Does weight really matter

[bladegeek]

Now before someone says......"train harder".....which is a no brainer....I am wondering if the weight of a bike really matters when it comes to hill climbing? Does the weight to money ratio make that big if a difference?

Example I can reduce the weight of my bike by 0.79lbs for $390....(changing the bar and fork) to a total weight 16.9 lbs.

So does less weight make you a better climber (which is the real question I am asking)? And is any weight change less then a pound worth the money?

 

[Bob Ross]

Not positive about hill climbing, but I know for a fact weight matters for stair climbing. Since I live on the 4th floor of a building whose elevator is constantly on the blink, it was trivial for me to justify the purchase of a <17lb bike this year...to replace a >30lb hybrid.

I would pay almost *anything* to shave several pounds off the weight of my bike!

 

[ScienceIsCool]

Now before someone says......"train harder".....which is a no brainer....I am wondering if the weight of a bike really matters when it comes to hill climbing? Does the weight to money ratio make that big if a difference?

Example I can reduce the weight of my bike by 0.79lbs for $390....(changing the bar and fork) to a total weight 16.9 lbs.

So does less weight make you a better climber (which is the real question I am asking)? And is any weight change less then a pound worth the money?

Well... When accelerating, F = m x a. So when coming out of a corner and going from 30 to 50 km/hr less mass will either let you use less force or accelerate quicker. Or a bit of both. Since the relationship is linear, the amount of performance you gain is equal to the amount of weight you lose.

In your case the weight loss is 0.8 lbs (I'll stick with Imperial for now). Let's say that your bike is 17 lbs, clothing, helmet, shoes, water bottles pump, etc, etc is another 5 lbs. Let's guess that your weight is 70 kg, er... 155 lbs. That means the total weight of the system is 177 lbs. Therefore, a 0.8 lb savings gives you a 0.8/177 = 0.4% boost in acceleration. Not a heck of a lot, but it's real. You might be able to get a similar boost for much less than $400.

And now climbing. The energy you expend going up a hill is equal to m x g x height gained. Power equals energy divided by time. So for a given power output, the time it will take you to climb a hill is equal to t = mgh/P. So your climbing performance should also improve by 0.4% or 14 seconds over an hour long climb. Or 1 second over a five minute climb. Again, not a heck of a lot.

Hope that helps...

John Swanson
www.bikephysics.com

 

[ScienceIsCool]

I was also going to calculate how much quicker you'd accelerate out of a corner, but kind of ran out of steam...

John Swanson
www.bikephysics.com

 

[Lucy_Aspenwind]

Well... When accelerating, F = m x a. So when coming out of a corner and going from 30 to 50 km/hr less mass will either let you use less force or accelerate quicker. Or a bit of both. Since the relationship is linear, the amount of performance you gain is equal to the amount of weight you lose.

In your case the weight loss is 0.8 lbs (I'll stick with Imperial for now). Let's say that your bike is 17 lbs, clothing, helmet, shoes, water bottles pump, etc, etc is another 5 lbs. Let's guess that your weight is 70 kg, er... 155 lbs. That means the total weight of the system is 177 lbs. Therefore, a 0.8 lb savings gives you a 0.8/177 = 0.4% boost in acceleration. Not a heck of a lot, but it's real. You might be able to get a similar boost for much less than $400.

And now climbing. The energy you expend going up a hill is equal to m x g x height gained. Power equals energy divided by time. So for a given power output, the time it will take you to climb a hill is equal to t = mgh/P. So your climbing performance should also improve by 0.4% or 14 seconds over an hour long climb. Or 1 second over a five minute climb. Again, not a heck of a lot.

Hope that helps...

John Swanson
www.bikephysics.com

Just to add to what John said, you could for example, go to analyticcycling.com, then using the same slope/grade/incline, see what difference in speed you'd obtain from the weight difference.

My personal opinion is that unless you are at a high level, you are much better off a) increasing your power output and/or b) losing weight of your body - it is much cheaper

 

[bladegeek]

Well... When accelerating, F = m x a. So when coming out of a corner and going from 30 to 50 km/hr less mass will either let you use less force or accelerate quicker. Or a bit of both. Since the relationship is linear, the amount of performance you gain is equal to the amount of weight you lose.

