Derivation of Wheel Relationship Formulas

[bagasme]

Hello,

Here in this thread I will derive formulas for relation between two wheels, either teethed (e.g. gears) or non-teethed.

In wheel relationship, we have three cases:

1. Two wheels at the same axle
2. Two wheels intersected in parallel (meshed)
3. Two wheels connected by a belt

We will examine each cases above on smooth (non-teethed) wheels first.

1. At the same axle

Figure 1 (source: fisikabc.com) shows two wheels connected to the same axle. Since both $A$ and $B$ are on same center, they have same angular accelleration:

$$\begin{align}
\omega_A & = \omega_B \nonumber \\
\frac {v_A} {r_A} & = \frac {v_B} {r_B} ~ (1)\nonumber
\end{align}$$

2. Intersected in Parallel

Figure 2 (source: fisikabc.com) show two wheels intersected each other in parallel. When wheel $A$ rotated counter-clockwise, wheel $B$ rotated clockwise, and vice versa. At the intersection point between two wheels, linear velocity of both wheels directed at the same direction, so:

$$\begin{align}
v_A & = v_B \nonumber \\
\omega_A \cdot r_A &= \omega_B \cdot r_B ~ (2)\nonumber
\end{align}$$

3. Connected by Belt

Figure 3 (source: fisikabc.com) shows two wheels connected by a belt. As wheel $B$ rotated, the belt rotated wheel $A$ at the same linear velocity direction as wheel $B$. Thus the formula is same as when both wheels intersected directly (2).

Relationship Involving Teethed Wheels (e.g. Gears)

We will now deriving equations that relate between angular velocity, radius, and number of teeth. $N_A$ and $N_B$ are number of teeth of wheel A and B, respectively.

From equation (2):

$$\begin{align}
\omega_A \cdot r_A & = \omega_B \cdot r_B \nonumber \\
\frac {2 \cdot \pi \cdot r_A} {t} & = \frac {2 \cdot \pi \cdot r_B} {t} \nonumber \\
r_A &= r_B \nonumber \\
\frac {r_A} {n_A} &= \frac {r_B} {n_B} \nonumber \\
r_A \cdot N_B &= r_B \cdot N_A \nonumber \\
\frac {r_A} {r_B} &= \frac {N_A} {N_B} ~ (3) \nonumber
\end{align}$$

Combining equation (2) and (3):

$$\begin{align}
\omega_A \cdot r_A & = \omega_B \cdot r_B \nonumber \\
\frac {\omega_A} {\omega_B} &= \frac {r_B} {r_A} \nonumber \\
\frac {\omega_A} {\omega_B} &= \frac {N_B} {N_A} \nonumber \\
\end{align}$$

We can conclude that angular accelleration is related to radius, but inverse related to number of teeth.

I hope that this thread will be useful. Please let me know any flaws and suggestions.

Cheers, Bagas

 

[kuruman]

When you say $\dfrac{2\pi r_A}{t}=\dfrac{2\pi r_B}{t}$, you are assuming that the two geared wheels have the same radius. This means that $N_A=N_B$ and $\omega_A=\omega_B$. Not an interesting case. What you want to say is that the number of teeth per unit length in A must be the same as in B otherwise the teeth will not mesh.

 

[Lnewqban]

... Figure 1 (source: fisikabc.com) shows two wheels connected to the same axle. Since both $A$ and $B$ are on same center, they have same angular acceleration velocity:

...

We can conclude that angular accelleration velocity is related to radius, but inverse related to number of teeth.

I hope that this thread will be useful. Please let me know any flaws and suggestions.

Thank you, Bagas.
I believe you meant angular velocity in both cases.

 

[bagasme]

When you say $\dfrac{2\pi r_A}{t}=\dfrac{2\pi r_B}{t}$, you are assuming that the two geared wheels have the same radius. This means that $N_A=N_B$ and $\omega_A=\omega_B$. Not an interesting case. What you want to say is that the number of teeth per unit length in A must be the same as in B otherwise the teeth will not mesh.

What I mean was cancelling $\frac {2\pi} {t}$ from both sides, so $r_A$ is not necessarily same as $r_B$.

 

[bagasme]

Since I can't edit my first post here, let me update derivation of (3).

From equation (2), we divide each sides by number of teeth of each wheels:

$$\begin{align}
\omega_A \cdot r_A &= \omega_B \cdot r_B \nonumber \\
\frac {\omega_A \cdot r_A} {N_A} &= \frac {\omega_B \cdot r_B} {N_B} \nonumber \\
\dfrac {\dfrac {2 \cdot \pi \cdot r_A} {t}} {N_A} &= \dfrac {\dfrac {2 \cdot \pi \cdot r_B} {t}} {N_B} \nonumber \\
\frac {r_A} {N_A} &= \frac {r_B} {n_B} ~(\text {cancel} ~ \dfrac {2 \cdot \pi} {t} ~ \text {on both sides}) \nonumber \\
r_A \cdot N_B &= r_B \cdot N_A \nonumber \\
\frac {r_A} {r_B} &= \frac {N_A} {N_B} ~ (3) \nonumber \\
\end{align}$$

 

[kuruman]

Since I can't edit my first post here, let me update derivation of (3).

From equation (2), we divide each sides by number of teeth of each wheels:

$$\begin{align}
\omega_A \cdot r_A &= \omega_B \cdot r_B \nonumber \\
\frac {\omega_A \cdot r_A} {N_A} &= \frac {\omega_B \cdot r_B} {N_B} \nonumber \\
\dfrac {\dfrac {2 \cdot \pi \cdot r_A} {t}} {N_A} &= \dfrac {\dfrac {2 \cdot \pi \cdot r_B} {t}} {N_B} \nonumber \\
\frac {r_A} {N_A} &= \frac {r_B} {n_B} ~(\text {cancel} ~ \dfrac {2 \cdot \pi} {t} ~ \text {on both sides}) \nonumber \\
r_A \cdot N_B &= r_B \cdot N_A \nonumber \\
\frac {r_A} {r_B} &= \frac {N_A} {N_B} ~ (3) \nonumber \\
\end{align}$$

Not better. Look at the second equation. You get it from the first by dividing the left side by $N_A$ and the right side by $N_B$. That's OK as long as $N_A=N_B$ which means that equation (3) says $\dfrac{r_A}{r_B}=1$. You have all the pieces but you are not sure how to put them together and you are going around in circles. You did not pay attention to what I suggested in post #2: What you want to say is that the number of teeth per unit length in A must be the same as in B otherwise the teeth will not mesh. Start from there.