# Gear problem

Hi
A bit hypothetical but, assuming two identical gears, the driver has
100mm pitch circle and the driven has a 100mm pitch circle, they are
spaced at 100mm centres. The driven is attached to a rubber wheel.
Correctly meshed, for one revolution of one gear the other will turn one
revolution, which is a distance of 100mm x pi = 314.15mm
If the gears centres are then altered to 102mm without altering the
pitch of either (or anything else) will the driven gear travel further
or less per revolution? Is there a formula to work it out?
Providing the gears will still mesh you will still get the same travel as in 314.15mm
Provided you can keep backlash out of the equation then the distances will be the same.
-- Regards,
John Stevenson Nottingham, England.
What will happen is that you will have an effective PCD of 102mm, but the teeth will now be the wrong shape for that PCD, the effect of which is that you will hear more noise from the train (assuming it is transmitting power and not movement as with clocks).
The noise arises because of stress in the teeth, with the teeth themselves bending backwards and forwards.
This, incidentally is one of the reasons for using involute teeth in power trains - the broad base of each tooth gives additional strength compared to the cycloidal teeth used in horology.
The driver/driven gears will still turn in the same ratio as before.
It depends on what accuracy you're looking for. The AVERAGE ratio between a pair of spur gears is uniquely defined by the ratio of the numbers of teeth in the wheels and is independent of variations in centre distance or errors in tooth shape provided sequential tooth contact is maintained.
In your case the average ratio will remain at 1:1 and will not change as the centre distance changes.
If you are looking for an accurate instantaneous velocity ratio life gets more complicated. Perfectly cut involute teeth have the interesting property that small changes in centre distance slightly change the effective pressure angle but the velocity ratio remains unaltered. Cycloidal teeth (mainly used in clocks) do not have this property and will only produce constant velocity when meshed at the correct centre distance. Centre distance errors will produce velocity errors at tooth frequency.
In either case eccentricity will result in instantaneus velocity errors at rotation frequency. 1% eccentricity in one of the two gears will produce 1% velocity variation. Eccentricity in the mating gear has a similar effect but this may add or subtract to the driving gear error.
Jim
Thanks for all the replies
Hang on fellars! It depends on which bit of the gear you are considering. A point on the pitch circle, i.e at 100mm diameter, travels 314.15mm The tip of a tooth goes further and the axis goes nowhere. What happened to the rubber wheel? I'll hazard a guess that it is driving something. It will rotate once for every turn of the driver. How far its rim will travel depends on its diameter. e.g. If 100 mm then it travels 314.15mm.
Cheers
Henry
"transmitting power and not movement as with clocks"
Is this some special physical process you have invented? Is it reliant on some special rare lubricant, perhaps made from the hide of virgin sheep? Is this lubricant Down and Out Systems latest product? Do tell us. When you have, maybe you can also revisit NMR and nuclear emissions.
That is just "colloquial usage", the sort of thing someone who doesn't really understand physics would say.
-- 73 Brian G8OSN
What's up, OM?
My complaint to your ISP causing a bitter taste and a fit of the sulks?
Bwahahahahahahahahahahahahahahahahaaaaaaa!!!!!
That is what I thought. One of those pseudo scientific pompous idiots who are no more than out of work tecnhicians who aspire to be engineers.
"Airy R. Bean, village idiot" wrote her usual load of crap in message news:c658ih\$882sg\$ snipped-for-privacy@ID-217727.news.uni-berlin.de...
Is it feeding time, or do you need your nappy changed? ...(_!_)...
Clearly Bean does not appreciate that a mechanical clock consumes energy in order for it to function! Amazing!

Regards,
Pete Keillor
In message , Gareth S Nemisis writes
Nah! It is a well known fact that clocks are driven by time. Any springs, weights or other apparent driving source are put there to fool the over-curious. A bit like God putting fossils in the rocks as a practical joke against Darwinists.
Read Richard Dawkins 'The Blind Watchmaker' if you don't believe me.
OK, time to come clean :-) The driven gear is attached to the back of a plastic wheel with a rubber tyre of about 200mm dia. The driver is attached to an electric motor. The driver is engaged with the driven gear by means of a very rudimentry hand lever with no fixed positon to marry the two pitch circles together, therefore the teeth can be engaged anywhere from the top of the tooth to the bottom of the tooth. This is what prompted the original post, I assumed that if the teeth were only just contacting on their tops then the contact would be on a larger diameter than that of the correct pitch circle thus making the gears travel a greater distance for one revolution than if they were meshing correctly at their respective pitch circles.
It would make no difference in the travel of the rubber tyre if the teeth just meshed or fully meshed. Only more wear and noise will be introduced when not correctly meshed.
The distance travelled with be a product of the teeth ratio between driver and driven gears and of the tyre diameter
(T1/T2) x pi D
Correct - 10/10
(assuming it is
Complete crap - 0/10 (How does the power get from the spring/weight to the pendulum/balance?)
Partially true but mainly crap - 2/10 (The noise mainly emanates from the contact of tooth on tooth which is minimised if the tooth 'depthing' is correct).
An almost completely misinformed opinion - 1/10 (The tooth base is dependent upon module or DP and in most cases cycloidal and involute have similar base dimensions - although cycloidal *could* have a narrower base. You will also find that involute tooth forms *are* used in modern horology, partly because the depthing is less critical and is therefore suited to mass-production methods)
Correct - 10/10
That's only 23 out of 50 - Must do better!
I think that, perhaps, you need to adopt a mature attitude to debate rather than the rather silly and infantile stance that you exhibit below.
Grow up, OM.
The purpose of the going train in a clock is to transmit movement and not power. The going train is stationary for most of the time, and the concept of power, as a rate of doing work is meaningless.
If you want to show yourself up to be a childish fool by arguing that some energy must come from the spring or the weights, then that is your prerogative.
Stupid Boy.
If you want to show yourself up to be a childish fool by arguing that malformed teeth are not subject to bending stresses, then that is your prerogative; text books on gear theory disagree with you.
Stupid Boy.
Bwaaaaahahahahahahahahahahahahahahahah!
What utter rubbish you utter. Maybe you might help us all out by telling us what the SI unit for the transmission of movement might be? The Bean perhaps?
It has clearly escaped (sic) your attention that the primary purpose of the going train is, as the name usefully prompts you, to keep the clock "going" - i.e., to supply the energy to the pendulum that is lost through friction, viscous drag, hysteresis in the suspension spring, .... etc., and thereby to keep the pendulum swinging at a (near) constant frequency. Remarkably, it supplies this energy at the same (average) rate as it is dissipated. As your elementary physics will tell you, power is energy per unit time. So the train very clearly transmits power.
Whether or not the transmission of power is continuous or discontinuous is massively irrelevant.
The transmission of movement (to the hands) is an incidental convenience - it saves the user from the significant inconvenience of having to count pendulum oscillations himself in order to tell the time.
Regards, Tony

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