Derivation of Wheel Relationship Formulas

In summary, this thread discusses the formulas for the relationship between two wheels, whether they are teethed or non-teethed. There are three cases of wheel relationships: two wheels at the same axle, two wheels intersected in parallel, and two wheels connected by a belt. The formula for the first case is ##\omega_A = \omega_B = \frac{v_A}{r_A} = \frac{v_B}{r_B}##. For the second case, the formula is ##v_A = v_B = \omega_A \cdot r_A = \omega_B \cdot r_B##. And for the third case, the formula is the same as the second case. The thread also discusses the relationship involving teethed
  • #1
bagasme
79
9
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

1582613635901.png


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

1582615070140.png


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

1582615692073.png


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
 
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  • #2
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.
 
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  • #3
bagasme said:
... 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.
 
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  • #4
kuruman said:
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##.
 
  • #5
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:

$$\begin{align}
\omega_A \cdot r_A &= \omega_B \cdot r_B \nonumber \\
\frac {\omega_A \cdot r_A} {N_A} &= \frac {\omega_B \cdot r_B} {N_B} \nonumber \\
\dfrac {\dfrac {2 \cdot \pi \cdot r_A} {t}} {N_A} &= \dfrac {\dfrac {2 \cdot \pi \cdot r_B} {t}} {N_B} \nonumber \\
\frac {r_A} {N_A} &= \frac {r_B} {n_B} ~(\text {cancel} ~ \dfrac {2 \cdot \pi} {t} ~ \text {on both sides}) \nonumber \\
r_A \cdot N_B &= r_B \cdot N_A \nonumber \\
\frac {r_A} {r_B} &= \frac {N_A} {N_B} ~ (3) \nonumber \\
\end{align}$$
 
  • #6
bagasme said:
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.
 
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Related to Derivation of Wheel Relationship Formulas

1. What is the purpose of deriving wheel relationship formulas?

The purpose of deriving wheel relationship formulas is to mathematically describe the relationship between the different variables involved in wheel motion, such as wheel radius, speed, and torque. This can help in understanding and predicting the behavior of wheels in various scenarios, such as in vehicles or machinery.

2. What are the key variables involved in wheel relationship formulas?

The key variables involved in wheel relationship formulas are wheel radius, speed, torque, and force. These variables are interrelated and their values can be used to calculate various other quantities, such as power and acceleration.

3. How are wheel relationship formulas derived?

Wheel relationship formulas are derived using principles of physics, such as Newton's laws of motion and the concept of rotational motion. These principles are applied to the specific geometry and mechanics of a wheel to develop the formulas.

4. What are the applications of wheel relationship formulas?

Wheel relationship formulas have various applications in engineering, particularly in the design and analysis of vehicles and machinery. They can also be used in robotics, physics experiments, and other fields that involve rotational motion.

5. Are there different types of wheel relationship formulas?

Yes, there are different types of wheel relationship formulas depending on the specific scenario and variables involved. For example, there are formulas for calculating the speed of a wheel based on its radius and angular velocity, or for determining the torque required to move a wheel at a certain speed.

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