It's not all Calories in vs Calories out (my theory, read first)

You don't know much about running Alistair Evian.

I do recall one 'fat' runner doing a sub 2.30 marathon, but his best time was 2.06 That was Antonio Pinto. He was about 20 pounds above his racing weight.

Chrome autofill settings wrote:

Are you really that stupid? wrote:

Running fast doesnt make people thin. Being thin enables people to run fast. Your whole post is idiotic from the get go.

Finally someone making some sense.

It's not that simple. Really the OP has asked a good question in an intelligent way and deserves some better answers.

Not a problem. I appreciate the discussion and being challenged on my claim.

Vertical component drops out for total work of the system, but not the work done by the runner against gravity. So my assumption is that work done by the individual against gravity is roughly the same in the 2 cases. Technically, you do less work in the vertical when running faster because you take fewer steps.

As for wind resistance, as you note later it should be a very small difference between the 2 scenarios. For simplicity's sake I feel we can ignore it.

I'm not a physiologist so feel free to disagree, but I assume that the biomechanical losses are roughly the same in the 2 scenarios. Why would a runner become significantly less efficient running 5:30 pace vs 7:00 pace? You could even make the case that a well-trained runner becomes more efficient at that pace.

Also, I don't agree that I'm assuming my conclusion. I am simplifying the problem with a set of assumptions (equal work against gravity, equal biomechanical losses, negligible increase to wind resistance) that lead to the conclusion. Scientists do this all the time. If the conclusion breaks down it will be because of the assumptions, and these can be examined and dissected and questioned, but I think mine are reasonable.

Let me pose this question to you: If you could accurately measure calories burned in both cases, how much difference would you consider significant? Suppose the runner goes 8 miles and burns 1000 calories at 7:00 pace and 1050 at 5:30 pace due to wind and perhaps a few other inefficiencies not accounted for in the simplified example. To me this is insignificant. Is the runner going to lose another 5 lbs by running 5:30 pace? Not likely.

comparing a runner doing 4 minute mile to him doing an 8 minute mile? Do you still think the calories burned are:

1, the same?

2, slightly more calories for a 4 minute mile vs an 8 minute mile?

3, a lot more calories?

BTW the answer is 3 a lot more calories. Think about it.

Jon Orange wrote:

comparing a runner doing 4 minute mile to him doing an 8 minute mile? Do you still think the calories burned are:

1, the same?

2, slightly more calories for a 4 minute mile vs an 8 minute mile?

3, a lot more calories?

BTW the answer is 3 a lot more calories. Think about it.

A lot is a general term. A guy running 140 miles a week and this on a 4K+ diet may not consider the amount of calories burned during his 4 min mile a lot compared to a mile jog. All relative.

Jon Orange wrote:

comparing a runner doing 4 minute mile to him doing an 8 minute mile? Do you still think the calories burned are:

1, the same?

2, slightly more calories for a 4 minute mile vs an 8 minute mile?

3, a lot more calories?

BTW the answer is 3 a lot more calories. Think about it.

This wasn't the original question. Of course as you go to extremes the problem becomes more complex and my assumptions break down. You could go even further: Walk 100m or all-out sprint 100m. Walking & sprinting are not biomechanically similar.

But running 7:00 pace and 5:30 pace are biomechanically similar. And so I claim my assumptions hold.

I think the difference between 7.00 and 5.30 if that was a hard pace for you (and that's an important distinction) would be about 3 or 4 % small, but significant.

Between a 4 minute mile and an 8 minute mile the difference would be about 25% more calories burned in the 4 minute mile and a lot more calories burned after the race too.

I'd agree that at the level we are taking 50kcal out of 1000kcal can be ignored and I don't take issue with your conclusion, just the argument to get there.

Honestly, I don't have a problem with what you are trying to do: treat the whole running person as one compound object and make a simple physics argument for why the energy will be the same regardless of the details. And in this case you came up with a nice one line argument to prove the result. I agree that physicists do this all the time.

The trouble is that to do it you had to hypothesize the existence of a force with certain properties so that it cancels out and disappears from your result. All of that is okay until someone looks closer and points out that no such force actually exists in the problem. At that point you've really got to let it go. It was a nice hypothesis but it doesn't fit with the experimental facts.

If you want to say, well I wasn't talking about the parallel direction but the vertical direction, you've got the new problem that it's not clear what D is because the motion is cyclical. You can probably figure out a way around this with the integral form of the work equation and I'm okay with that although it's starting to get pretty complicated now and at the end of the day you still have to assume that losses are independent of velocity to get all of this to drop out. Which is fine--I think they are roughly independent of velocity--I'm just saying that it is an assumption is all, an assumption that happens to hold for people running but doesn't hold for horses or kangaroos running or people biking, skating or holding boxes in the air.