In your case the weight loss is 0.8 lbs (I'll stick with Imperial for now). Let's say that your bike is 17 lbs, clothing, helmet, shoes, water bottles pump, etc, etc is another 5 lbs. Let's guess that your weight is 70 kg, er... 155 lbs. That means the total weight of the system is 177 lbs. Therefore, a 0.8 lb savings gives you a 0.8/177 = 0.4% boost in acceleration. Not a heck of a lot, but it's real. You might be able to get a similar boost for much less than $400.

And now climbing. The energy you expend going up a hill is equal to m x g x height gained. Power equals energy divided by time. So for a given power output, the time it will take you to climb a hill is equal to t = mgh/P. So your climbing performance should also improve by 0.4% or 14 seconds over an hour long climb. Or 1 second over a five minute climb. Again, not a heck of a lot.

Hope that helps...

John Swanson
www.bikephysics.com

Ok....Heck of an answer. Made me use my college physics for the first time in years. Thanks for the help and I get your point. Thanks again.

 

[graphixgeek]

Professional level racing (or any sport, or any level), at one of its most basic levels is about having the advantage over your opponents in order to win. Yeah the advantage may not be much but an advantage is an advantage. Having a bunch of little advantages can lead up to one large one. Now for us mere mortals...it's a different purpose all together...

 

[sogood]

I understand that weight of wheel/tyre makes a greater difference when it comes to climbing and accelerating given that it's a rotating mass. Hence there are so called climbing wheels etc. Within limits, having weights in this department is more productive in terms of outcome.

 

[Strid]

And now climbing. The energy you expend going up a hill is equal to m x g x height gained. Power equals energy divided by time. So for a given power output, the time it will take you to climb a hill is equal to t = mgh/P. So your climbing performance should also improve by 0.4% or 14 seconds over an hour long climb. Or 1 second over a five minute climb. Again, not a heck of a lot.

Hope that helps...

John Swanson
www.bikephysics.com

You need evaluate the contribution of power you need to ride in two steps. One is drag (wind, resistance and so on) which goes at Fdrag = k * v, so power you need to overcome this is Pdrag = k * v² (no effect from weight savings here!) and then the contribution from climbing which is Pclimb = m * g * sin (climbing angle) * v

I will just do an example here (all numbers are made up!).

I only drive 400 watts no more no less. On a flat run, this equals 12m/s. So as I pedal along with 400 Watts at constant speed, my power (Pmy) is equal to the drag resistance (Pdrag), no power used for climbing or descenting here.

Pmy = Pdrag = k * (12 m/s)² = 400 W, <=> k = 400 W / (12 m/s)² = 2.8 kg
so the drag coefficient, k, for me and my bike is roughly = 2.8 kg

Lets say that I climb at a 10% climb = about 5.7 degrees, again at 400 watts, and I weigh 75 kg and we're on approximately earth, so g = 10 m/s².

Then my power output goes as

Pmy = Pdrag + Pclimb = k * v^2 + m * g * sin ( a ) * v
<=>
400 W = 2.8 kg * v² + 75 kg * 10 m/s² * sin(5.7 degrees) * v
<=>
0 = 2.8 kg * v² + 75 kg * 10 m/s² * sin(5.7 degrees) * v - 400 w

Now, this is a quadratic equation, easily sovled, and the result is

v = 4,575 m/s = 16,46 km/h

Now, I shave off 2 kgs of my bike and water bottles, my speed is nowgiven by:

0 = 2.8 kg * v² + 73 kg * 10 m/s² * sin(5.7 degrees) * v - 400 w

This gives v = 4,667 m/s = 16,80 km/h

This is an 2% increase in speed by shaving off two kgs.

Its the correct way to do it, unless I did calculations errors - the principle should be correct.

 

[Strid]

One thing I don't know to make this more correct, what is a typical power output (watts) at, say, 40 km/h for a cyclist?


