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How much does weight mean when going uphill?

Let's say I'm going up a mountain with 1000 m of elevation and it's quite steep: 10%. Let's also say I can maintain 250 watts. If I add 1 kg of weight to the bike, can I determine how much slower (in time) I will be? How many seconds will I lose?

Please show me the formula and calculation of this, and feel free to elaborate on the whole subject.

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I think this belong to physics.stackexchange.com –  heltonbiker Nov 7 '12 at 15:57
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@heltonbiker: It's on-topic for this site, so I wouldn't migrate it. A differently phrased version of the question would likely be appropriate there. –  freiheit Nov 7 '12 at 19:35
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4 Answers

up vote 11 down vote accepted

Presuming you are doing a standing start and coming to a complete stop at the top of the hill. The simple requirement is you need energy to move your from the bottom to the top. Most of the energy required will be to raise potential energy of the payload (you and the bike). Essentially you will be creating kinetic energy (moving the bike) by converting chemical energy in your body. There will be loses due to heat, friction with the road surface and air resistance.

Disregarding them for the moment (they are not negligible, but complicate the calculation).

Potential Energy (PE) = m * g * h

Where:

m = mass

g = gravitational acceleration

h = height

PE is proportional to m, so a 10% increase in mass will increase the PE by 10%. Meaning you will need 10% more kinetic energy to get to the top of the same hill.

Power is work done (energy) divided by time:

P = W / t

Where:

P = Power in watts

W = work done or energy in Joules

t = time to do work.

If your power is constant we can rearrange the equation to get

P = (m * g * h)/ t

becomes:

t = (m * g * h) / P

so with constant power, gravity and hill height your time will increase proportional to increase in mass, given by the above equation.

If there is no wind air resistance will become less relevant the slower you are going. Friction will increase due to increase in weight. The steepness of the hill is theoretically irrelevant in this calculation. You are gaining the same amount of Gravitation Potential energy when you the same mass to the same height. So it shouldn't theoretically matter whether the hill is 10% or double as long and 5%.

However, as you are generating the energy you need to create it from chemical energy, and there is only so much you can generate at one time. Your muscles will become inefficient and therefore on steeper hills you may require more energy than less steep ones. So on a steeper hill wind resistance might become less relevant but power (how much energy you can put out over time) to weight ratio will become the most relevant factor.

What I am trying to point out in the last paragraph is that the energy you require to put into your body and for your body to convert into forward motion is not the same as the simple kinetic energy required to get you to the top of the hill. However all things being equal a change in mass will have the same affect on the time as I stated in the equations.

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Whilst the physics here is correct, if you want to know the answers,and play with scenarios, you can't beat bikecalculator.com –  THEMike May 1 '13 at 6:52
    
My answer was more about highlighting the relationships between the contributing factors. Bikecalculator.com is great though for getting more precise answers. –  robthewolf May 1 '13 at 7:43
    
Oh absolutely, I think your answer (and some of the additional information on the thread) was great. But, also, nice to point to the tool so people can then play with the model and understand it was all I was saying. –  THEMike May 1 '13 at 10:59
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If I on one ride add 1 kg of weight to the bike, how much slower (in time) will I be?

Assuming that you and your bike mass 100kg (in round numbers), an extra 1kg causes a 1% increase in weight, i.e. a 1% increase in the potential energy associated with climbing the hill.

If your power output is constant, that implies a 1% increase in time.

However some of your power output is going to overcome wind and rolling resistance, not potential energy. If only half of your power is going to potential energy (which depends on weight), and half is constant (independent of weight), I think that would imply a 0.5% increase in time.

Does that make sense, I don't think so without saying how many watt I output, lets say I make 250W

What I wrote above is unaffected by your total power; the change is relative rather than absolute: i.e. it's 1%, no matter what it's 1% of.

Part of the hand-waving in my statement is "If your power output is constant": which is true if you have fine control over your gears (so that you can, like, you know, gear down by 1% in order to adjust to the 1% increase in the weight-and-therefore-effort).

The change isn't actually linear: for example if it were a 1000kg increase in weight, i.e. 1000% instead of 1%, you'd need to gear down so far that you'd be going so slowly that you couldn't stay upright on a two-wheeled bike. For relatively small increases in weight, though, I expect the difference in effort (and therefore also, via the hand-waving outlined above, the difference in the duration) to be approximately linear.

