Gentlemen, I need help. Attaching the photograph of a geared / belt driven drive mechanism.

Please study the attached photograph carefully. As per my reckoning, the diameter of the drive (smaller) gear is approximately the radius of the driven (bigger) gear. If the smaller gear spins at 10,000 RPM, at what RPM will the larger gear spin? Intelligent Estimates / Guesstimates will be appreciated.

As per my reckoning, it should be around 2500 RPM.

Would appreciate a prompt reply.

Regards.

p.s. I miss Sandy now.

the RPM of larger gear would be 5000RPM

2pi R /2pi 2R =0.5

Your transmission ratio should be given my the ratio of number of teeth on the driven to number of teeth on the driving pulley.

EDIT: You can use the formula, (n1)/(n2)=(d2)/(d1)

Where n1, n2= Rotational speed of driving pulley and driven pulley respectively and d1, d2 is the diameter of the driving and driven pulleys respectively

Or you can use basic math like "rastsaurabh" and Sundaram sir.

5000

RPMx3.14x2xD=10000x3.14xD

RPM=10000x3.14xD/3.14x2xD=5000

My 4999th post here - a milestone on this forum, for anybody.

Sandy old friend, Saurabh -  

Iamahuman - It is belt driven so the number of teeth is unavailable and redundant.

You did say "geared/belt drive" which led me to think of something like this.

Ironical that my 5000th post on this forum would concern the number 5000. So be it.

I am taking the above data to be sacrosanct as prescribed by the pundits. At whatever RPM the DRIVE gear runs, it will generate half the revs on the DRIVEN gear, provided that the diameters of the two gears are 1:2. Therefore, at a 1: 4 ratio, I can hope for a 2500 RPM if the drive gear spins at 10,000 RPM. Right?

iamahuman - thanks buddy, I had mentioned geared / belt driven and NOT geared belt driven.  Grateful to you for having taken all the trouble for me. Thanks!

Thank you www.rcindia.org, thank you Anwar Sahab.

At whatever RPM the DRIVE gear runs, it will generate half the revs on the DRIVEN gear, provided that the diameters of the two gears are 1:2  Provided the module of the two gears is the same. 

Congratulations on 5000 posts!  You have now spent 121 days, 23 hours and 18 minutes online 

Answers are correct, the bigger gear will run half the speed of the smaller one. The maths goes like this:

1. Assuming the belt is non extendable and there is no slipping, in a given time time t it will travel same distance x over both the pulleys (gears).
2. Lets translate this x into circular movement, x = beta*R1 = alpha*R2 where beta is angle moved by smaller and alpha is angle by bigger, R1 and R2 corresponding radii.
3. Divide it with the time t, (beta/t) * R1 = (alpha/t) * R2
=> RPM_smaller * R_smaller = RPM_bigger * R_bigger

I guess this sketch may explain it better. (sorry for poor handwriting, been ages since I held a pen)



Also, when talking about circular motion, always use consistent terms. Never mix up radius and diameter, it is the easiest way to land into "divided by two" error.

Hope this helped.

 

was in not easier to say circumference of one gear multiplied by its RPM is equal to circumference of second gear multiplied by its RPM assuming there is no slippage in the gear or belt   

lol yeah, it is always easier to see distance and have a "ohyeah" moment and then come to speed from it. I guess I did complicate it though. 

actually in case of v belt pulleys - just
compare the dia and get the dividing / or multiplying factor.

I feel I've been thoroughly educated. Thanks everyone.

Quote from: VC on November 27, 2013, 11:47:00 PM
p.s. I miss Sandy now.

VC dada I missed that in first read.  . I am deeply honored. Thanks.

I am always available to you.  

Note: I never knew about the "/ m e" (without spaces) tag that was available in the forum software... example above.

 
"/me"

* this is a star in red text

Apart from * and text highlight to red, the purpose of tag?

It is from IRC where you use "/me <status>" for  acknowledgements. If you use Skype you can use /me command there too. It is used to write sort of things that you want to broadcast to the participating members but do not expect an answer to it. It may be a thank you note or a general commentary.

I am a dinosaur, no, a fossil.

386 in the world of i7

include me too!!!!

  Calculator

Sorry, didn't mean to hurt someone. I just go somewhat longer when explaining. 

Quote from: VC on November 28, 2013, 12:18:04 AM
Ironical that my 5000th post

after 5000 post, ones post count reduces....that is ironical...until it reaches 'Shunya' Then ...you have reached Nirvana/Moksha 

That is because a member with multiple threads and posts got deleted, which means some of the threads he started would also get deleted taking the post count of others who had participated in those threads.

Consider this a one off thing, hopefully should not happen again.  Such people are "karoron mein ek"