For goodness sake test2, the numbers aren't the same at different paces.

Simple High school physics equations don't cut it. It's far more complicated than that.

If physicists had the answers we wouldn't need coaches, training would all be done by numbers and formulae. We would all be following them like robots.

Running is not like that.

Jon Orange wrote:

For goodness sake test2, the numbers aren't the same at different paces.

Simple High school physics equations don't cut it. It's far more complicated than that.

If physicists had the answers we wouldn't need coaches, training would all be done by numbers and formulae. We would all be following them like robots.

Running is not like that.

You seem not to notice that I'm arguing against the simple equations not for them. This whole back and forth started with me agreeing with your post that it wasn't as simple as W=F D. I'm using equations to show that simple equations are insufficient.

The biggest difference is actually surface area to volume ratio. Skinny people have higher surface area to lose heat, so to maintain body temperature when resting, they need to have a higher metabolic rate.

Kidding

chupacabra wrote:

Think about it in terms of lifting the heavy box. You can lift the box slowly or quickly, but either way you accomplish the same amount of work. In the box case you are working against gravity. In the running case, you are working against more complicated forces which can ultimately be modeled as summed in the negative direction. It doesn't matter how fast or slow you move your mass, it takes the same amount of work to get it from A to B.

Your final conclusion is correct, that calories are about the same per mile. But you keep using faulty reasoning to get there. Speed only doesn't matter for external work applied to the box (or a runner). At constant velocity the external work is zero. Opposing forces equal applied forces and a=0.

But there is internal work being done to apply that force. A higher speed will increases the opposing forces and increase internal work required. Meanwhile external work is zero in both cases. You said it yourself, at the extremes it becomes obvious you'd have to do more work (expend more energy) to overcome external forces. So in the context of energy expenditure it absolutely does matter how fast you move the box.

It just so happens that with running the changes in external forces are small enough to be negligible, or we make up for them with better efficiency.

By saying that work done is F*d and then saying that F is solely m*a, you're saying that once velocity is constant you're not doing any more work since a is zero. *facepalm*

There is certainly a difference in wind resistance, even with small variations in velocity. Wind resistance is proportional to the square of the relative velocity between the body and the wind, therefore if your velocity is only 20% greater than someone else's, that means your wind resistance will be 44% greater! (1.2*1.2=1.44)

I'm not arguing for either side, specially cause of the physiology component that I know nothing about (not only heat production, but repairing muscle, Etc). I just posted cause I wanted to point out how stupid your "physics" argument was.

PS. That is internal and external with respect to the system being analyzed, not say a spring force vs applied force.

test2 wrote:

Jon Orange wrote:

For goodness sake test2, the numbers aren't the same at different paces.

Simple High school physics equations don't cut it. It's far more complicated than that.

If physicists had the answers we wouldn't need coaches, training would all be done by numbers and formulae. We would all be following them like robots.

Running is not like that.

You seem not to notice that I'm arguing against the simple equations not for them. This whole back and forth started with me agreeing with your post that it wasn't as simple as W=F D. I'm using equations to show that simple equations are insufficient.

Ok, sorry, misread your post.

I'm assuming you meant the facepalm for yourself, since obviously there are friction forces working against a runner. If this were not true your feet would slip out from under you.

In my model, starting & stopping energies are too small to matter, work against gravity is roughly equivalent, biomechanical energy is roughly equal. This only leaves work against friction and hence the point mass model becomes useful. a will always be positive at constant v when working against friction.

The problem with % is that it ignores scale. The change in work against wind resistance is very small, so even if it is 44% greater, perhaps that only accounts for 20 additional calories. This fails to explain the weight loss posters are saying they experience at greater work intensity.

I'll accept your apology for calling my physics stupid :)

The "internal" factor that makes the biggest difference is aerobic vs. anaerobic metabolism. The drastically reduced efficiency of anaerobic metabolism at higher speeds will certainly increase your calorie consumption, though not by an amount that is likely to have any effect on body composition. Extending your cool down by a short distance could easily have the same effect.

I'm not clear on what you mean by internal and external work. External would be me pulling on a box, internal would be the mechanics of my muscles, bones tendons, etc?

I'm not arguing that constant velocity is all that matters. The way I rearranged the work equation looks that way, but it's not what's really going on. My previous post sums up my argument pretty well. I'm modeling this problem in a certain way that makes W=Fd workable (pun intended). Perhaps writing it as W=mvd/t is misleading -- what I'm really going for is dropping pieces of the total work of the system that are equivalent across the 2 scenarios in order to simplify the problem.

If there are errors in my model, it's in the assumptions.