How many watts is a 1lbs weight loss worth?

nickd01

Is there a way of working it out?

If I dropped a combination of 1lbs of weight off me and the bike - how many watts of power would it "add" ?

Anonymous

On the flat at a steady speed, probably none.

DaveyL

It wouldn't add any power, you need to train to do that.

It would require you to use slightly fewer Watts when going up hills at a given speed, or you would go slightly faster for the same number of Watts.

You can work it out on analyticcycling.com

ColinJ

Plug the numbers into this calculator and work it out.

antikythera

Given that

1 calorie = 4.186 8 wattsecond
and
One pound of body fat contains approximately 3,500 calories

Then the maths suggests that you need to generate approximately

244 Watts/minute

It's all theory on my part and its likely that biochemistry, biophysics, nutrition, your fitness level etc will skew the calcuation.[/quote]

nickd01

ColinJ wrote:

Plug the numbers into this calculator and work it out.

Thats useful, thanks.

If I drop the rider weight by 5lbs, and bike weight by 2lbs on a 5 mile 5% climb it seems to give around a 0.5mph advantage.

That's enough to justify some new bling!!

Pokerface

nickd01 wrote:

ColinJ wrote:

Plug the numbers into this calculator and work it out.

Thats useful, thanks.

If I drop the rider weight by 5lbs, and bike weight by 2lbs on a 5 mile 5% climb it seems to give around a 0.5mph advantage.

That's enough to justify some new bling!!

Or a diet.

Anonymous

Pokerface wrote:

nickd01 wrote:

ColinJ wrote:

Plug the numbers into this calculator and work it out.

Thats useful, thanks.

If I drop the rider weight by 5lbs, and bike weight by 2lbs on a 5 mile 5% climb it seems to give around a 0.5mph advantage.

That's enough to justify some new bling!!

Or a diet.

Or both.

dennisn

NapoleonD wrote:

Pokerface wrote:

nickd01 wrote:

ColinJ wrote:

Plug the numbers into this calculator and work it out.

Thats useful, thanks.

If I drop the rider weight by 5lbs, and bike weight by 2lbs on a 5 mile 5% climb it seems to give around a 0.5mph advantage.

That's enough to justify some new bling!!

Or a diet.

Or both.

Just go for the bling. Much easier. Of course you could go on a diet and use the money you save by not eating to buy even more bling. Decisions, decisions.

Pokerface

My bike is currently underweight and I'm now in the process of 'fattening' it up to make it UCI legal.

But at the same time, trying to get my own fat arse down a few sizes. :oops:

dennisn

Pokerface wrote:

My bike is currently underweight and I'm now in the process of 'fattening' it up to make it UCI legal.

But at the same time, trying to get my own fat ars* down a few sizes. :oops:

I've got you beat. Both myself AND my bike are overweight.
You know, I think I could be a good dieter and peel of the pounds if I could only convince myself that I really didn't want those beers, margarita's, and Mexican food. At 61 years I sort of view cycling as something fun to do AND something that helps me keep the excess blubber, from the above mentioned beer etc., somewhat at bay. Lately though
this theory has not been working all that well. Somethings got to give and I don't want it to be my belt.

nickd01

Pokerface wrote:

nickd01 wrote:

ColinJ wrote:

Plug the numbers into this calculator and work it out.

Thats useful, thanks.

If I drop the rider weight by 5lbs, and bike weight by 2lbs on a 5 mile 5% climb it seems to give around a 0.5mph advantage.

That's enough to justify some new bling!!

Or a diet.

Yep, a couple of kg's off to hit 70kg would be perfect for me but that'll be easy to loose as the summer looms and not nearly as much fun as buying bling!

Team Banana Spokesman

e = mc ^2

if you have 2 riders with the same w per kg and same bike but one is 80kg and one is 60kg: would a 1kg drop off the bike feel the same to both?

dennisn

Team Banana Spokesman wrote:

e = mc ^2

if you have 2 riders with the same w per kg and same bike but one is 80kg and one is 60kg: would a 1kg drop off the bike feel the same to both?

Only in a leap year. Due to the gravitation forces that occur during that extra day, the space - time continuum is not the same as a normal day and this affects string theory in a way that's very hard to follow. I can't explain it but I believe it.

Alex_Simmons/RST

Team Banana Spokesman wrote:

e = mc ^2

if you have 2 riders with the same w per kg and same bike but one is 80kg and one is 60kg: would a 1kg drop off the bike feel the same to both?

It depends or more than just mass.

stokepa31

being a fat boy. I can tell you it makes a huge difference in a non scientific way. ive cycled from 22.5 stone down to 18.5.

at 22 stone at the end of a summer of riding regularly i could average about 14mph over 20m.

last Sunday at 18.5 stone i averaged 15.1mph over 60 miles.

this is full of holes in terms of all the other variables but it feels like night and day. this should of course not mean that you cant spend shed loads of cash on new lighter bling

G-Wiz

Interesting, from what I can see from Alex's chart I should sod the diet, spend all my time getting my power up, then concentrate on matching the skinny climbers up the descents, and leaving them for dust on the other side.

Sounds like a lot more fun :-)

What does CdA stand for? I have a nasty feeling it could be the spanner in the gears of my plan.

Alex_Simmons/RST

Coefficient of Drag Area (Cd - a unitless measure of the drag characteristics of shapes through air) multiplied by the Effective Frontal Area (A - the size of object presented to the wind in square metres) to give you CdA, which is a measure of how slippery you are through the air. Units are m^2,

The lower your CdA, the less power is required to travel at same speed.

The larger you are, the bigger your CdA.