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.
This result, e.g. an ounce of weight at the rims is like adding two ounces of frame weight is true regardless of the mass of the cyclist/cycle, wheel radius or rider torque. To see this, note that
$$\begin{align*}
\frac{da}{dm} &=\frac{-8\tau}{R(4m+M)^2}\\
\frac{da}{dM} &=\frac{-2\tau}{R(4m+M)^2}
\end{align*}$$
Adding a small amount of mass $\delta m$ to the frame changes the acceleration by $\delta m \frac{da}{dM}$ while adding half this amount to each of the wheels changes the acceleration by $\frac{1}{2} \delta m \frac{da}{dm}$. The ratio of acceleration changes is
$$\frac{\frac{1}{2} \frac{da}{dm}}{\frac{da}{dM}} = 2$$
regardless of the other parameter values. It's not hard to see that this result is true for unicycles and trikes as well (i.e. doesn't depend on the number of wheels on the cycle).
Best Answer
Aha, I got it. :) Most resources on bicycle physics dismiss gyroscopic effects because those are too weak.
Here, we need them:
The wheels turn with angular velocity $\omega = v/r$ and hence have an angular momentum which we approximately can write as $L = mr^2 \omega$. In cylindrical coordinates, $L$ has a component in $z$-direction which stays constant in time, and a component in $r$ direction which changes with time. Let us look at the radial component:
$\vec{L}_r = mr^2 \omega \cos \theta \vec{e}_r$. The change of it over time involves a torque: $$\vec{T} = mr^2 \omega \Omega \cos \theta \vec{e}_r$$ where $\Omega = v/R$ is the angular frequency of the overall circular path. Using this, and $r\omega = v$, we obtain $$\vec{T} = mr \frac{v^2}{R} \cos \theta \vec{e}_r$$
Hence, for the total torque to be $0$, we need $$MgL \sin \theta = \frac{v^2}{R}\left( 2mr + ML\right) \cos \theta$$ which immediately yields the desired result.