I’ve been looking into the AM shifter transmissions, but have no idea how to calculate the 2:56:1 or 4:1 into my calculations (using JVN design calc). Can someone explain the two to me, and if possible the gearing that goes on in there? This is probably really simple, but I’ve spent 3 days trying to figure this out.

Thanks!

The 2.56:1 has final reductions of 10.67:1 and 4.17:1.
The 4:1 has final reductions of 10.67:1 and 2.67:1.
If I remember correctly, the difference comes from flipping a 28:35 to a 35:28, or vice versa.

Andrew,

The spread is determined by changing the gears inside the transmission. On JVN’s calc on the 2 speed section, replace the numbers with the different gear sets available from AM.

If your still a little lost, feel free to hit me up via PM/IM.

-RC

Those are the ratios between the high and low gears. The ratios that you would want to put in the calculator are under gearing on the product page.
Low Gear: 10.67:1
High Gear: 2.67:1

Low Gear: 10.67:1
High Gear: 4.17:1

I sorta understand the spread, but the gears are my main problem. I’m looking at this picture (http://www.andymark.com/PhotoDetails.asp?ShowDESC=N&ProductCode=am-0001), and don’t see what gears are used. I see the two gears in the picture, but I don’t think those are the only ones.

``````
AM Shifter AM-0957

Drive PN    Driven PN   Drive NT    Driven NT   SET RATIO   O'ALL RATIO
AM-0018    AM-0024       12              40            3.33
AM-0451    AM-0022       15              48            3.20        10.67
AM-0139    AM-0140       35              28            0.80         2.67

AM Shifter AM-0001

Drive PN    Driven PN   Drive NT    Driven NT   SET RATIO   O'ALL RATIO
AM-0018    AM-0024       12              40           3.33
AM-0451    AM-0022       15              48           3.20          10.67
AM-0025    AM-0023       28              35           1.25	     4.17

``````

Also, here is a render of AM-0001 with the gear teeth shown.

(40/12)(48/15)=10.667
(40/12)
(35/28)=4.167

12T input gear (x2 - CIM)
15T small cluster gear (output side)
28T medium cluster gear (motor side)
35T small output gear (motor side)
40T large cluster gear (center of cluster - direct input from 12T)
48T large output gear (output side)

(40/12)(48/15)=10.667
(40/12)
(28/35)=2.667

12T input gear (x2 - CIM)
15T small cluster gear (output side)
28T small output gear (motor side)
35T medium cluster gear (motor side)
40T large cluster gear (center of cluster - direct input from 12T)
48T large output gear (output side)

the only thing that matters is the final reduction of each of the two speeds.

Those two gears are the output gears. They’re free-spinning. The shifter dog couples one or the other to the hex-shaped output shaft. There are two other relevant gears on the “power shaft” that engage the output gears; they are hidden directly below what you can see.

There are two or three other gears that transfer the output of the motor(s) to what I’ve called the power shaft. They provide a first-stage reduction from the motor speed, but they aren’t involved in the shifting.

I would also caution you to look at your output power vs. speed with the two ratios. The 4:1 has a pretty big power trough between the peak powers. If your system spends a lot of time in this zone it can eat a battery alive.

I must say, that is a beautiful rendering.

Thanks everyone! I understand it now!