Remember this is a somewhat crude calculation, but it shouldn't matter too much that you've got friction from moving parts and rotational friction in the wheels and so on. Most of these go as F = k * v² and so they are by nature built into our k value.

DUH, did a small mistake .. will correct tomorrw...

 

[TrekDedicated]

Yes it does.

I have lost 23 lbs thanks to my jaw surgey, and have now started riding again. HUGE DIFFERENCE.

Add some lighter wheels, lighter brakes, and I am one fast mo-fo (even after 3 months of no riding)

 

[ScienceIsCool]

You need evaluate the contribution of power you need to ride in two steps. One is drag (wind, resistance and so on) which goes at Fdrag = k * v, so power you need to overcome this is Pdrag = k * v² (no effect from weight savings here!) and then the contribution from climbing which is Pclimb = m * g * sin (climbing angle) * v

I will just do an example here (all numbers are made up!).

I only drive 400 watts no more no less. On a flat run, this equals 12m/s. So as I pedal along with 400 Watts at constant speed, my power (Pmy) is equal to the drag resistance (Pdrag), no power used for climbing or descenting here.

Pmy = Pdrag = k * (12 m/s)² = 400 W, <=> k = 400 W / (12 m/s)² = 2.8 kg
so the drag coefficient, k, for me and my bike is roughly = 2.8 kg

Lets say that I climb at a 10% climb = about 5.7 degrees, again at 400 watts, and I weigh 75 kg and we're on approximately earth, so g = 10 m/s².

Then my power output goes as

Pmy = Pdrag + Pclimb = k * v^2 + m * g * sin ( a ) * v
<=>
400 W = 2.8 kg * v² + 75 kg * 10 m/s² * sin(5.7 degrees) * v
<=>
0 = 2.8 kg * v² + 75 kg * 10 m/s² * sin(5.7 degrees) * v - 400 w

Now, this is a quadratic equation, easily sovled, and the result is

v = 4,575 m/s = 16,46 km/h

Now, I shave off 2 kgs of my bike and water bottles, my speed is nowgiven by:

0 = 2.8 kg * v² + 73 kg * 10 m/s² * sin(5.7 degrees) * v - 400 w

This gives v = 4,667 m/s = 16,80 km/h

This is an 2% increase in speed by shaving off two kgs.

Its the correct way to do it, unless I did calculations errors - the principle should be correct.

I think you over-complicated the calculation, but we arrived at the same conclusion. You can deal strictly with energy and do the m x g x h thing. If 0.8 pounds (~0.4 kg) makes a 0.4% difference and the relationship is linear then I would expect 2 kg to make a 2% difference. Power doesn't vary with weight - that's the key...

John Swanson
www.bikephysics.com

 

[Strid]

I think you over-complicated the calculation, but we arrived at the same conclusion. You can deal strictly with energy and do the m x g x h thing. If 0.8 pounds (~0.4 kg) makes a 0.4% difference and the relationship is linear then I would expect 2 kg to make a 2% difference. Power doesn't vary with weight - that's the key...

John Swanson
www.bikephysics.com

Anyways, although I did some serious units mixups, my point is that it isn't totally linear.

More like f (x) = a + b x linear. loose N% of x and f doesn't increase by N%, because you still get a contribution from a. a is sort of drag and has nothing to do with mass and b is what you have to overcome by climbing. f is your speed and x is your mass.
This is just an analogy.

Even if you were had no mass, you'd still have to overcome wind resistance.

 

[ScienceIsCool]

Anyways, although I did some serious units mixups, my point is that it isn't totally linear.

More like f (x) = a + b x linear. loose N% of x and f doesn't increase by N%, because you still get a contribution from a. a is sort of drag and has nothing to do with mass and b is what you have to overcome by climbing. f is your speed and x is your mass.
This is just an analogy.

Even if you were had no mass, you'd still have to overcome wind resistance.

Tja, Strid! Hur ar det? (I'm just leaning Swedish...) I agree 100% with your analysis.

