# How much effect does the mass of a bicycle tire have on acceleration?

There are claims often made that, eg, "An ounce of weight at the rims is like adding 7 ounces of frame weight." This is "common knowledge", but a few of us are skeptical, and our crude attempts at working out the math seem to support that skepticism.

So, let's assume a standard 700C bicycle tire, which has an outside radius of about 36cm, a bike weighing 90Kg with bike and rider, and a tire+tube+rim weighting 950g. With the simplifying assumption that all the wheel mass is at the outer diameter, how much effect does adding a gram of additional weight at the outer diameter have on acceleration, vs adding a gram of weight to the frame+rider?

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## 2 Answers

A few simplifying assumptions:

• I'm going to ignore any rotational energy stored in the bike chain, which should be pretty small, and wouldn't change when you change bike tires
• I'm going to use 50 cm for the radius of the bike tire. This is probably a little big, and your bike will likely have a different radius, but it makes my calculations easier, so there. I will include a formula nonetheless.
• I'm going to assume that the rider provides a fixed torque to the wheels. This isn't strictly true, especially when the bike has different gears, but it simplifies our calculations, and, once again, the torque provided won't vary when you change the weight profile of the tire

OK, so now, let's analyze our idealized bicycle. We're going to have the entire $m$ of each of the two wheels concentrated at the radius $R$ of the tires. The cyclist and bicycle will have a mass $M$. The cycle moves forward when the cyclist provides a torque $\tau$ to the wheel, which rolls without slipping over the ground, with the no-slip conditions $v=R\omega$ and $a=\alpha R$ requiring a forward frictional force $F_{fr}$ on the bike.

Rotationally, with the tire, we have:

$$\begin{align*} I\alpha &= \tau - F_{fr}R\\ mR^{2} \left(\frac{a}{R}\right)&=\tau-F_{fr}R\\ a&=\frac{\tau}{mR} - \frac{F_{fr}}{m} \end{align*}$$

Which would be great for predicting the acceleration of the bike, if we knew the magnitude of $F_{fr}$, which we don't.

But, we can also look at Newton's second law on the bike, which doesn't care about the torque at all. There, we have (the factor of two comes from having two tires):

$$\begin{align*} (M+2m)a&=2F_{fr}\\ F_{fr}&=\frac{1}{2}(M+2m)a \end{align*}$$

Substituting this into our first equation, we get:

$$\begin{align*} a&=\frac{\tau}{mR}-\frac{1}{m}\frac{(M+2m)a}{2}\\ a\left(1+\frac{M}{2m} +1\right)&=\frac{\tau}{mR}\\ a\left(\frac{4m+M}{2m}\right)&=\frac{\tau}{mR}\\ a&=\frac{2\tau}{R(4m+M)} \end{align*}$$

So, now, let's assume a 75 kg cyclist/cycle combo and a 1 kg wheel, and a 0.5 m radius for our wheel. This gives $a=0.0506 \tau$. Increasing the mass of the cyclist by 1 kg results in the acceleration decreasing to $a=0.0500 \tau$. Increasing the mass of the wheels by 0.5 kg each results in the acceleration decreasing to $a=0.0494$, or roughly double the effect of adding that mass to the rider/frame.

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Using the parameters you provided, and distributing the one gram evenly amongst the tires, you get the same result--putting the weight at the rims will slow the bike by roughly twice as much as you would by adding the weight to the rider. – Jerry Schirmer Dec 24 '11 at 16:40
When you say "roughly double", I gather you're figuring that much mass per wheel, so a total additional mass of 1g added to the wheels is equivalent to adding 4g to the non-rotating mass?  And does this take into account the energy required to accelerate the added mass linearly, or does that bump it up to 5:1 (assuming my first assumption is correct)? – Hot Licks Dec 24 '11 at 20:40
Ok, read through that again and I think I understand it -- it looks like you've got a factor of two for the two tires, and you've entered the tire mass both into the angular and linear components, so the final answer is 2x -- a gram of added weight at the radius of the tires is equivalent to 2 grams on the frame (or the rider), with all factors accounted for. – Hot Licks Dec 25 '11 at 15:14
Right.  The weight is either added to the rider or distributed evenly to both tires. – Jerry Schirmer Dec 26 '11 at 16:42
And a bit of a thought experiment confirms this:  It would take as much force to accelerate a mass on a stationary wheel to a given tangential velocity as it does to accelerate the mass to the same linear velocity.  But once accelerated to the tangential velocity it (of course) takes the same amount of force again to accelerate the (otherwise massless) wheel to the same linear velocity.  So to accelerate simultaneously to the same tangential and linear velocity takes twice as much force as either alone. – Hot Licks Dec 26 '11 at 19:04
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You have to put angular momentum into the spinning wheels.

The energy of a rotating object is = I w^2 /2

Where I is the moment of intertia, which for a ring is I = mr^2 /2
And w is the angular velocity in rad/s

Essentailly this is wasted energy since it has to be generated in addition to the1/2 m v^2 of the rider+bike.
And since to accelerate you need to increase the angular velocity quickly you have to put a lot of energy into the angular rotation quickly and since you are limited in how much power you can provide this limits the rate of change of 'w'

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