# 60 miles ~3400 calories burned correct???

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There is no way 3400 kcal was burned. 3400 kcal = 34 miles of walking, jogging, or running. Nobody could run 34 miles in 3.5 hours (the time it took the OP to ride).

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Correct. But then the non fit rider will be generating more force; ergo more kcals spent. But the basic equation of W=FxD applies assuming all thing being equal. The more fit person will have an easier time doing the task because of conditioning; lowr HR, higher stroke volume and ejection fraction to pump more blood in a given time=greater ease in performing work. In the end the kcals used will be the same. If lance rides at 35 mph and the potato at 15 mph, well now you have introduced a time component and you can measure energy usage in terms of power(watts) (or as I prefer:kcals/min). Lance will be at 400 watts (guess) and the potato will be at 170 watts. In the end Lance will finish his 5 mile ride in less than half the time of the potato but both will burn the same kcals because lance will be at 20 kcals/min(guess) and the potato at 9 kcals/min.

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Steve, corect no one could do 34 mile in 3.5 hrs. Thats not what matters. Its total kcals burned. If you decide to walk 34 miles it will take 8.5 hrs (at 4mph) and 3400 kcals ( ~7kcals/min). If you run at twice the speed (8mph) it will take half as long but your kcals/min will double (4hr15min at 14 kcals/min). Its still the same amount of work performed-3400 kcals. So, if you COULD run 34 miles in 3.5 hrs it would still be 3400 kcals, only the kcals/min would change. Bototm line- for someone to ride 60 miles and burn 3400 kcals in 3.5 hrs ( 14 kcals/min) is very reasonable. ( FYI- my prevoius post this AM was in response to Danno and CbCf. I'm new to this forum and not used to how you would reply directly to someone else.)

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Originally Posted by

**pacemaker**Correct. But then the non fit rider will be generating more force; ergo more kcals spent. But the basic equation of W=FxD applies assuming all thing being equal. The more fit person will have an easier time doing the task because of conditioning; lowr HR, higher stroke volume and ejection fraction to pump more blood in a given time=greater ease in performing work.

??? Assuming that the unfit person weighs the same as the fit person, the force would be the same whether the person is fit or not. The unfit person will have a smaller engine (heart, muscle, conditioning) and it would 'feel' lke he/she is pushing harder, but in reality, the force applied would be the same. Its like a lawn-mower engine that is used to push a 100 kg mass up a hill in one minute would use the same force as a car engine to push the mass up in the same time.

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Originally Posted by

**CdCf**If we assume they're absolutely identical in weight, size and shape, use the same position on the bike, and pedal at the same cadence, on identical bikes, then they will have to expend the same minimum amount of energy to travel any given distance.

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That's not what I said in my post. Read it again if you didn't understand it.

Perhaps you're confused by the word "shape". I use it to describe the form of a body, not the physical condition of an athlete.

Also note the word "minimum"...

That is, if they both were powered by a 100% efficient power source, the two outwardly identical riders would use the same amount of fuel.

The difference in energy expended by the two, is all down to relative efficiency levels within the respective bodies. That's impossible to measure using something like a power meter. That meter will only register the minimum power (and thus, energy) required. There's no way for it to measure that the unfit rider burns 10% more energy in total, or whatever the difference might be.

Perhaps you're confused by the word "shape". I use it to describe the form of a body, not the physical condition of an athlete.

Also note the word "minimum"...

That is, if they both were powered by a 100% efficient power source, the two outwardly identical riders would use the same amount of fuel.

The difference in energy expended by the two, is all down to relative efficiency levels within the respective bodies. That's impossible to measure using something like a power meter. That meter will only register the minimum power (and thus, energy) required. There's no way for it to measure that the unfit rider burns 10% more energy in total, or whatever the difference might be.

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Originally Posted by

**pacemaker**If lance rides at 35 mph and the potato at 15 mph, well now you have introduced a time component and you can measure energy usage in terms of power(watts) (or as I prefer:kcals/min). Lance will be at 400 watts (guess) and the potato will be at 170 watts. In the end Lance will finish his 5 mile ride in less than half the time of the potato but both will burn the same kcals because lance will be at 20 kcals/min(guess) and the potato at 9 kcals/min.

