# Two Stage Epicyclic Gear-box - Finding the Gear ratio

• Engineering
• Master1022
In summary, the conversation discusses a question about the method for attempting a two-stage epicyclic gear-box. The solution involves considering the linear speeds of the gears and dividing out the final radius at the end to get the final ωout. The mistake made by the asker was assuming that ωout = ωS2, when in fact it is determined by both S2 and A2 through P2. The solution presented involves setting up simultaneous equations and solving for unknowns to find the gear ratio.
Master1022
Homework Statement
Find the gear ratio when Brake 1 is applied, the clutch is disengaged and Brake 2 is
released.
Relevant Equations
$$\frac{\omega_1}{\omega_2} = \frac{r_2}{r_1} = \frac{N_2}{N_1}$$
Hi,

Just have a question about the method in attempting this two-stage epicyclic gear-box.

We are given the following information about the number of teeth: S1 = 48, P1 = 28, S2 = 96, and P2 = 48

My attempt:
1. Work out the number of teeth for the two ring/annulus gears

From geometry, we can find that A1 = 104, A2 = 192

2. Find the ratio between the input and output (gear ratio)
So the input is $\omega_{S_1} = \omega_{S_2} = \omega_{in}$ and $\omega_{out} = \omega_{S_2}$

Then I will start the method of considering motion with the carrier and then relative to the carrier

- I considered everything relative to C2

Here is where I went wrong and I am looking for the reason why:
- Then we know that the total $\omega_{A1} = 0$ as brake 1 is applied, thus I set the motion of A1 relative to C2 as $-\omega_{C_2}$ (so that it sums to 0)
Why was it wrong for me to consider both stages at once (i.e. by relating C2 and A1 directly as I have done)?
- From there, I just used the gear equations to get that $\omega_{S} = \omega_{C_2} \left( 1 + \frac{N_{A_1}}{N_{S_1}} \right)$. From there we can simply divide out the speed of the carrier (output) and get the gear ratio.

The solution considered the whole gearbox in two stages and did:
(overview: set up some simultaneous equations and solve for unknowns)
1. Consider speed of second stage relative to C2
2. Set an arbitrary speed of A2 relative to C2 (we now have a total of $\omega_{C_2} + \omega_{Q}$)
3. Now consider the first stage of the gearbox (the speed of C1 is the same as A2)
4. Now we can use the A1 has no speed condition to get relative speed of A1
5. Then we can get relative and total speeds of P1 and S1
(now we have an equation for S1 in terms of $\omega_{C_2}$ and $\omega_{Q}$)
6. To find $\omega_{Q}$, we once again consider the second stage of the gearbox and use relative motion of A2 to work our way down to an equation for S2 in terms of $\omega_{C_2}$ and $\omega_{Q}$
7. Set S1 = S2 and solve for $\omega_{Q}$ in terms of $\omega_{C_2}$
8. Substitute it back in get our ratio.

So I understand this method, but am wondering why I cannot directly consider speed of A1 relative to C2?

I realize that I have mainly provided a qualitative description of the method, but I don't really know how to write it out as the method just involves filling in tables - I will provide any clarification that is required,

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Are you still seeking an answer for this? I see that it was posted some weeks ago.

I worked this out from first principles and came up with ωout = ##\frac{31}{57}##ωin.

Your mistake was in the assumption that ωout = ωS2. The annular gear A2 is turning as a result of the movement from the first stage, and so C2 will have a movement that is determined from both S2 and A2 through P2. To solve the problem, I found it easier to consider linear speeds rather than angular speeds, and then divide out the final radius at the end to get the final ωout.

So plate C1 is moving at a linear speed (at the points where it connects to the axles of P1) that is halfway between the at-rest A1 and the turning S1, with the P1 spinning between them. The resulting ωC1 (which does equal ωA2 as you indicated), is that linear speed divided by the radius at the axis of P1. That gives us ωA2 = ##\frac{24}{76}##ωin. We then have C2 at the axle of the P2 gears moving at a linear speed that is halfway between 96ωin, from S2, and 192ωA2, from A2. Dividing that result by the radius of 144 at the P2 axle for C2 yields the ωout that I calculated above.

Hope this helps!

Last edited:
Master1022
@anothermike - thank you so much for taking the time to reply and help me out! I apologise for not responding for so long, but I have taken such a long break from the subject (so I just started remembering the concepts now whilst getting back into it).

anothermike said:
I worked this out from first principles and came up with ωout = ##\frac{31}{57}##ωin.

Your mistake was in the assumption that ωout = ωS2.

Yes, that is true, I am not sure why I thought that.

anothermike said:
The annular gear A2 is turning as a result of the movement from the first stage, and so C2 will have a movement that is determined from both S2 and A2 through P2. To solve the problem, I found it easier to consider linear speeds rather than angular speeds, and then divide out the final radius at the end to get the final ωout.

