# How many sled dogs does it take to warm a house? What if they're running?

January 25, 2011 10:40 AM Subscribe

I'm looking for the amount of heat output by an Iditarod sled dog, running.

I find an estimate that an Iditarod sled dog burns 10,000-12,000 calories per day during the race (blog). A large proportion of that will be expended as mechanical energy, just to move the dog and sled forward. Average weight for a long-distance racing sled dog is 25-30kg (Wikipedia).

A previous question asked after the heat output of a human. The range is pretty wide, as it is based on the activity of the human (93-700W). Is there a veterinary reference for numbers like this?

This is for a lecture on climate and its influences on building design, with references to the PassivHaus and Living Building standards, that will be presented in Alaska. We would like to express the energy required to heat an average home in the geographically relevant (if absurd) measure of "sled dog running days".

I find an estimate that an Iditarod sled dog burns 10,000-12,000 calories per day during the race (blog). A large proportion of that will be expended as mechanical energy, just to move the dog and sled forward. Average weight for a long-distance racing sled dog is 25-30kg (Wikipedia).

A previous question asked after the heat output of a human. The range is pretty wide, as it is based on the activity of the human (93-700W). Is there a veterinary reference for numbers like this?

This is for a lecture on climate and its influences on building design, with references to the PassivHaus and Living Building standards, that will be presented in Alaska. We would like to express the energy required to heat an average home in the geographically relevant (if absurd) measure of "sled dog running days".

(Also that assumes a flat, frictionless vacuum, which I think approximates Alaska, right? Maybe throw in some extra mass to account for friction, wind and uneven terrain.)

posted by auto-correct at 11:01 AM on January 25, 2011

posted by auto-correct at 11:01 AM on January 25, 2011

Amory Lovins' home took into account the heat that the dog would produce. The link mentions a 50 watt dog, but that dog isn't usually running in the house, obviously. But it gives you a jumping off place, and maybe a place to follow up with.

posted by ldthomps at 11:10 AM on January 25, 2011 [1 favorite]

posted by ldthomps at 11:10 AM on January 25, 2011 [1 favorite]

the above approach is almost certainly too simplistic. by neglecting friction, you're saying that all the dog has to do is accelerate itself to its running speed, then it can do no work at all to maintain its velocity for the whole day.

also, you can't multiply kinetic energy by time to get energy again. you need to find the average power (in watts) expended by the running dog not lost to heat and multiply that by the time spent running.

probably the best bet is to find some rough estimate of the fraction of energy lost to heat for typical animal locomotion.

posted by alk at 11:11 AM on January 25, 2011

also, you can't multiply kinetic energy by time to get energy again. you need to find the average power (in watts) expended by the running dog not lost to heat and multiply that by the time spent running.

probably the best bet is to find some rough estimate of the fraction of energy lost to heat for typical animal locomotion.

posted by alk at 11:11 AM on January 25, 2011

Wouldn't ALL of the energy expended, including the energy that's initially expended as mechanical energy, be converted to heat energy in short order? It's been twenty-some years since physics class.

posted by jon1270 at 11:13 AM on January 25, 2011

posted by jon1270 at 11:13 AM on January 25, 2011

*Amory Lovins'*[..]

*50 watt dog*

thanks,

**ldthomps**, I'd heard of Avery's statement about his 50-100W adjustable dog, but never had a hard reference for it. Now I wonder what breed it is.

posted by Prince_of_Cups at 11:19 AM on January 25, 2011

The SPKennel blog reckons the racing sled at 300-400lbs (135-180kg) for a team of 16 dogs.

posted by Prince_of_Cups at 11:30 AM on January 25, 2011

posted by Prince_of_Cups at 11:30 AM on January 25, 2011

Another angle on this problem is to just model the dog as a 30kg bag of 100 degree water that never cools down. I realize that this is a thermodynamic impossibility, but it does make the system more simple (in a way that is realistic for a first order approximation).

This avoids the thermodynamic problem of accounting for food and work and friction.

The question becomes "how effectively does the rest of the room conduct or convect that heat away from the dog?"

posted by milqman at 12:12 PM on January 25, 2011

This avoids the thermodynamic problem of accounting for food and work and friction.

