Gear Cutting

[Darren]

I was reading through Johns page the other day on gear cutting, I sure have not fully absorbed it all as yet, but I'm glad to see the cutters are standardised to aprox sizes to reduce their number.

If we have two wheels with say 60 teeth and one is twice the diameter of the other. Then the teeth on the larger wheel would obviously need to be larger.

How do you accommodate for this cutter wise. Do you need two sets of cutters?

http://www.metalwebnews.com/howto/gear/gear1.html

[raynerd]

Darren from my understand based mainly on research on cyclodial gears, and I am guessing you are attempting to cut involute gears, the module, which dictates in most part the teeth form is based on the pitch circle diameter.

Module = Pitch dia  /  number of teeth

So you said 60 teeth on two wheels, one twice the size of the other, so lets say a 60mm Pitch diameter and 120mm Pitch circle diameter wheel.

Gear 1 - 60 teeth with a 60mm PCD
Module = 60/60 = 1.0    

So Gear 1 would need a 1.0 mod cutter


Gear 2 - 60 teeth with a 120mm PCD
Module = 120/60 = 2.0

So gear 2 would need a 2.0 mod cutter.


So my understanding is that you would need two cutters a Mod 1 cutter and a Mod 2 cutter and even then you would need to ensure that since they are involute cutters, they are the correct angle for the number of teeth you are cutting, or in that range.

Hope that helps and hope and is correct!

Chris

[Darren]

Thanks for that Chris, that's one piece of the jigsaw clarified.

So you work out the module first and then select the cutter number from the eight sizes as in the carts.

I was mixing up the module with size (for No' of teeth) 

[bogstandard]

Darren,

Methinks you are getting confused.

When working with involute cutters, there are actually 8 cutters in each set, you select which cutter you require for the number of teeth needed on the gear. So say you require 23 teeth on the gear, you would use #5 cutter, or for 46 teeth #3.

8 cutters make up a set for one sized tooth.

To make a larger tooth, you will require another set of 8.

Myford use 20 DP for their standard sets of gears, so you would buy the full set of 8 cutters to make a set of change gears. It would be the same if your lathe or whatever used say 14 DP. Another set of cutters to your growing collection.

It is for that reason I will soon be experimenting with gear hobs. One hob of say MOD 0.8 will cut the whole range of teeth. Whereas for involute, you will require 8 cutters.


John

[raynerd]

Thanks for that Chris, that's one piece of the jigsaw clarified.

So you work out the module first and then select the cutter number from the eight sizes as in the carts.

I was mixing up the module with size (for No' of teeth) 

Darren, as John said above, there are eight cutters within each module or more accurately -"diametral pitch". So take the 1.0 mod cutter, once you have decided that you need this module cutter you then need to choose the correct one for the number of teeth you are cutting:

http://www.rdgtools.co.uk/acatalog/_1_MODUL_E_GEAR_CUTTERS.html

The cutter approximation method uses a radius to approximate the involute form on a single tooth tool. The circular approximation is placed at the pitch line of the tooth and the difference in the two profiles becomes more pronounced towards the tip but is still small enough to be disregarded until eventually it must be taken account for and a different cutter used. With a cylodial cutter, such as those I am looking to form, they can be used over a much wider range - infact 20 teeth and above, but a new cutter is still needed for pinions, 20 leaves and below and as the number or teeth reduce, the range of the cutters use also reduces.


Chris

[old-biker-uk]

Of course if you want involute gears and own a shaper it a piece of cake.....
http://myweb.tiscali.co.uk/remark/pages/workshop/gear%20cutting.html
Mark

[raynerd]

 

I have considered this method for cutting a pinion on a clock but it would need supporting underneath....
Great find.

Chris