Just want to confirm, but is the carrier velocity (when in the standard epicyclic gear train) always the average of the two surrounding velocities? This makes sense intuitively as the velocity should vary linearly across the planetary gear.

anothermike said:
So plate C1 is moving at a linear speed (at the points where it connects to the axles of P1) that is halfway between the at-rest A1 and the turning S1, with the P1 spinning between them. The resulting ωC1 (which does equal ωA2 as you indicated), is that linear speed divided by the radius at the axis of P1. That gives us ωA2 = ##\frac{24}{76}##ωin. We then have C2 at the axle of the P2 gears moving at a linear speed that is halfway between 96ωin, from S2, and 192ωA2, from A2. Dividing that result by the radius of 144 at the P2 axle for C2 yields the ωout that I calculated above.

Hope this helps!

Yes, thank you very much! Your method is so much faster than mine - I hadn't ever thought about it this way! I will be sure to use this method going forward.

In case anyone else reads this, I did retry the problem and use a tabular method twice. Once was considering all of the '1' gears relative to C1. That yielded an expression for $\omega_{S_1} = f(\omega_{C_1})$. Then I used a second tabular method to analyse the '2' gears relative to C2 - an extra variable needs to be introduced in this part to form a system of equations. You can then get an algebraic solution from there which yields the same answer as @anothermike 's post. However, @anothermike 's method is so much simpler.

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anothermike said:
Your mistake was in the assumption that ωout = ωS2. The annular gear A2 is turning as a result of the movement from the first stage, and so C2 will have a movement that is determined from both S2 and A2 through P2. To solve the problem, I found it easier to consider linear speeds rather than angular speeds, and then divide out the final radius at the end to get the final ωout.

I think I have seen a method similar to this before (i.e. where we consider velocity profiles). Is this method only applicable in situations where there is a ring gear present? For example, I have seen some problems where there is just a sun gear and a planet gear and was unable to see how this method could be applied.

Master1022 said:
Just want to confirm, but is the carrier velocity (when in the standard epicyclic gear train) always the average of the two surrounding velocities? This makes sense intuitively as the velocity should vary linearly across the planetary gear.

Yes, it is. If you think about it, the rim of the carrier gear is traveling with a constant speed relative to its axis, and where it makes contact with the other surrounding gears, one velocity is $v_c + \omega r$, and the other is $v_c - \omega r$. It thereby follows that $v_c$ would be the average of the two.

Last edited:
Master1022
Master1022 said:
I think I have seen a method similar to this before (i.e. where we consider velocity profiles). Is this method only applicable in situations where there is a ring gear present? For example, I have seen some problems where there is just a sun gear and a planet gear and was unable to see how this method could be applied.
I believe this method would be applicable in all situations, though perhaps unnecessary. If you post one of those sun gear and planet gear problems, I'll take a look and see if I can apply it.

Master1022

## 1. What is a two stage epicyclic gear-box?

A two stage epicyclic gear-box is a type of gear system that uses two sets of gears to transfer power from one source to another. It consists of an outer ring gear, an inner sun gear, and multiple planet gears that rotate around the sun gear. This design allows for a compact and efficient gear system with high gear ratios.

## 2. How does a two stage epicyclic gear-box work?

In a two stage epicyclic gear-box, the outer ring gear is fixed while the sun gear and planet gears rotate. As the planet gears rotate, they also rotate around the sun gear, creating a gear reduction between the input and output shafts. The gear ratio is determined by the number of teeth on each gear and the arrangement of the gears within the gear-box.

## 3. What is the gear ratio of a two stage epicyclic gear-box?

The gear ratio of a two stage epicyclic gear-box is determined by the number of teeth on the sun gear, planet gears, and outer ring gear. The gear ratio can be calculated by dividing the number of teeth on the sun gear by the number of teeth on the planet gears, and then multiplying that result by the number of teeth on the outer ring gear.

## 4. How do you find the gear ratio of a two stage epicyclic gear-box?

The gear ratio of a two stage epicyclic gear-box can be found by counting the number of teeth on each gear and using the formula mentioned in the previous question. Alternatively, the gear ratio can also be determined by measuring the speed of the input and output shafts and calculating the ratio between them.

## 5. What are the advantages of a two stage epicyclic gear-box?

Two stage epicyclic gear-boxes offer several advantages over other types of gear systems. They are compact and can provide high gear ratios, making them suitable for use in a variety of applications. They also have low noise levels and high efficiency, making them ideal for use in precision machinery. Additionally, their design allows for smooth and continuous power transfer, reducing wear and tear on the gears.

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