# Gears - RPM x No of teeth

Icetray
Hi guys,

I was doing my tutorial and given the angular velocity of the first gear, we were required to calculate the angular velocity of the 8th gear given just the no. of teeth for all the gears.

Now I know that there must be a faster way of calculating the angular velocity of the 8th gear then solving for the angular velocity each gear from 1 - 8. (i.e. using the speed ratio formula $\frac{N1}{N2}$ = $\frac{rpm2}{rpm1}$)

Question 1: Is there a faster away? (I think there's something like all the gears must move at the same tangential velocity or something right?)

I also noticed that when you multiplied the rpm of each gear by the number of teeth, you get a fixed number (lets call it x) for all the gears. This means that to find the rpm of the 8th gear all I have to do is ($\frac{x}{N8}$.

Question 2: What is this relationship? Is it an actual relationship or does it just happen to be a unique thing for the question that I am doing?

voko
The number of teeth is proportional to the circumference. Circumference L = dN, where d is the arc length due to one tooth. So when you multiply angular velocity by N, you get a value proportional to tangential velocity, because tangential velocity is the product of angular velocity and circumference L. As you observed, the tangential velocity of two connected gears must be equal (otherwise they would be slipping at the point of contact, thus breaking their teeth). That is why you get the constant. So if you have a number of interconnected gears, you just need to compute this constant given the RPM and teeth of one gear, and the RPM of any other gear can be obtained by dividing the constant by its number of teeth.

One caveat, though. They must all be interconnected through their teeth, not through anything else, like a common axle.

Icetray
The number of teeth is proportional to the circumference. Circumference L = dN, where d is the arc length due to one tooth. So when you multiply angular velocity by N, you get a value proportional to tangential velocity, because tangential velocity is the product of angular velocity and circumference L. As you observed, the tangential velocity of two connected gears must be equal (otherwise they would be slipping at the point of contact, thus breaking their teeth). That is why you get the constant. So if you have a number of interconnected gears, you just need to compute this constant given the RPM and teeth of one gear, and the RPM of any other gear can be obtained by dividing the constant by its number of teeth.

One caveat, though. They must all be interconnected through their teeth, not through anything else, like a common axle.

Thanks for the reply voko! Anyways is there a special name for this proportionate value?

I was attempting another problem and this time it's on planetary train gears but I'm a little lost. This is the question: I'm managed to solve for the ring gear using:
$\frac{nr - nc}{ns-nc}$ = $\frac{Ns Np2}{Np1 Nr}$

and I got nr = 266 2/3 RPM (can anyone help me confirm this? My tutorial comes with no answer ): )

Assuming my first part is correct, what do I do next to find the angular velocity of the other gears?

Homework Helper
Gold Member
Hi guys,

I was doing my tutorial and given the angular velocity of the first gear, we were required to calculate the angular velocity of the 8th gear given just the no. of teeth for all the gears.

Now I know that there must be a faster way of calculating the angular velocity of the 8th gear then solving for the angular velocity each gear from 1 - 8. (i.e. using the speed ratio formula $\frac{N1}{N2}$ = $\frac{rpm2}{rpm1}$)

Question 1: Is there a faster away?

Skip the ones in the middle...

N1/N2 * N2/N3 * N3/N4 * N4/N5 * N5/N6 * N6/N7 * N7/N8 = N1/N8

voko
The product of gear speed and number of teeth is sometimes called gear mesh frequency.

Regarding your problem, I think it could be analysed easier if we look at in the reference frame whose origin is at the axis and that is rotating together with the carrier. In this frame $n_S' = -n_C$, $n_R' = n_R - n_C$ where the primed speeds are in the rotating frame. Then $n_R' = \frac {n_S' N_S} {N_R}$, and $n_R = n_R' + n_C = \frac {n_S' N_S} {N_R} + n_C = \frac {-n_C N_S} {N_R} + n_C = n_C (1 - \frac {N_S} {N_R})$.

Icetray
The product of gear speed and number of teeth is sometimes called gear mesh frequency.

Regarding your problem, I think it could be analysed easier if we look at in the reference frame whose origin is at the axis and that is rotating together with the carrier. In this frame $n_S' = -n_C$, $n_R' = n_R - n_C$ where the primed speeds are in the rotating frame. Then $n_R' = \frac {n_S' N_S} {N_R}$, and $n_R = n_R' + n_C = \frac {n_S' N_S} {N_R} + n_C = \frac {-n_C N_S} {N_R} + n_C = n_C (1 - \frac {N_S} {N_R})$.

First off, thank you for the replies guys! And I apologize for thelate reply.

Hmmm... so Veko, what you're saying is that for the internal gears (everything minus the ring gears), we can take their relative velocities and use them to solve using the spur gear formulas?

i.e. $\frac{ns'}{np1'}$=$\frac{Np1}{Ns}$

Then np1 = np1' + ns'?

voko
For the first planet, I would consider that for one round of the carrier, it has to go over $N_s$ teeth of the sun. So its speed should be $n_c \frac {N_s} {N_p1}$.

If you look at the problem in the co-rotating frame, then $N_{p1}' = - N_s' \frac {N_s} {N_p1}$, which yields the same result.

Icetray
For the first planet, I would consider that for one round of the carrier, it has to go over $N_s$ teeth of the sun. So its speed should be $n_c \frac {N_s} {N_p1}$.

If you look at the problem in the co-rotating frame, then $N_{p1}' = - N_s' \frac {N_s} {N_p1}$, which yields the same result.

I managed to solve for the speeds of nr and np2 but I'm having issues with finding the speed of np2 now.

This is what I did (ccw +ve):

$\frac{np2 - nc}{ns - nc}$ = (- $\frac{Ns}{Np1}$)(-$\frac{Np1}{Np2}$

$\frac{np2 +200}{0 + 200}$ = (- $\frac{50}{25}$)(-$\frac{25}{20}$

I get np2 = 300 rpm (ccw) but the answer given is 700rpm (ccw).

voko
The speed of the second planet is very simply related to that of the first planet, as they are fixed with respect to each other. So $$n_{p2} = -n_{p1} \frac {N_{p1}}{N_{p2}} = -n_c \frac {N_s} {N_{p2}}$$

Icetray
The speed of the second planet is very simply related to that of the first planet, as they are fixed with respect to each other. So $$n_{p2} = -n_{p1} \frac {N_{p1}}{N_{p2}} = -n_c \frac {N_s} {N_{p2}}$$

Thanks for the quick reply but I still don't get the answer with your method. ): Anyways, can I ask if the method I used is wrong? Can I set the sun as my input and the planet 2 as my output or this that wrong? ):

voko
I am not sure what method is "mine" and what method you used.

Icetray
$$n_{p2} = -n_{p1} \frac {N_{p1}}{N_{p2}} = -n_c \frac {N_s} {N_{p2}}$$

Oh I was referring to this when I said your method and this:
$\frac{np2 - nc}{ns - nc}$ = (- $\frac{Ns}{Np1}$)(-$\frac{Np1}{Np2}$

When I was referring to my method.

Was my approach wrong? Sorry for asking so many questions!

voko
Perhaps you need to explain how you get your result. It is not obvious for me.