View Full Version : One gear rev travel distance how to ?.

Ken_Shea

05-10-2007, 10:20 AM

I have three gears, all 20 pitch.

One 8-tooth pinion and one 15 and 8-tooth combo on a common shaft.

The 8-tooth pinion drives the 15-tooth gear, at the opposite end of the 15 tooth gear shaft is another 8 tooth that meshes with a rack gear.

What I am trying to calculate is the distance traveled with one full revolution of the 8 tooth pinion.

How would I go about this?

Is the pitch circle diameter used or the gear OD ?

Thanks

Ken

Use the pitch circle. It is equivalent to two smooth wheels in contact.

Well...subject to correction, I think it works out like this:

The pitch circle circumference of a gear is ((# of teeth)/(diametral pitch))*pi.

The 20dp 8 tooth pinion has a pitch circle circumference of .4*pi inches.

The 20dp 15 tooth gear has a pitch circle circumference of .75*pi inches.

So one revolution of the 8-tooth pinion will produce a .4/.75 revolution of the 15-tooth gear...and of the other 8-tooth pinion on the other end of the shaft.

.4/.75 is .533333......

So the output pinion will rotate on its circumference (.5333*.4*pi) inches, or 0.670 inches.

Ken_Shea

05-10-2007, 10:47 AM

That helps Evan,

But I find I need more help, just about the time I think I have the calculation method figured out, my mind just goes blank beyond a point.

I know the PD of both gears, what do I do with them :D

Ken_Shea

05-10-2007, 10:51 AM

SGW,

You posted as I was replying, that is the additional information I was needing. Looks right to me.

Thanks

Ken

Lew Hartswick

05-10-2007, 10:22 PM

To heck with all the calculating. Just measure it. :-)

...lew...

Ken_Shea

05-11-2007, 12:39 AM

Lew,

Yes, I thought about that but there is no dial and to turn exactly one revolution from start to stop would be only a guess not to mention excessive wobble in the pinion shaft to complicate that even further.

I do like that often overlooked simple approach, but in this case it does not give me what I wanted to know.

Thanks

Ken

BadDog

05-11-2007, 01:40 AM

Why not just use tooth ratios? 8/15*8 = 4.2666... That's 4.2666... teeth on the drive pinion for each rev of the input pinion. Then all you need is the pinion's teeth per inch.

Ken_Shea

05-11-2007, 02:01 AM

BadDog,

My mind was just going blank with all the calculations I was coming up with, SGW's reply gave me the answers so I went with that. That is not to say that there are not simpler approaches to the same result.

In your example I had tried the same approach but was unable to follow through with what I needed.

Always being open to other solutions, can you expand on your approach more ?

Thanks

Ken

BadDog

05-11-2007, 02:29 AM

One rotation of the input pinion moves 8 teeth. That means 8 teeth out of 15 on the driven gear go by. So that's 8/15. Since it's connected by shaft, the same ratio of teeth on the 8 tooth drive gear go by. So that's (8/15)*8, or 4.2666... (repeating) teeth moving by the rack.

So, if you had 10 teeth per inch on the rack, you would have moved it 0.4267" (rounded), assuming a perfect system. If you had 8 teeth per inch, that would be 4.267/8 or 0.5334".

Forrest Addy

05-11-2007, 02:37 AM

BadDog: don't forget Pi and the effect of diametral pitch in your calculations.

BadDog is quite correct. One may use just the ratio of the input pinion teeth to the driven teeth on the rack. As the rack has an infinite diameter the pitch circle of the rack cannot be calculated and the teeth per inch is used instead.

Ken_Shea

05-11-2007, 09:52 AM

Two solutions, same answer :)

Thank you very much guys.

Ken

Lew Hartswick

05-11-2007, 10:17 AM

BadDog: don't forget Pi and the effect of diametral pitch in your calculations.

OH! and the phase of the moon, the relative humidity, the color of

your underwear, and the age of the milk in the frig. :-)

...lew...

JCHannum

05-11-2007, 11:37 AM

One rotation of the input pinion moves 8 teeth. That means 8 teeth out of 15 on the driven gear go by. So that's 8/15. Since it's connected by shaft, the same ratio of teeth on the 8 tooth drive gear go by. So that's (8/15)*8, or 4.2666... (repeating) teeth moving by the rack.

So, if you had 10 teeth per inch on the rack, you would have moved it 0.4267" (rounded), assuming a perfect system. If you had 8 teeth per inch, that would be 4.267/8 or 0.5334".

That will work if the number of teeth on the rack is an even number of teeth per inch akin to a thread. Since gear terminology is based on circular pitch, it is still necessary to use the circular pitch and pi to determine the teeth per inch of the rack and the linear distance moved.

Ken_Shea

05-11-2007, 12:07 PM

That will work if the number of teeth on the rack is an even number of teeth per inch akin to a thread. Since gear terminology is based on circular pitch, it is still necessary to use the circular pitch and pi to determine the teeth per inch of the rack and the linear distance moved.

That is correct and what I did, and I suspect that BadDog was implying the same thing.

For others that may be following this thread and are as confused as I was, maybe I am the only one !

Here is how the travel was arrived at using BadDog's method.

A 20 Diametral Pitch gear has a Circular Pitch of .157

So take 1" / .157 = 6.369 Teeth per Inch.

Take his 4.267 / 6.369 = a travel distance of .670 rounded.

Same as SGW's result only simpler.

BadDog

05-11-2007, 02:44 PM

Exactly. I didn't imply anything about how to determine the teeth per inch of the rack since that is easily calculated (as Ken shows), or possibly even just gathered from docs in some cases. I just tried to describe a simple and intuitive way to calculate linear displacement of any rack once TPI is known. In general, most gear train solutions can also be simplified by just using rotational ratio (gear ratios, easily calculated by counting teeth) and putting off conversion to circumferential till the last step as needed.

Unless I'm missing something, why make things harder than they have to be? Pitch circle and all the rest are important when making a gear and determining shaft spacing, but the ratios control driver/driven rotation.