Say FL rides a 30-mile course at 30 mph. Kreuzotter says it takes 499 watts x 3600 seconds = 1796 kJ.

Say SP rides the same 30 miles at 15 mph. Kreuzotter says it takes 87 watts x 7200 seconds = 626 kJ.

This makes sense. It costs energy to move through air, and the cost goes up faster than the speed does.

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How about if they both were riding on the moon?

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If they were both riding on the moon, we can disregard the atmosphere (although the moon technically has one, it's so feeble that it's completely negligible).

Then, all that is left is internal friction and rolling resistance.

Both increase pretty much linearly with speed, so if a rider moves twice as fast, he will need to exert twice the power.

However, double the speed, and you only have to travel for half the time. So, if the slower rider pushes 100 W going 25 mph for one hour, he will have used 0.1 kWh (100 W = 0.1 kW, and one hour makes it into kWh) of energy to overcome the mechanical resistance. And about 0.5 kWh in total, since the body has an efficiency of about 20% when cycling.

The faster rider pushes 200 W for a speed of 50 mph, but reaches the destination in only half an hour. He too has only used up 0.1 kWh, or 0.5 kWh in total.

Imagine then what vacuum tunnels could do for train travel!

Then, all that is left is internal friction and rolling resistance.

Both increase pretty much linearly with speed, so if a rider moves twice as fast, he will need to exert twice the power.

However, double the speed, and you only have to travel for half the time. So, if the slower rider pushes 100 W going 25 mph for one hour, he will have used 0.1 kWh (100 W = 0.1 kW, and one hour makes it into kWh) of energy to overcome the mechanical resistance. And about 0.5 kWh in total, since the body has an efficiency of about 20% when cycling.

The faster rider pushes 200 W for a speed of 50 mph, but reaches the destination in only half an hour. He too has only used up 0.1 kWh, or 0.5 kWh in total.

Imagine then what vacuum tunnels could do for train travel!

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Originally Posted by

**pacemaker**Steve, corect no one could do 34 mile in 3.5 hrs. Thats not what matters. Its total kcals burned. If you decide to walk 34 miles it will take 8.5 hrs (at 4mph) and 3400 kcals ( ~7kcals/min). If you run at twice the speed (8mph) it will take half as long but your kcals/min will double (4hr15min at 14 kcals/min). Its still the same amount of work performed-3400 kcals. So, if you COULD run 34 miles in 3.5 hrs it would still be 3400 kcals, only the kcals/min would change. Bototm line- for someone to ride 60 miles and burn 3400 kcals in 3.5 hrs ( 14 kcals/min) is very reasonable. ( FYI- my prevoius post this AM was in response to Danno and CbCf. I'm new to this forum and not used to how you would reply directly to someone else.)

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Terry, so basically your saying in your last post that the SP will only use 149 kcals (626kj) and FL will use 427 kcals (1796kj) to accomplish the same amount of work. Impossible! Besides you have converted watts to kj and you cannot do that. Watts =FxD/time, kj is simply FxD. It does cost energy to move through air but since both riders are moving through the same air over the same distance the energy cost is the same; the difference is that FL rides in one hour and SP in two. FL will encounter more wind resistance because he is going faster but that will be offset by the fact that he will be riding for a shorter time. In the end both riders cover 30 mile at the same kcal cost whether it is done in one hour or ten. Kreuzotter is wrong, Isaac Newton is right.

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Originally Posted by

**pacemaker**Terry, so basically your saying in your last post that the SP will only use 149 kcals (626kj) and FL will use 427 kcals (1796kj) to accomplish the same amount of work. Impossible! Besides you have converted watts to kj and you cannot do that. Watts =FxD/time, kj is simply FxD. It does cost energy to move through air but since both riders are moving through the same air over the same distance the energy cost is the same; the difference is that FL rides in one hour and SP in two. FL will encounter more wind resistance because he is going faster but that will be offset by the fact that he will be riding for a shorter time. In the end both riders cover 30 mile at the same kcal cost whether it is done in one hour or ten. Kreuzotter is wrong, Isaac Newton is right.