The question becomes "how effectively does the rest of the room conduct or convect that heat away from the dog?"

posted by milqman at 12:12 PM on January 25, 2011

First,

Second, the way to calculate this is to find out what they eat, and how much, when on the trail. If the dogs maintain their weight during the race, then for each calorie out, a calorie goes in.

posted by Chocolate Pickle at 12:24 PM on January 25, 2011 [1 favorite]

__all__of the energy expended by the dog eventually becomes heat.Second, the way to calculate this is to find out what they eat, and how much, when on the trail. If the dogs maintain their weight during the race, then for each calorie out, a calorie goes in.

posted by Chocolate Pickle at 12:24 PM on January 25, 2011 [1 favorite]

As Chocolate Pickle says above, if the dogs don't gain or lose weight over the race then all the calories consumed become heat one way or another. As a refinement, some of those calories heat the air (wind resistance), the snow, the tracks of the sled, etc., and so wouldn't be detected as heat

To first order approximation the amount of mechanical energy transferred out is negligible compared to the heat output. If everything is steady state then 10,000 calories consumed (really kcal) means about 450W is expended (average) over a day, which gives you at least an upper bound., and given "normal" efficiencies (somewhere there's probably an zoology text that lists the mechanical efficiencies of animals) you're looking at at least 300 or 350W radiated as heat by the dog.

posted by range at 12:38 PM on January 25, 2011

*emitted by the dog*as a technical matter -- the dog is radiating some amount of direct heat output, and some (much smaller) fraction of the energy consumed by food is transferred to the sled, which in turn heats the snow via friction, etc.To first order approximation the amount of mechanical energy transferred out is negligible compared to the heat output. If everything is steady state then 10,000 calories consumed (really kcal) means about 450W is expended (average) over a day, which gives you at least an upper bound., and given "normal" efficiencies (somewhere there's probably an zoology text that lists the mechanical efficiencies of animals) you're looking at at least 300 or 350W radiated as heat by the dog.

posted by range at 12:38 PM on January 25, 2011

50 watts is not far off for a dog at rest; I think you might get 60 or even 75 from a sled dog. Essentially, a hard-working dog produces about as much heat as an incandescent light bulb.

First of all, I think it's safe to assume that practically all of the caloric energy a dog consumes is converted to heat, unless you're talking about converting it to potential energy (going uphill). The dirt, snow, and the dog itself will all eventually come to a stop thanks to friction, which converts the kinetic energy into heat.

Greyhounds are not sled dogs, but you may find it informative anyway - here's a page describing the science of a racing Greyhound.

It claims a greyhound burns 100 Kcals during a 30 second race, which sounds completely plausible to me. 100 Kcals == 116.2 Watt/hours == 418.4 kJ. Greyhounds are large dogs, up to 100 lbs, and I imagine that a racing Greyhound's peak energy consumption is the maximum

One caveat - that page nonsensically claims that 100 Kcals == "100,000 watts of waste heat energy". The scientific howler in that statement makes the whole page suspect, in my mind. A watt is a unit like horsepower; it describes instantaneous power output, not energy. Since 1 horsepower == 745.7 Watts, a 100 kW greyhound would be quite the monster - a 134 horsepower monster, to be exact!

Next time you see my extremely lazy greyhound powering a car at highway speeds, please take pictures. His average speed, by my calculations, is barely above 0 mph.

posted by LightStruk at 12:44 PM on January 25, 2011

First of all, I think it's safe to assume that practically all of the caloric energy a dog consumes is converted to heat, unless you're talking about converting it to potential energy (going uphill). The dirt, snow, and the dog itself will all eventually come to a stop thanks to friction, which converts the kinetic energy into heat.

Greyhounds are not sled dogs, but you may find it informative anyway - here's a page describing the science of a racing Greyhound.

