Training for weight loss



wolfgang

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Feb 19, 2004
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Dominic Sansom said:
It may not be for everyone but I found the South Beach Diet was the only one I could stick to. I lost 16kg in 12 weeks. Never got hungry. Cycle 200-300km per week and 3-4 days a week in the gym for boxing (sparring) and weights.

It's worth checking out. The only problem I have with the diet is the use of artificial sweeteners so I use sugar (only in tea and coffee) instead. I'm on stage 3 of the diet and still loose abot 1-2kg a month.

You might want to try Stevia, a natural low-cal sweetener. I get mine at Whole Foods and use it mainly in my oatmeal.
 

frenchyge

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Apr 3, 2005
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Orange Fish said:
Right, if you're putting out more power, you'll put out more calories.
Ok, we're mixing apples and oranges here and I think that's causing some confusion.

Calories are a unit of *energy.* Power (watts) is energy *per unit of time.* More of one does not equal more of the other, because one is dependent upon time and the other is not.

energy is to power as distance is to speed.

You can expend 1000 calories at 50 watts or at 250 watts. The only difference is that it takes longer at 50 watts. You can ride 20 miles at 5mph or at 20mph. The only difference is that it takes longer at 5mph.
 

dhk

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frenchyge said:
Ok, we're mixing apples and oranges here and I think that's causing some confusion.

Calories are a unit of *energy.* Power (watts) is energy *per unit of time.* More of one does not equal more of the other, because one is dependent upon time and the other is not.

energy is to power as distance is to speed.

You can expend 1000 calories at 50 watts or at 250 watts. The only difference is that it takes longer at 50 watts. You can ride 20 miles at 5mph or at 20mph. The only difference is that it takes longer at 5mph.
Yes, but you'll burn more total calories riding the 20 miles @ 20 mph vs. 5 mph. At 5 mph, all you're doing is overcoming rolling resistance, which doesn't take much output. After all, most little kids go 5 mph on a level with no problem.

If 20 mph takes 200 watts of power, 5 mph is on the order of 12 watts, according to a velo calculator I use. If we accept that kjoules = kcals, then 200w for an hour would burn 720 kcals, while 12 watts for 4 hours (to cover the same distance) would burn 173 kcals. That's only 24% of the calorie expenditure.

Obvious conclusion is that going slowly on a bike is a much more energy efficient way to cover the miles.

ps: Can anyone describe a Jaffa cake...perhaps we have something close?
 

wilmar13

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frenchyge said:
If your car gets 20 miles per gallon, it takes a gallon to go 20 miles. Doesn't really matter how fast or slow you drive those 20 miles.

The only difference would be any difference in motor efficiency from travelling at different speeds. I can't really speak from experience, but what I recall from other threads is that the human body's efficiency is constant enough that it's not a big effect.

Ummmm no. It doesn't work that way chief, you are forgeting air resistance is a squared component in the drag calculation. Go twice as fast get there twice as fast sure, but you need four times the energy.
 

frenchyge

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dhk said:
Yes, but you'll burn more total calories riding the 20 miles @ 20 mph vs. 5 mph. At 5 mph, all you're doing is overcoming rolling resistance, which doesn't take much output. After all, most little kids go 5 mph on a level with no problem.

If 20 mph takes 200 watts of power, 5 mph is on the order of 12 watts, according to a velo calculator I use. If we accept that kjoules = kcals, then 200w for an hour would burn 720 kcals, while 12 watts for 4 hours (to cover the same distance) would burn 173 kcals. That's only 24% of the calorie expenditure.
Yep, you're right. The energy required is equal to the resistance x distance travelled. Since the resistance is much higher at higher speed, but the distance is the same, more energy (kcals) is used during the faster ride even though the duration is shorter.

My previous agreement that "20 miles was 20 miles" was incorrect. :eek:
 

wilmar13

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frenchyge said:
You can expend 1000 calories at 50 watts or at 250 watts. The only difference is that it takes longer at 50 watts. You can ride 20 miles at 5mph or at 20mph. The only difference is that it takes longer at 5mph.

First part is true, but second is not. You will expend more energy riding at a higher rate of speed for less time over the same distance. You will burn more calories riding 20 miles at 20 miles an hour than 20 miles at 10 mph.
 

EoinC

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Feb 9, 2004
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wilmar13 said:
First part is true, but second is not. You will expend more energy riding at a higher rate of speed for less time over the same distance. You will burn more calories riding 20 miles at 20 miles an hour than 20 miles at 10 mph.
You are correct.
 

