How Accurate Is Calorie Counting Through Physical Activity?

[krete77]

Hey guys, this isn't a homework question or anything. I ended up lugging a lot of wood the other day and decided I would try to figure out on my own how to find out how many calories I burned.

I have a ton of data, but I'm just going to show you what I have done via this picture; hopefully you can see it.

So the answer I came up with for the total amount of work done on a medium sized log was 977J. When I convert this to calories (977J*.239cal) I get 233cals. There's no way I burned 239 calories lifting one log .3m up, and walking it 3m away.

I used the 1 joule = 0.239005736 calories.

What am I doing wrong?

 

[Doc Al]

There's no simple way of calculating the energy burned, since that's a biological function, not simple physics.

Some points: You are only doing work against gravity when you lift the log, not when you carry it horizontally. Also, food calories are actually Kcals = 1000 calories. So one calorie of energy is really only 0.001 Kcals.

 

[krete77]

Thanks for your reply.

I know there isn't an simple way, I'm just trying to get a ballpark figure..just for self completeness really.

Surely there must be some sort of work being done when I'm moving it since all muscles are contracted. I suppose that's more of a metabolic function rather then simple work being done?

 

[Doc Al]

Surely there must be some sort of work being done when I'm moving it since all muscles are contracted. I suppose that's more of a metabolic function rather then simple work being done?

There's certainly work being done within your muscles to maintain the tension as you move horizontally. But no work is being done on the log against gravity. (Mechanical work is Force*Displacement*cosθ, where θ is the angle between the force and the displacement. As you move horizontally, the angle is 90° and thus the cosine is 0.)

 

[krete77]

You're right. I remember that.

So, since there is no work being done as I move in the x direction, I could just take that out of the equation and get a ballpark figure, no?

 

[Doc Al]

So, since there is no work being done as I move in the x direction, I could just take that out of the equation and get a ballpark figure, no?

I would say no. For example: Stand there hold that log for an hour. No work is done on the log. So what would your ballpark figure be for the calories burned?

 

[krete77]

I would say no. For example: Stand there hold that log for an hour. No work is done on the log. So what would your ballpark figure be for the calories burned?

Good point.

Say I stayed in one spot. I lifted the log repetitively 150 times.

Eliminating the x direction, would the formula now give me a ball park estimate?

 

[mrspeedybob]

2 points.

You not only lifted the log, but also various parts of your body. Your arms at least, If you had to squat down to pick up the log then most of your body weight came up with it.

The human body is pretty inefficient. Whatever number you get for the actual work done, the energy you expended to do it will be several times more.

 

[krete77]

So is my approach a bad one? any idea on a new one?

Should I come up with a multiplier for energy, and use that in conjunction with the number I get from work done?

 

[mrspeedybob]

Your best estimate will come from an experiment replicating the activity while wearing a piece of equipment like this...
http://www.google.com/products/cata...=X&ei=2aY3UMOKNef22gXv5IGwCQ&ved=0CIsBEOUNMAI

 

[krete77]

I'm a body builder.. so I know a little about those things. They generally get very bad reviews and are never accurate. Waste of $$ imo.

Back to the formula on hand, anyone else have any suggestions? I'm going to keep working it until I find someway to get a ballpark figure.

 

[mrspeedybob]

Here's an experiment you could do with a little investment of time and money.

Install air conditioning equipment in you garage to maintain constant temperature and humidity. Build a small room in the center of your garage. Install several accurate thermometers in the room which can feed data to a computer. With constant outside conditions the temperature inside will vary with the amount of heat being generated inside. Since any energy you expend while in the room will be converted to heat this will give you a good measurement of that energy. If the room is well insulated the temperature difference will be greater, this will make any measurement error less significant. Since you will need oxygen, ventilation should be a controlled to be constant across all experiments. Sit in the room for some period of time (longer is better) and let the computer record temperature data. Now put an electrical resistive type heating device in the room and adjust it's power output until the temperatures recorded by the computer match what they were when you were in there. This will tell you how much heat you were generating while at rest. Now let the room cool back down to ambient temperature, go back in, and move your logs, record the temperature data. Now put the heater back in and modulate the power so that the temperature at every point in time during the experiment matches what it was while you were moving logs. Once again the power delivered to the heater will tell you how much heat you generated. Convert all your units to Kilo-calories and you will know how many calories you burned at rest and how many you burned moving the logs.

 

[krete77]

Here's an experiment you could do with a little investment of time and money.

Install air conditioning equipment in you garage to maintain constant temperature and humidity. Build a small room in the center of your garage. Install several accurate thermometers in the room which can feed data to a computer. With constant outside conditions the temperature inside will vary with the amount of heat being generated inside. Since any energy you expend while in the room will be converted to heat this will give you a good measurement of that energy. If the room is well insulated the temperature difference will be greater, this will make any measurement error less significant. Since you will need oxygen, ventilation should be a controlled to be constant across all experiments. Sit in the room for some period of time (longer is better) and let the computer record temperature data. Now put an electrical resistive type heating device in the room and adjust it's power output until the temperatures recorded by the computer match what they were when you were in there. This will tell you how much heat you were generating while at rest. Now let the room cool back down to ambient temperature, go back in, and move your logs, record the temperature data. Now put the heater back in and modulate the power so that the temperature at every point in time during the experiment matches what it was while you were moving logs. Once again the power delivered to the heater will tell you how much heat you generated. Convert all your units to Kilo-calories and you will know how many calories you burned at rest and how many you burned moving the logs.

That's an outstanding response. It's a shame no one else could even come close to what you replied with. If there were some way of giving you a virtual high five, I would do it in a heartbeat.

I will save this and in a few years down the road when I have some free time, I'll give this a shot.

The only negative that comes to mind from this experiment is from waiting a few years, in which my metabolism will most likely be slower than it is right now. Thus, burning a little less then I would normally. I'll figure out a constant to multiply my result by in order to account for lost time.

Thanks for your response.

Any other less genius responses?