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Steve, Correct, work would be more for the faster rider when calculated in terms of kcals/min. But time is only an issue if your discussing Power, this concerns Work, and Force and Distance are the only variables in the equation. In the end two riderss of the same wt travelling 30 miles will use the same energy to do so regardless of time.

PS: sorry for the Kreuzotter comment. I was out of line.

PS: sorry for the Kreuzotter comment. I was out of line.

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Originally Posted by

**pacemaker**Terry, so basically your saying in your last post that the SP will only use 149 kcals (626kj) and FL will use 427 kcals (1796kj) to accomplish the same amount of work. Impossible!

Besides you have converted watts to kj and you cannot do that. Watts =FxD/time, kj is simply FxD.

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Originally Posted by

**pacemaker**In the end two riderss of the same wt travelling 30 miles will use the same energy to do so regardless of time.

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Originally Posted by

**pacemaker**Steve, Correct, work would be more for the faster rider when calculated in terms of kcals/min. But time is only an issue if your discussing Power, this concerns Work, and Force and Distance are the only variables in the equation. In the end two riderss of the same wt travelling 30 miles will use the same energy to do so regardless of time.

PS: sorry for the Kreuzotter comment. I was out of line.

PS: sorry for the Kreuzotter comment. I was out of line.

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I still haven't read here the rational behind the difference between the nutritional value of a calorie and the energy calorie value -2 completely different things.

V

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OK, I'm still not convinced. Let me pose a slightly different scenario than the two rider one and show me the math so we can put this thread to bed. I ride today for 30 miles in 90 minutes. Tomorrow I ride the same route in 120 minutes. Show me how the amount of kcals used would be different in the two rides. Assume the rider is 160#. Also FYI: Joules can be converted to Watt-hours but not directly to watts. One is a measure of work the other power, not the same thing. Another thing: I'm looking at energy tables for running & cycling from McArdle & Katchs Exercise Physiology, Energy, Nutrition and Human Performance. Energy for a 71 kg man to run a 11min mile is ~10 kcals/min, to run a 5.5min mile uses ~20 kcals/min; hes running twice as fast and burning twice the kcals, not 4x the kcals as was postulated in one of the posts. The tables for cycling show the same increase in kcals based on increases in speed. Cycling at 5.5 mph uses 4.5 kcals/min, at 9.4 mph it is 7.1 kcals/min, 16 mph 12 kcals/min, looks fairly linear to me. Anyway, I look forward to your replys. Thanks.

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Damon Rinard did a nice write-up on the resistance forces in cycling. Rather than writing it up again, I direct you to Damon's article:

https://damonrinard.com/aero/formulas.htm

It's very clear, complete with graphs.

https://damonrinard.com/aero/formulas.htm

It's very clear, complete with graphs.

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Physics lesson follows:

First, a few basic concepts...

Force = N

Distance = m

Time = s

Speed = distance / time = m / s = m/s

Work = force x distance = N x m = Nm = J

Power = force x speed = force x distance / time = N x m / s = Nm/s = J/s

but also...

Work = power x time = J/s x s = J

When you move through air, you encounter resistance. I'm sure you're familiar with that.

The force encountered can be described with the following formula:

F = Cd * A * rho * 0.5 * v^2

F = force (N)

Cd = coefficient of drag

A = frontal/cross-sectional area (m^2)

rho = air density (kg/m^3)

v = speed through the air (m/s)

If we look closer at the variables, we find that all but "v" can be considered constant for any given rider, at least if the rider wishes them to be over the course of a ride, and there's no significant change in altitude.

"v", however, varies as the speed squared, and it's obvious that's the major concern.

So, if we put reasonable values into the formula, we end up with the force encountered at a given speed. But force doesn't require work to oppose it (if you are in doubt about that, just consider that the ground opposes your vertical force all the time, without using up any energy in doing so...). Instead, we must find the power required to move through the air at the given speed.