MichaelB

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Feb 10, 2004
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Andy/RST If you cycle 20 miles at either 20mph or 10mph you will use the same amount of energy (kcals) except that if you cycle at 10mph you will take twice as long to do it (obviously). Andy[/QUOTE said:
Thats not correct actually. If we ignore power absorption from the tyres and bearings which, at 10mph is grossly inaccurate, then we only have aero losses. Aerodynamic drag forces are proportional to the square of the velocity. The power requirement to overcome a force is equal to the force multiplied by the velocity and hence the power requirement to overcome parasitic drag forces is cubic P ~ v^3

That means that the power to overcome the parasitic drag experienced when moving an object through a fluid increases by a factor of 8 when you double your speed (2^3 = 8).

This means although you take half the time when cycling at 20 as opposed to 10mph the energy required is many factors higher.
 

wilmar13

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MichaelB said:
Aerodynamic drag forces are proportional to the square of the velocity. The power requirement to overcome a force is equal to the force multiplied by the velocity and hence the power requirement to overcome parasitic drag forces is cubic P ~ v^3

Wait a minute is it square or cube? I am too lazy to look it up, but I initially wanted to say it was cubed as well, but then remembered that in the equation for drag velocity is only a squared component...

Either way there is not a linear relationship between speed and calories burned, but in a simplistic model there is for energy and calories, therefore if Andy changes it to somethine like: "ride one hour at 300 watts vs. two hours at 150watts the calories burned will be the same" we will all be on the same page.
 

MichaelB

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wilmar13 said:
Wait a minute is it square or cube? I am too lazy to look it up, but I initially wanted to say it was cubed as well, but then remembered that in the equation for drag velocity is only a squared component...

Either way there is not a linear relationship between speed and calories burned, but in a simplistic model there is for energy and calories, therefore if Andy changes it to somethine like: "ride one hour at 300 watts vs. two hours at 150watts the calories burned will be the same" we will all be on the same page.


Yeah thats a better way of looking at it.

Drag Force = Cd * 1/2 * [density] * A * v^2

P = F.v

so,

P = Cd * 1/2 * [density] * A * v^3

Dont forget the tyres though as that makes it more difficult.
 

frenchyge

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wilmar13 said:
First part is true, but second is not. You will expend more energy riding at a higher rate of speed for less time over the same distance. You will burn more calories riding 20 miles at 20 miles an hour than 20 miles at 10 mph.
That's all true, but none of it contradicts what I said in that post. I was just pointing out that speed and distance are only related by time. Certainly there are other differences like cadence, gearing, heartrate, etc. that would be involved in propelling the bike at 20mph vs 5mph.

I don't believe I said anything wrong in *that* post (although my other previous post in agreement with Andy was incorrect).
 

wilmar13

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frenchyge said:
That's all true, but none of it contradicts what I said. I don't believe I said anything wrong in *that* post (although my other previous post was incorrect).

Yeah you are right...your post I quoted was technically correct actually :eek:
 

jbieryjr

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Apr 15, 2004
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frenchyge said:
Ok, we're mixing apples and oranges here and I think that's causing some confusion.

Calories are a unit of *energy.* Power (watts) is energy *per unit of time.* More of one does not equal more of the other, because one is dependent upon time and the other is not.

energy is to power as distance is to speed.

You can expend 1000 calories at 50 watts or at 250 watts. The only difference is that it takes longer at 50 watts. You can ride 20 miles at 5mph or at 20mph. The only difference is that it takes longer at 5mph.



Frenchyge,
You are close. The defn is correct, but the example is not. A good formula that I use is:

W = Cv [K1 + {K2(Cv+Cw)(Cv+Cw)} + {10.32Em(s/100 + 1.01a/g)}]
Where:

W = power in watts
1 W = 1 joule/sec
69.78W = 1000 calories/min = 1 kilocal/min = 1 Calorie/min
1 Calorie = 4186 joules
Cv = speed of cyclist in meters/sec
1 mph = .447 meters/sec
1 mph = 1.609 kilometeres/hr
K1 and K2 are constants
Cw = headwind in meters/sec
Em = mass of cyclist and bicycle in kg
1 pound = .4536 kg
s = slope or grade in %
a = acceleration of the bicycle in meters/(sec)(sec)
g = gravitational accel = 9.806 m/sec-sec at sea level

From Bicycling Science by Frank Whitt and David Wilson, p.157

So, Using this formula. An average speed on flats burn calories approximately:

5 mph - 7 Cal/mile - 37 Cal/hr
20 mph - 37 Cal/mile - 742 Cal/hr

if you multiple 5mph x4 miles. 37 cal/hr (148cal/4hrs) does not equal 1 hr at 20 mph (742 cal/hr).
 

wilmar13

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jbieryjr said:
Frenchyge,
You are close. The defn is correct, but the example is not. A good formula that I use is:

W = Cv [K1 + {K2(Cv+Cw)(Cv+Cw)} + {10.32Em(s/100 + 1.01a/g)}]
Where:

