[JerryFlyGuy]

Can anyone offer advice oh how to calculate the forces driving a pinion out of the rack? I have a pivot style set up designed in which the pinion is forced into the rack via two gas shocks, I'm currently providing 210# force on the gear into the rack. There is, I'm sure much geometry to get the proper calculations [as far as the angle between the pivot location in relation to the pinion and how close to perpendicular or not this relation is to the rack..thats probably clear as mud I know but....] If anyone has anything they can offer I'd be a happy camper. I can probably also get you some Cad views of what my configuration is to better illustrate the geometry.

JerryFlyGuy

[NC Cams]

Thre isn't that much geometry involved for a rough calculation.

Go to a local library or college engineering department and borrow a copy of the following textbook:

Design of Machine Elements by M.F. Spotts,

Chapter 10 shows you how to calculate the forces on and by gears.

Basically,

Fr = Ft * tan x

where:
Ft is tangential force applied to the pinion and
x is the pressure angle of the gear and
Fr is the separation force.

There is a bit of a loss due to friction but this is probably close enough for government work....

[Vaughan Sage]

Sounds right to me, if pressure angle geometry of gear teeth is 20deg & angle of friction at the gear teeth contact rubbing is 10 deg then you have 30deg total, so spread apart force is about half the tangential thrust in axis direction.

[JerryFlyGuy]

HAHA.. well that sounded like a really nice.. {insert cool sounding, internal combustion powered vehicle here} as it zoomed over my head :-) Anyone got a sketch or drawing to illustrate it? I haven't gotten to the library ta take a look for that book as of yet..

Jerry

[NC Cams]

Getting the book will make things clear - all the sketches you'll ever need with respect to gear design are in there. The formula was copied from same.

In some respects, CAD has oversimplified the design process. Drawing the pictures is but half of the design effort - the force calcs (IE the math) compromises the balance and perhaps the most critical part of the design effort.

In this case, all you have to do is to plug in the numbers into the formula (you already have the force, all you need is the pressure angle of the gear - without it, you're screwed).

The only "much geometry" you need to know is how to get input "tan x" into a calculator. Once you do this, the formula will spit out the separation force (in lbs) that you asked for.

This can't be made any more simple.

[JerryFlyGuy]

Nc, if I'm understanding you correctly, using the above formula.

Ft= the force that is being output at the pinion [ at the interfacing pinion Pitch Diameter] in my case ~320lbs.

I'm using 20Deg pressure angle on my Pinion and rack so...

Tan(20)=0.3639702

therefore..

Fr=320 x 0.3639702
Fr=116.5 lbs?

now would this force vector be perpendicular to the face of the rack?
If so, then I have ample force with my gas shock. [see pic below]

I would agree that the that CAD has created many, many changes in the industry. I would also agree that the requirment for calculations and solving equations has never changed and is still required. However it's in the understanding of the terminoligy that I lack. As you've stated, having the book would clarify things greatly. Being able to see the physical interaction rather than read technical description is definatly easier for at least myself. However, having said that I did infact search for that book at my local library without any luck what-so-ever. [just checked it online] Hency my quandry..

I've included below a top view sketch of my geometry to better illustrate and confirm that what I'm calculating is infact going to work.

Jerry

[NC Cams]

Yes, 116.5 lbs would be normal (perpendicular in X direcion) force applied to the pinion via the separation force that occurs between it and the rack.

You'd have to do some tangent work with the pinion shaft and pivot shaft angle to find the force normal to the Y component of the 4.79" length.

THis force (not the 116.5) would have to be reacted by the force of the shock.