a 7:00/mile split so long as it’s his local Olympic distance race.
His heat production would be: 4 x 66kg x 3.81m/s = 1,005 watts.
I have not yet been able to find a tidy equation relating an athlete’s cycling power output to heat generation. However, studies estimate comparable values, so for the sake of argument we’ll say that our hypothetical athlete is generating 1,000 watts of heat energy throughout the entire course of a triathlon. That’s 1,000 joules of heat energy every second. It only takes 4 joules to raise the temperature of 1 gram water one degree Celsius. So every four seconds, an athlete produces enough heat energy to raise the temperature of a liter (1,000g ≈ 1 L) of water one degree. That means if you could concentrate all the heat energy you produce while running and beam it into a 1-liter pot of water, you could bring it to boiling from room temperature in about five minutes. That’s some serious heat!
Obviously, we don’t want to boil our blood or internal organs, so the body has to get rid of the heat like a sailor bailing water out of a sinking ship. How well can it do that? The answer is “it depends.” Let’s look at each of the four ways our bodies can get rid of heat to figure out how well each will work.
First, let’s try Radiation.
To determine the amount of heat our bodies can dump via radiation, we use the Stefan-Boltzman equation, which is:
Heat (watts) = (Emissivity of skin) x (Stefan-Boltzman constant) x (Surface area of skin) x [(Skin surface temperature in Kelvin)4 – (Air temperature in Kelvin)4]
Don’t get riled up by the math! It’s what the equation shows us that matters. (I generously do all the math for you in my book FASTER.) Using our hypothetical athlete on a 90⁰F day and assuming his skin temperature elevates to 100⁰F, we find:
Heat = .97 x 5.67×10-8 x 2 x [ 3104 – 3054] = 67 watts
1,005 watts minus a meager 67 watts = one very hot athlete!
Okay, so radiation isn’t going to help much in staying cool. (In fact, we actually radiate less heat when we race than when we just sit on the couch watching television.)
Did you notice in the formula how the rate of heat exchange depends on the outside temperature? When it gets hotter, the temperature difference totally kills your ability to vent the heat. We’ll have to try something else. Unfortunately, conduction is out the window because there’s nothing cool against our body (yet).
Next let’s try Convection.
We’ll have to look at convection in two cases—running and biking—because your relative velocity changes the convection coefficient. For cycling, let’s say our cyclist can keep up a good 20mph speed into a 5mph headwind, giving him a 25mph relative wind velocity. We use the following equation to calculate the convection coefficient:
Convection coefficient = 10.45 – v + 10 x v1/2
After converting his speed to meters per second, we get
Convection coefficient = 10.45 – 11.2 + 10 x 11.21/2 = 32.7
For running at a speed of 7:00/mile into that same headwind, the coefficient is 29.
Our equation for convective heat transfer is:
Rate of convection = (Skin surface area) x (Convection coefficient) x (Skin temperature – Air Temperature)
This means our rate of heat dissipation by convection on the bike is 327 watts and on the run it’s 290 watts. Now we’re getting somewhere! So far, we’re up to getting rid of 394 joules of heat every second.
But we still have 606 joules left to deal with! This is a serious problem. So serious, in fact, the phrase “no sweat” no longer applies. It’s going to take a lot of perspiration if we want to avoid expiration.
Transpiration (aka sweating like a pig)
In hot conditions, the human body will sweat out between 1500-2000 grams of water per hour (http://www.ncbi.nlm.nih.gov/pubmed/9282540). But all that waterworks only helps if the water is evaporating.
Our bodies have no way to measure evaporation rates, so we keep pouring out the sweat regardless how much is already sitting on your skin waiting to be vaporized. While that puddle you make on the floor during spin class does serve to carry away some heat, it’s a negligible amount. What really counts is how much liquid gets sucked up into the air, and that depends on three factors:
- the speed of the air flowing over you,
- the ambient vapor pressure of the air,
- and the pressure at your skin.
Vapor pressure is a little complicated. It depends on the temperature and dew point outside. It would take too long to wander through a meteorological discussion, so instead I’ll simply provide you the link to a handy vapor pressure calculator, courtesy of the National Weather Service. For our purposes, we’ll assume a moderate dew point of 55⁰F (let’s say the outside temp is still 90⁰F). That gives us an ambient vapor pressure of 11mm Hg. For our athlete whose skin is 100⁰F, we assume a vapor pressure of 47mm Hg. Remember that our relative wind velocities while biking and running are 11.2m/s and 6m/s, respectively.
Given that, we can finally use the following equation to calculate the evaporative cooling rate.
E = 11.9vair0.6(Psk-Pa)
During cycling, that equates to 1,825 watts of heat. For running, we get 1,255 watts.
Suck it, heat! You’re dissipated!
Combined with radiation and convection, this evaporative cooling rate vastly exceeds our requirements to dissipate heat energy. Outstanding! Thanks to sweat, our hypothetical athlete doesn’t die.
This stands to reason, as the majority of athletes in the majority of races don’t keel over from heat exhaustion. But not all athletes and courses are built the same. For this overview of heat dissipation, we’ve used pretty favorable conditions. 90 degrees is about as hot as it will get at most races, but a 55-degree dew point is well within the “comfortable” region. In other words, it’s a dry heat.
What happens when we turn the humidity up? As you might expect, it becomes much more difficult to get that sweat to vaporize. In conditions such as Hawaii in October, our hypothetical athlete’s heat losses due to evaporation drop to only 1200 watts (cycling) and 1000 watts (running), respectively. That’s barely keeping up with heat production, especially when you consider temperatures on certain roads through the lava fields can soar over 100⁰F.
That’s not the only place, either. Athletes can find extreme heat and humidity throughout the country during the summer, and depending on your conditions, it can cause a slow buildup of heat that will ultimately cramp your style, not to mention your legs. The question is at what point the environment overcomes our efforts, and what physiological penalties we start paying.
We’ll look at that in the next installment.
If you’re interested in getting faster, you’ll be fascinated by FASTER: Demystifying the Science of Triathlon Speed. In Faster, astronautical engineer and triathlon journalist Jim Gourley explores the science of triathlon to see what truly makes you faster—and busts the myths and doublespeak that waste your money and slow down your racing. With this knowledge on your side, you can make simple changes that add up to free speed and faster racing.