Since force times speed equals power, we have to multiply the result of the formula by the speed again, to get the power requirement. By doing this, we have effectively cubed the speed, "v", in the formula. And we find that the power required to move through air varies as the speed cubed. Which means that if we double the speed, the power required goes up by 2^3, or a factor of eight!

That's not the whole story, though, since air is not the only opposing factor. Internal mechanical resistance in the bike and the rolling resistance of the tyres is another factor. But this factor increases roughly linearly with speed, so that double the speed means the power required is doubled as well.

The total power required at any given speed is the sum of the two factors - air resistance and rolling resistance/mechanical losses. At low speeds, air resistance is the smaller factor of the two, and rolling resistance the greater. At some speed, air resistance takes over as the major factor, because it increases as v^3, while rolling resistance as v^1. Don't forget, though, that even if rolling resistance is no longer the greater factor, it still continues to increase as the speed increases - it doesn't diminish!

When reasonable values for the two formulas are used (for speeds between 12 and 25 mph), and the speed vs power is plotted, the resulting curve can be approximated/paralleled with a curve following v^2.5. This means that for an increase in speed by 50%, the increase in power required is 1.5^2.5, which means that the power demand increases by roughly 180 %.

(I've ignored hills for now - assume this is only for flat ground...)

So, to answer your question, if you ride 30 miles in 90 minutes one day, and riding at 20 mph takes 200 W (for example), you use 1080 MJ, or 260 kcal to travel that distance. Multiply by five to get the energy used up by the body - in this case about 1300 kcal.

The next time, it takes 120 minutes to ride 30 miles. Using my factor above, the slower speed of 15 mph requires

(15/20)^2.5 ~ 0.5

as much power, so that's 100 W in this example. 100 W for two hours is 720 MJ, or about 170 kcal. 860 kcal for the body.

So, 15 mph for two hours takes 860 kcal, while 20 mph for one and a half hour takes 1300 kcal.

Did that clear things up?

(Never mind if the numbers used aren't 100% accurate - their relative proportions and the principles of it all still are!)

First, a few basic concepts...

Force = N

Distance = m

Time = s

Speed = distance / time = m / s = m/s

Work = force x distance = N x m = Nm = J

Power = force x speed = force x distance / time = N x m / s = Nm/s = J/s

but also...

Work = power x time = J/s x s = J

When you move through air, you encounter resistance. I'm sure you're familiar with that.

The force encountered can be described with the following formula:

F = Cd * A * rho * 0.5 * v^2

F = force (N)

Cd = coefficient of drag

A = frontal/cross-sectional area (m^2)

rho = air density (kg/m^3)

v = speed through the air (m/s)

If we look closer at the variables, we find that all but "v" can be considered constant for any given rider, at least if the rider wishes them to be over the course of a ride, and there's no significant change in altitude.

"v", however, varies as the speed squared, and it's obvious that's the major concern.

So, if we put reasonable values into the formula, we end up with the force encountered at a given speed. But force doesn't require work to oppose it (if you are in doubt about that, just consider that the ground opposes your vertical force all the time, without using up any energy in doing so...). Instead, we must find the power required to move through the air at the given speed.

Since force times speed equals power, we have to multiply the result of the formula by the speed again, to get the power requirement. By doing this, we have effectively cubed the speed, "v", in the formula. And we find that the power required to move through air varies as the speed cubed. Which means that if we double the speed, the power required goes up by 2^3, or a factor of eight!

That's not the whole story, though, since air is not the only opposing factor. Internal mechanical resistance in the bike and the rolling resistance of the tyres is another factor. But this factor increases roughly linearly with speed, so that double the speed means the power required is doubled as well.

The total power required at any given speed is the sum of the two factors - air resistance and rolling resistance/mechanical losses. At low speeds, air resistance is the smaller factor of the two, and rolling resistance the greater. At some speed, air resistance takes over as the major factor, because it increases as v^3, while rolling resistance as v^1. Don't forget, though, that even if rolling resistance is no longer the greater factor, it still continues to increase as the speed increases - it doesn't diminish!