John Swanson
www.bikephysics.com

 

[alienator]

I understand that weight of wheel/tyre makes a greater difference when it comes to climbing and accelerating given that it's a rotating mass. Hence there are so called climbing wheels etc. Within limits, having weights in this department is more productive in terms of outcome.

The difference that a wheel's moment of inertia makes on acceleration is almost nothing. Human's--including McEwen, Boonen, or whoever--don't accelerate fast enough for the difference to be noticeable.

Even when climbing, a "heavy" wheel does not cause any significant performance difference until the slope gets very steep.

 

[BikingBrian]

All of this scientific mumbo-jumbo is interesting and certainly a good thought experiment. However....I have found in actual experience that the weight of a bicycle almost never makes any difference to race results. The top riders are generally strong enough that, for example, a .4% difference doesn't matter.

 

[rayhuang]

Ive read this before and I believe it to be true from seat of the pants riding. The heavier YOU are , the heavier a bike you can ride and not be disadvantaged. The lighter you are then it stands that the more a light weight bike can help. When I was 130+ lbs racing, my 20 to 21 lb bikes always felt like bricks to me. I did as many climbing road races as I could. Now I weigh 164 lbs (down form 175 in July) and my 17 to 18 lb (depending on which wheels are on it) bike is never an issue. When I am suffering or just get shelled on a climb-its only my legs that whimper against the cranks-it never feels like I am trying to haul the bike up the climb with me.

NOw-if I get down to 150 lbs and race a lot next year-I may very well go the extra mile and shave my bike down to 15.5 to 16 lbs which I think is very doable (tires, Cranks, alum rings, Dura-ace cassette, handle bars, seat and seat post should do it).

Ray
p.s. sorry-not very scientific.

 

[dhk2]

Anyways, although I did some serious units mixups, my point is that it isn't totally linear.

More like f (x) = a + b x linear. loose N% of x and f doesn't increase by N%, because you still get a contribution from a. a is sort of drag and has nothing to do with mass and b is what you have to overcome by climbing. f is your speed and x is your mass.
This is just an analogy.

Even if you were had no mass, you'd still have to overcome wind resistance.

Believe you've got an error in the drag portion of your calculations. Drag force is a factor of cdA and v**2, not a constant. As a result, power to overcome drag is a function of the v**3, the cube of the velocity.

Ignoring the small drag due to tire rolling resistance, at 12 m/sec, believe 400 watts of power would equate to a drag force of 33.3 newtons, or 3.4 kg. At 6 m/sec, the drag force would only be 1/4 of 3.4 kg, and the drag power required would only be 1/8th of 400w, or 50w. This cube factor for drag explains why most all riders can easily reach 6m/sec (13.5 mph) on a level, but only a relative few can reach 12 m/sec (27 mph).

But for steep climbs at low speeds, eg, the 4.5 m/sec (10 mph) in your example, drag power would be down to 21w, small enough relative to 400w that it can be ignored in favor of just the rate of change of potential energy as John suggests.

 

[Rhubarb]

Does weight make a difference? Hell yeah!

I don't know how much your going to get out of 0.7lbs but I recently bought a new bike. I went from a 12kg bike to a 7 kg a bike and the difference is out of this world! Hills which were hard work, well are still hard work but I am going so much faster!

If your looking for some minor weight gains might be best looking so see if you can trim some fat of yourself. Of course you might be tip top in that regard.

 

[dhk2]

Does weight make a difference? Hell yeah!

I don't know how much your going to get out of 0.7lbs but I recently bought a new bike. I went from a 12kg bike to a 7 kg a bike and the difference is out of this world! Hills which were hard work, well are still hard work but I am going so much faster!

If your looking for some minor weight gains might be best looking so see if you can trim some fat of yourself. Of course you might be tip top in that regard.

5 kg certainly is a big drop. If you and your old bike weighed 100 kg, you've dropped 5%. On a steep climb at 10 mph, 5% faster means 10.5 mph...that's a big deal.

Regardless of numbers, the 7 kg bike is going to feel a lot more responsive underneath you; like it's ready to jump whenever you step down on the gas.