It claims a greyhound burns 100 Kcals during a 30 second race, which sounds completely plausible to me. 100 Kcals == 116.2 Watt/hours == 418.4 kJ. Greyhounds are large dogs, up to 100 lbs, and I imagine that a racing Greyhound's peak energy consumption is the maximum

*any*dog could do, since they are the fastest dogs. Greyhounds only manage that for 40 seconds, tops. If a steady, sustainable run is about 2/3 the speed of an all-out sprint, then 75 Watts is a good guestimate for a sled dog's heat output.One caveat - that page nonsensically claims that 100 Kcals == "100,000 watts of waste heat energy". The scientific howler in that statement makes the whole page suspect, in my mind. A watt is a unit like horsepower; it describes instantaneous power output, not energy. Since 1 horsepower == 745.7 Watts, a 100 kW greyhound would be quite the monster - a 134 horsepower monster, to be exact!

Next time you see my extremely lazy greyhound powering a car at highway speeds, please take pictures. His average speed, by my calculations, is barely above 0 mph.

posted by LightStruk at 12:44 PM on January 25, 2011

Okay, so I'm looking for energy radiated as heat by the sled dog.

There are some good answers and directions for calculation so far, but I'm interested in more, if anyone has it.

posted by Prince_of_Cups at 2:02 PM on January 25, 2011

There are some good answers and directions for calculation so far, but I'm interested in more, if anyone has it.

posted by Prince_of_Cups at 2:02 PM on January 25, 2011

*Okay, so I'm looking for energy radiated as heat by the sled dog.*

Neglecting net change in altitude, it's all heat eventually, due to friction, right? Just measure the food the dogs eat and the amount of weight (fat) they lose.

posted by DarkForest at 3:09 PM on January 25, 2011

*Okay, so I'm looking for energy radiated as heat by the sled dog.*

This will be a very small number. At physiological temperature, dogs are not very efficient block body emitters. I think you'll find most of the heat loss is convective and (from the mouth) evaporative.

posted by mr_roboto at 3:41 PM on January 25, 2011

Okay, I'm just a liberal arts student, and I get antsy when we start talking about watts and stuff. I freely admit, I do not understand what a watt is.

But surely it would factor into this equation that a normal dog's temperature is 101 degrees, with 102 being undesirably hot (i.e. a fever)?

The question then becomes, how much heat does a running husky have to dump, in order to burn 10,000 calories, but maintain a temperature of 101 degrees F.

posted by ErikaB at 5:40 PM on January 25, 2011

But surely it would factor into this equation that a normal dog's temperature is 101 degrees, with 102 being undesirably hot (i.e. a fever)?

The question then becomes, how much heat does a running husky have to dump, in order to burn 10,000 calories, but maintain a temperature of 101 degrees F.

posted by ErikaB at 5:40 PM on January 25, 2011

Thank you, everyone. These responses were helpful to me as I researched and re-defined the question.

I was able to contact the Director of the Comparative Exercise Physiology Laboratory at Oklahoma State University (movie of horse on treadmill and description of their Mobile Alaskan Laboratory). He responded that a sled dog can generate up to 14kWh per day under Iditarod conditions.

posted by Prince_of_Cups at 12:17 PM on January 26, 2011

I was able to contact the Director of the Comparative Exercise Physiology Laboratory at Oklahoma State University (movie of horse on treadmill and description of their Mobile Alaskan Laboratory). He responded that a sled dog can generate up to 14kWh per day under Iditarod conditions.

posted by Prince_of_Cups at 12:17 PM on January 26, 2011

This thread is closed to new comments.

KE = 0.5*m*v^2. You'd have to look up the running speed of the dog. This is an energy quantity, so multiply it by the average time spent running per day to get energy per day. Remember to work in meters, kg and seconds to stay consistent.

Then convert to total daily energy expenditure of the dogs to KJ (upper limit =~ 50 000 KJ). Subtract from this the energy put into running, and whatever's left over should be heat output.

One other thing to watch out for is the mass term in the first equation. The dog isn't just pulling itself, it's also pulling it's share of the sled. So m = m_dog + m_load/(# of dogs).

Caveat: I'm doing this like a first year physics problem, not a metabolism problem. I have no idea if it will give a sensible answer.

posted by auto-correct at 10:57 AM on January 25, 2011