# Gears and shafts problem

1. Oct 29, 2014

### mattjones

1. The problem statement, all variables and given/known data

A gear train is to be designed to transmit power from shaft 1 to shaft 2 with the following condition:

Shaft 1 (driven by an electric motor) rotates at 100 revs min–1.

Shaft 2 is to rotate at 400 revs min–1 in the opposite direction to shaft 1 against a load of 200 Nm.

All shafts have a frictional resistance of 5 Nm.

Question: Determine the input power required at shaft 1

The gear train I have put together consists of a drive gear, two idlers and the output gear in a linear form.

Can someone point me in the right direction, any help would be appreciated!

Thanks

Matt

2. Oct 29, 2014

### OldEngr63

Just to be clear, Matt, you used 2 idlers, but no compound gears? It might help if you could post a sketch of your train to make completely clear what you have.

3. Oct 30, 2014

### billy_joule

The problem statement does not say anything about the input and output shaft arrangement so there is no need for idler gears to get the input and output shafts coaxial. you only need two gears - one on each shaft.

calculate input power by working back from the output power. I'd assume gearing losses are ignored.
The trickiest part is dealing with the 5 Nm friction for each shaft - it will give a different power loss for each shaft as their rpm is different.

4. Oct 30, 2014

### OldEngr63

As billy_joule has observed, there is no need for idler gears, but the OP says (first post) that he has used two idler gears. This indicates additional shafts and therefore more friction power losses. This is why I asked for clarification regarding idlers versus compound gears.

5. Nov 3, 2014

### mattjones

Thanks for the replies, here is the question in full:

A gear train is to be designed to transmit power from shaft 1 to shaft 2 with the following condition:

(i) Shaft 1 (driven by an electric motor) rotates at 100 revs min–1.

(ii) Shaft 2 is to rotate at 400 revs min–1 in the opposite direction to shaft 1 against a load of 200 Nm.

(iii) The centre of shaft 1 is 300 mm from shaft 2.

(iv) The minimum number of teeth on any gear is 15, all gears must have a multiple of 5 teeth and have a module of 2 mm.

(v) The maximum number of gears permissible is 4 gears and the diameter of the maximum gear must be minimised.

(vi) The centres of the gears should lie on a line which is as close to straight as possible.

(vii) All shafts have a frictional resistance of 5 Nm. Carry out the following:

(a) Design a gear train which satisfies the above criteria. Use a sketch to illustrate your design, label the gears A, B, C, etc. from the driver to the driven gear, and state the number of teeth on each gear.
(b) Determine the input power required at shaft 1.
(c) Specify the efficiency of the gear train as a percentage.
(d) Determine an equation for the efficiency of the gear train in terms of the load (torque) on shaft 2 (all other factors remaining constant).

6. Nov 3, 2014

### mattjones

It's a very basic sketch but here's what I have so far

File size:
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7. Nov 3, 2014

### OldEngr63

mattjones: It looks like you have assumed that you can get gears of any diameter, which is true in theory, but not possible with standard gears. You need to take a look at possible diametral pitch values and the associated gear sizes that result. Do they line up in a straight line as you have shown? Not too likely.

Items (iv) and (v) essentially dictates the first gear have 60 T and the last gear have 15.

On re-reading the problem, I see that the module is specified to be 2 mm, so the diametral pitch is fixed right there. It looks like the only flexibility you have is whether to use the 15/30 combination for the two middle gears (the idlers) or perhaps a 10/10 pair. This whole problem is fairly academic, that is, constrained in such a way as to force a unique solution. In real life, the module/diametral pitch choice would be up for grabs, and the multiple of 5 constraint would never be there (more about that below).

The requirement that all tooth numbers be a multiple of 5 (Item (iv)) forces a bad design. This means that there will be a common factor on all the tooth numbers, and consequently the same tooth pairs will engage time after time resulting in accelerated wear. In good gear design, common factors are to be avoided.

8. Nov 3, 2014

### billy_joule

Your current arrangement has around half the required spacing. .. Forget 15 and 60 that won't work if you want your gears lined up. There is no solution for the idler gears tooth counts to get the 300mm spacing with 15 and 60.

9. Nov 3, 2014

### OldEngr63

I did not check to see if the gears were large enough to span the gap; I assumed that they were. Evidently, this was a bad assumption.