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A Joule is a Newton-meter and is also a Watt-second. Gravity is about 9.81 Newtons/kilogram.

Raising 1 pound 1000 feet would be raising 0.4536kg 304.8 meters. So that would be 9.81 * 0.4536 * 304.8 = 1356 Joules, or 1356 Watt-seconds.

Your peak sustained energy output is probably in the general range of 300 watts (and "cruising" would be somewhere around half that), so you'd have to use all your energy for about 4.5 seconds to raise that one pound 1000 feet. (Or, to put it in perspective, about 19 minutes to raise a 250 pound bike+rider 1000 feet.)

For your assumed 250 watts, this would be 5.4 seconds for one pound or 22.6 minutes for 250 pounds. This would produce a speed, on the 10,000 foot distance, of about 5 mph. (Note that dropping much below roughly 200 Watts will produce a speed too slow to stay upright, especially given that the slower you go the more energy you must expend trying to stay upright.)

Of course, this is ignoring wind and rolling resistance losses, and hence the time required to "cover the distance" on level ground. Rolling resistance would be about the same as on level ground, but wind resistance would be less, since you're moving slower, and wind resistance is generally the larger of the two. So you need to add to the above times maybe 1/2 or 2/3rds of the time it would take you to cover the same distance on level ground. For a 10% grade that would be the time to cover 10,000 feet or about 1.9 miles. At 15 mph that would be about another 7.5 minutes, so add maybe half that.

Rats -- Just realized the question was in kg and meters...

"If I on one ride (of 10,000 meters at 10%) add 1 kg of weight to the bike, how much slower (in time) will I be (assuming 250 watts output)?":

That would be 9.81 * 1kg * 1000 meters * = 9810 watt-seconds. At 250 watts that's 38.84 seconds additional time due to adding 1kg.

And it occurs to me... that one could use the same calculations backwards in order to roughly calculate wattage output, given a weight, an average speed, and an average slope. This would likely be more accurate than many other wattage estimation schemes.

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I like your mixture of metric and imperial units ;-) –  Benedikt Bauer Nov 7 '12 at 12:59
    
@BenediktBauer -- Necessary, since Watts are metric and pounds/feet are imperial. Besides, it's easier to find the formulae expressed in metric. –  Daniel R Hicks Nov 7 '12 at 14:04
    
@BenediktBauer -- Oops!! -- Just realized what you were trying to say. –  Daniel R Hicks Nov 7 '12 at 16:58
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As already pointed out by the other answers, an additional kilogram is rather negligible when concerning purely the additional potential energy you need. But there are other factors where it may have a more or less larger effect.

Firstly your body does not necessarily respond linearly to higher load. As long as you are in a region where you can do the uphill without noticeable fatigue you will just be a bit slower with the same power output. But if the uphill will bring you to your limits even without the additional weight, every additional gram will bring you to your limit a bit earlier and make it more hard to go on.

Also mass is playing an important role every time you have to accelerate it in some way. Some things I think about are:

  • Smoothness of the uphill: The approximation which concerns only the potential energy works the better, the smoother the uphill is. If you go up on a constantly inclined, paved slope where you can assume to ride with nearly constant speed, there shouldn't be much of a difference. However, if you go mountain biking on a way where you have lot of "action" on the bike (climb up steps, try to keep traction on loose ground, switch between steeper and flatter sections), it can be different.

  • "Place" of the additional weight: It can make a difference where you add the weight. The largest effect will be if you have additional weight on rotating parts (wheels, drive train). But also a heavier bike itself will make it feel less responsive which will make your muscles go tired faster if you have to do a lot of bike handling stuff during the uphill (therefore it is again related to the smoothness of the uphill). If you had the additional weight in your backpack instead, it would have a much smaller effect.

Conclusion: Just from the additional potential energy point of view it may not make much of a difference but if you concern bike handling or fatigue close to your limits, it will have a larger effect.

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Surely "mass" not "ass". ;) –  James Bradbury Nov 27 '13 at 9:27
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@JamesBradbury of course it should have been "mass", but "ass" would fit as well with some people ;-). I will correct it. –  Benedikt Bauer Nov 27 '13 at 11:00
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Additional weight on rotating parts (vs weight on the frame) only affects acceleration (and then only slightly), and one does not generally accelerate much going up a serious hill. –  Daniel R Hicks Nov 27 '13 at 12:08
    
...and helps you by slowing deceleration. –  James Bradbury Nov 27 '13 at 13:38
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