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Adjusting Gear Sizes


Date: 07/14/2000 at 10:47:55
From: Rob
Subject: Proper gear tooth size

Hello,

I am in need of some help, and since you were very helpful to me on a 
question I asked previously, I thought I'd come back for more help.

Setup: Some material (foam tape on a white liner) is to be die-cut 
into 4" long pieces. To accomplish this, the material goes through a 
station that has two rollers positioned vertically. The top roller 
makes a 4" cut that goes through the foam tape and not the liner. The 
bottom roller is connected to the driveshaft of the machine and has a 
94-tooth, 2.9375" pitch diameter.

The material is then fed into two more rollers that are positioned 
vertically. The bottom roller of this set is 3.310" diameter. The 
bottom roll gear of this set and the bottom roll gear of the other set 
are connected via a free-spinning transfer gear, and all three gears 
have a 94-tooth, 2.9375" pitch diameter with a 1/8" circular pitch.

Before the material gets to the second set of rollers, the liner is 
stripped off. A new liner wraps around the second bottom roller with 
about a 120-degree wrap angle. The idea is to have the new liner 
moving fast enough to create a gap between the parts. In this case the 
part size needs to be 4" and the gap 1/8".

Question: What gear (how many teeth) size do I need to use for the 
second bottom roller to achieve the required gap between parts? I know 
that only the transfer gear size will have to change to accommodate 
the second bottom roller's new dimensions. Everything else should stay 
the same.

Calculations: I have made several calculations, but I keep getting 
more confused. If I assume a web speed of 40 feet per minute, that 
gives me 52 rpm for the first bottom roller and 46.19 rpm for the 
second bottom roller. The second bottom roll gear can be calculated at 
4342 teeth/min. by multiplying 46.19 rpm by 94 teeth per revolution. 
Then things get really screwy when I calculate 40 fpm as 480 inches 
per minute. 40 feet of material gives me ten 4" pieces, and since the 
pieces must be separated by 1/8", I need an extra 1.125 inches. I came 
up with 91.4 teeth, but I don't know how. Could you please help me 
with this? I will be forever indebted to you.

Thanks, 
Rob


Date: 07/14/2000 at 13:19:51
From: Doctor Rick
Subject: Re: proper gear tooth size

Hello Rob. I hope I can help you again.

I'm not quite sure I understand the problem fully. The picture, I 
think, looks something like this:

                                  ---
                              //--   --\\
                             /           \
             -----          |             |
           /       \       |               |
          |         |      |               |
         |           |     |               |
         |           |      |             |
          |         |        \           /
           \       /          \\--   --//
 ======>     -----                ---          ======>
 --+-------+-------+-------+-------++-------++-------++-------++-----
             -----                ---
           /       \          //--   --\\
          |         |        /           \
         |           |      |             |
         |           |     |               |
          |         |      |               |
           \       /       |               |
             -----          |             |
                             \           /
                              \\--   --//
                             \    ---
                              \ 
                               \ 
                                \ 
                                 \

I don't quite see how the liner is stripped off and the new one 
applied, but I think what you're saying is that the parts, which came 
every 4 inches on the first liner (moving with the first set of 
rollers), will be spaced one every 4 1/8 inches on the second liner 
(moving with the second set of rollers.) This produces a gap of 1/8" 
between the 4" parts. Is that right?

What this means is that the second liner must be moving at a speed 
4.125/4 = 1.03125 times the speed of the first liner.

As it stands, with the same number of teeth on all three rollers, the 
speed of the second liner is 3.310"/2.9375" = 1.12681 times the speed 
of the first liner. This is because the two rollers have the same 
rotational speed, and the linear velocity of their surfaces is in 
proportion to their diameters. At this speed ratio, the gap between 4" 
parts will be 4"(1.12681-1) = 0.5074". Is this correct?

If I'm with you so far, you want to reduce the speed ratio from 
1.12681 to 1.03125. To do this without changing the diameter of the 
second roller, you must reduce its rotational speed by a factor of 
1.03125/1.12681 = 0.91519.

Since the original tooth ratio is 1:1, the tooth ratio that you want 
is 0.91519:1, or 1:1.09266. Having more teeth on the second roller 
gear (by this latter factor) will make it turn more slowly, reducing 
the gap from 1/2 inch to 1/8 inch. With 94 teeth on the transfer gear, 
this comes out to 94*1.09266 = 102.71 teeth on the second roller gear.

Can you make do with the nearest integer number of teeth? Let's work 
out what we get with 103 teeth on the second roller gear. For each 
turn of the first roller, the second roller turns 94/103 = 0.91262 
times. Its surface moves 1.12681 times faster than it would if its 
diameter were the same as the first, so its surface moves 
0.91262 * 1.12681 = 1.02835 times as fast as the surface of the first 
roller. When the first liner moves 4", the second liner will move
 4" * 1.02835 = 4.1134". The gap is 0.1134", rather than 0.125" - 
it's a little too small now.

As in the last time we corresponded, you'll have to decide whether 
this is good enough, or whether you will have to change the number of 
teeth on the transfer gear as well in order to get closer to the 
correct ratio.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Physics
High School Physics/Chemistry

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