When reasonable values for the two formulas are used (for speeds between 12 and 25 mph), and the speed vs power is plotted, the resulting curve can be approximated/paralleled with a curve following v^2.5. This means that for an increase in speed by 50%, the increase in power required is 1.5^2.5, which means that the power demand increases by roughly 180 %.

(I've ignored hills for now - assume this is only for flat ground...)

So, to answer your question, if you ride 30 miles in 90 minutes one day, and riding at 20 mph takes 200 W (for example), you use 1080 MJ, or 260 kcal to travel that distance. Multiply by five to get the energy used up by the body - in this case about 1300 kcal.

The next time, it takes 120 minutes to ride 30 miles. Using my factor above, the slower speed of 15 mph requires

(15/20)^2.5 ~ 0.5

as much power, so that's 100 W in this example. 100 W for two hours is 720 MJ, or about 170 kcal. 860 kcal for the body.

So, 15 mph for two hours takes 860 kcal, while 20 mph for one and a half hour takes 1300 kcal.

Did that clear things up?

(Never mind if the numbers used aren't 100% accurate - their relative proportions and the principles of it all still are!)

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I love BF when there is civilised, intelligent debate.

As posted earlier, I wouldn't know where to start to figure these things out applied to the human body...

Thanks for the back and forth. It's good reading.

As posted earlier, I wouldn't know where to start to figure these things out applied to the human body...

Thanks for the back and forth. It's good reading.

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Thanks CdCf. I'm almost convinced. The crow is thawing, but I'm not willing to cook it just yet. You mentioned in your post that you would multiply the 260 kcals by 5 for energy used by the body; why? I couldnt determine where the 5 factor came from. Also someone please tell me how to insert part of someone elses quote so I can better direct my post. I'm such a geek, I'm still going to test this out myself on the road later this week. If I can borrow the portable calorimeter at work I can also better test the accuracy of my HRM at the same time.

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Originally Posted by

**pacemaker**Thanks CdCf. I'm almost convinced. The crow is thawing, but I'm not willing to cook it just yet. You mentioned in your post that you would multiply the 260 kcals by 5 for energy used by the body; why? I couldnt determine where the 5 factor came from.

Originally Posted by

**pacemaker**Terry, so basically your saying in your last post that the SP will only use 149 kcals (626kj) and FL will use 427 kcals (1796kj) to accomplish the same amount of work. Impossible!

**wind-resistance**. That's something a lot of people forget, especially those not involved in performance bike-racing. If you're not testing riders in the 30-50mph range, you won't be able to see the non-linear increase in power needed to overcome wind-resistance.

Here's all the equations and measurements distilled down to an easy to see graph from Damon Rinnard's site:

While trying to compare biking to running in the 5-10-15mph range shows fairly linear increases in power-requirements, that's because air-resistance hasn't built up very much. At 9.3mph/15kph, you need about 25-watts with roughly 1/2 to overcome rolling-resistance and 1/2 of that going to fighting wind-resistance.

Double that speed to 20mph/32kph and you'll need 175-watts of power with the majority it going to fight wind-resistance. That's a power-increase factor of 7x to go 2x as fast. I deal with 200mph cars all the time and can tell you that it required a substantial amount of extra power to hit 200mph compared to 100mph...

*Last edited by DannoXYZ; 02-04-06 at 01:36 PM.*

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If we use the graph to look up the power levels for 32 km/h (20 mph) and 24 km/h (15 mph) used in my example, we see that they're 180 W and 90 W respectively.

90 / 180 = 0.5

Very close to the 0.49 predicted by my (v1/v0)^2.5 formula.

90 / 180 = 0.5

Very close to the 0.49 predicted by my (v1/v0)^2.5 formula.

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Originally Posted by

**pacemaker**You mentioned in your post that you would multiply the 260 kcals by 5 for energy used by the body; why? I couldnt determine where the 5 factor came from..

DannoXYZ has already explained why, though.

I assume that the ~20% efficiency of the body is for mainly fat-burning aerobic work, i.e. low-intensity cycling at around 50-65% of working heart rate.