Haw to Calculate Spline with profile shift X 2

[makary]

I need to use a 24/48 involute spline with 60 teeth, but need to get 65.0 mm OD (a bit larger than what you get with a standard 60T-24/48).
To get the 65mm OD I need to use profile shift X - same technique as with a gear. Can anyone explain how to calculate the necessary X factor? For a gear it would be simple but for a spline I don't know what I should shift - the deddendum of the 24 DP, or addendum of the 48 DP?
Or something else yet?

My calculations, when shifting the deddendum show that X=0.21
I got another opinion, that the X should be 0.4134. No idea how though. Are they applying this shift to the addendum (48 DP)? Can someone explain how this shift is accomplished, and consequently how the X is calculated?

 

[tbuelna]

markary,

You didn't specify if that 65.0mm OD was for an internal or external spline, or what type of fit the spline was.

If, by chance, you want a fillet root, side fit, 30deg pressure angle, external involute spline, with a standard 65.0mm (2.5591 inch) major diameter, then you should specify a DP of 23.83692. You can calculate PD, addendum, etc. based on that DP value.

Good luck,
Terry

 

[makary]

Thanks Terry, you are correct in your assumtions: external involute, side fit, 30deg, the root is flat but should make no difference, and 65.0 mm (2.559") major is needed.

If DP needs to be 23.8237 than may X shift factor calulation should be correct, and X = 0.210 for shifting the 24 DP.
So I still don't understand what the 0.4134 represents. This number was given to me by a spline cutting place as the required X shift factor to get the 24/48 spline to a 65.0 mm major.
Possibly we speek two different languages, beacuse I recalculate english size to metric, then do standard calculations on metric gears and go back to english. I would not want that to result in getting me wrong splines.

 

[dinjin]

It is a minor change but I get an X factor of .20866.

If the od equals (60 plus 1.00 plus .41732)/24 yields
2.559055 inches or 65mm. Then X = .41732/2 = .20866.

Maybe .21 is close enough. The 24/48 simply means that the addendum and dedendum are only one half the normal values.
If it were a full depth tooth the formula would be
(60 plus 2.00 plus 2X)/24). If you use the module system,
24 Dp equals 1.0583333 Module.
Then the od equals (60 plus 1.00 plus .41732) x 1.0583333
equals 65mm.

 

[Spurs]

As a footnote to the discussion of X factor - it is often misunderstood that X factor is really another way of saying "tooth thickness".

It is a commonly used practice in Europe to design by X factor because the theoritical practice based on standard center distances and without allowing for backlash, the X factor of the pinon would be the same magnitude but opposite sign as the gear.
ie at these standard conditions you could have a pinion with X= + .125 and a gear with X= -.125.

The formula for X factor based on tooth thickness is:

X =(tn/mn-Pi/2)/(2TAN(PAn))

where tn = the normal circular tooth thicness
mn is the normal module
PAn is the normal pressure angle

There is a practical problem in how people use X factors. Generally people believe a gear can have only a single X factor, but almost everyone would agree that a gear may not have a single tooth thickness. We generally have a tolerance on tooth thickness but for some reason we rarely see a tolerance on X factor.

If you specify a dimension over pins with a tolerance or a tooth span with tolerance, you have achieved the same thing as defining an X factor; without the ambiguity of what the allowable tolerance is.