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Helical Gear Questions

CharlieBiler

Hot Rolled
Joined
Mar 11, 2005
Location
Connellsville PA
To All,
I am very rusty in the gear calculation department. I need to make gears with the following data.
Lead=1.8264"
Normal D.P.=14.
Pitch Dia.=0.980"
Helix Hand= Left
Helix Angle=59degrees, 19 minutes, 18 seconds
Pressure Angle=14.5 degrees involute

I need to know the diameter of the blank to hob this gear. Does anyone out there remember how to calculate the formulas for helical gears?

I thought that the outer diameter was the number of teeth plus two divided by the diameteral pitch. That would be (7+2)/14. This gives me a bogus answer of 0.6429 inches. The gear has to be near 1.2 inches in diameter. I have a sample of an old gear in my hand and a print from 1982. What am I reading wrong? Working backward I add two to the teeth and take the answer and divide it by my rough measurement of 1.2 inches. That gives me a diametral pitch of 7.5 . Again that is a bogus answer.

Have pity on a poor old fool. I have lost my mind and can nor remember my first year apprentice formulas. I am sunk.

Charlie Biler
 
If the pinion is only 7-teeth, it's likely enlarged or long adendum design. So, standard proportions do not apply. Do you have a spec for circular tooth thickness?
All this is addressed in the Machinery's Handbook.
 
I'll post a follow up for Charlie:

I spoke with Charlie is morning and he mentioned this problem. I rarly work with gears but I had a feeling that the helical gears did not use the same formulas as spur gears. A quick Google search confirmed this. See http://www.engineersedge.com/gear_formula.htm

Addendum = 1 / Normal D.P.
= 0.0714285

OD = Pitch Dia + 2 Addendum
= 1.12285714285

Which is close to his measured 1.2 inches

Ted
 
PD= Num teeth /(normal DP * Cos (helix angle rel to axis))

For this I got 0.9793, using 59.3 deg as rough helix angle.

Adding 2 addendums 0.9973 + 2(0.07142) = 1.12215
 








 
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