W = power in watts
1 W = 1 joule/sec
69.78W = 1000 calories/min = 1 kilocal/min = 1 Calorie/min
1 Calorie = 4186 joules
Cv = speed of cyclist in meters/sec
1 mph = .447 meters/sec
1 mph = 1.609 kilometeres/hr
K1 and K2 are constants
Cw = headwind in meters/sec
Em = mass of cyclist and bicycle in kg
1 pound = .4536 kg
s = slope or grade in %
a = acceleration of the bicycle in meters/(sec)(sec)
g = gravitational accel = 9.806 m/sec-sec at sea level

From Bicycling Science by Frank Whitt and David Wilson, p.157

So, Using this formula. An average speed on flats burn calories approximately:

5 mph - 7 Cal/mile - 37 Cal/hr
20 mph - 37 Cal/mile - 742 Cal/hr

if you multiple 5mph x4 miles. 37 cal/hr (148cal/4hrs) does not equal 1 hr at 20 mph (742 cal/hr).

Just make sure you don't use this as gospel unless you derived K1 and K2 for your specific situation. For a larger rider the values will be different that a smaller rider... even better is just to have a power meter so you don't have to mess around with making calculated guesses... then you only have to guess about your efficiency... ;)
 

frenchyge

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Apr 3, 2005
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jbieryjr said:
Frenchyge,
You are close. The defn is correct, but the example is not. A good formula that I use is:

<snipped>

if you multiple 5mph x4 miles. 37 cal/hr (148cal/4hrs) does not equal 1 hr at 20 mph (742 cal/hr).
Yeah. The point I was trying to make was that: just as talking about speed does not imply anything about the distance travelled, neither does talking about power imply anything about the amount of calories (energy) expended. You need to consider the time involved in the activity as well.
 

mises

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May 27, 2005
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biker-linz said:
I just thought you guys might be interested in these studies:

Tremblay, A., J. A. Simoneau, and C. Bouchard. Impact of exercise intensity on body fatness and skeletal muscle metabolism. Metabolism. 43:814-818, 1994.
Yoshioka, M., E. Doucet, S. St-Pierre, N. Almeras, D. Richard, A. Labrie, J. P. Despres, C. Bouchard, and A. Tremblay. Impact of high-intensity exercise on energy expenditure, lipid oxidation and body fatness. Int J Obes Relat Metab Disord. 25:332-339, 2001.

Basically high-intensity training is more effective for reducing fat mass than is low intensity exercise which uses the same number of calories. Just another nail in the ol' 'fat-burning' intensity myth:).

L.
I will second that. I have tried it both ways and adding more intervals of varying lengths has way more effect. Too much low intensity lowers the levels of HGH, Testosterone, etc. and that makes your fat problems worse. High intensity does the opposite, promoting anabolic hormones.
 

jbieryjr

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Apr 15, 2004
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wilmar13 said:
Just make sure you don't use this as gospel unless you derived K1 and K2 for your specific situation. For a larger rider the values will be different that a smaller rider... even better is just to have a power meter so you don't have to mess around with making calculated guesses... then you only have to guess about your efficiency... ;)


You are so right. Don't for get EM (mass of cyclist and bike) also has a great impact on watts. I wish I could afford a power meter, but some of use still use calculations to get a idea of where our training is.
 

jbieryjr

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Apr 15, 2004
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frenchyge said:
Yeah. The point I was trying to make was that: just as talking about speed does not imply anything about the distance travelled, neither does talking about power imply anything about the amount of calories (energy) expended. You need to consider the time involved in the activity as well.


Good response. Try to use Zone training with HR and watts for weight loss or cardiovascular training. This includes Lactic Acid threshold training as well. Incorporating each defined zone will get you better gains for overall performance.
 

wilmar13

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Nov 30, 2003
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jbieryjr said:
You are so right. Don't for get EM (mass of cyclist and bike) also has a great impact on watts. I wish I could afford a power meter, but some of use still use calculations to get a idea of where our training is.

Well the EM in the formula you used will only matter for climbing for if you are sprinting from 0-20mph where you could generate meaningful acceleration values...on flat road notice that unless you adjust the K values for your aero drag that you can change your weight all over the place and it has no effect on the result... that is why I said make sure the K values are applicable to you...

Yeah there is nothing wrong with calculating power as even the cheapest option is several hundred dollars, of course those that cough up 1K for new wheels may have been better off with the power meter ;)
 

Sidetrack

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May 30, 2005
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dhk said:
ps: Can anyone describe a Jaffa cake...perhaps we have something close?
Just had my first Jaffa cake. I had to laugh when I was strolling through my local high end grocery store (Larry's Markets in Seattle) and saw that in their International foods section that along with the asian, mexican, indian and kosher they had added a "British Food" section

I picked up a carton of Jaffa cakes and ate them in the MINI on the way home :)

to describe them... 2inch round sponge cake with a dollop of an orange-gelatin filling on top then a layer of chocolate over that. I don't have the nutrition information with me at the moment.