How do I caculate the output RPM

I am having trouble trying to picture how the arrangement you describe is connected schematically. The 50:1 transmission is pretty straight forward. It is reduced the output to 1/50th the input speed or multiplies the speed by 50. I don't know which way it is connected.

The sprockets are another matter. The correct way to determine dimeter is the pitch diameter. I assume this is a chain? The pitch diameter is going to be the distance of a line from the radius of one of the sprocket concave sections where the roller of the chain sits, through the radius of the sprocket hub itself, and to the next roller chain radius. If the sprocket has an odd number of teeth, you have to calculate where that point would be. Do this for both sprockets.

If the pitch diameter is given for both sprockets, then the ratio between the two circumferences of each sprocket determines the change in RPM. If both are the same size the ratio is 1:1. You can calculate the circumference multiplying PI by the diameter.

Since you are calculating a ratio you get Ratio = (d1 * PI) / (D2 * PI).

PI factors out, so it really is just the ratio of the two pitch diameters or D1/D2.

However, forget everything I just said about sprockets! The easy way to do this is to simply count the number of teeth on each sprocket!!! The ratio of the number of each sprocket is the magic number you want. If the large sprocket has 50 teeth and the small one 30, the the ratio is 50:30 or 1.67:1.

Does that help?

Good to see we are all still able to do homework.

All above answers are correct. Easy way out isn't it.

Nobody gives the figures for reverse gearbox.

1000*50*(4.2/8.7)=a very large and unlikely number, I go with TVP45

24,137.93 RPM

Sorry...must be late!...I meant to say 103,571.439 RPM

8.7 / 4.2 will give you a ratio of 2.071...multiply this with the motor revolution of 1000 RPM multiplied with gearbox ratio of 50 to 1

By gosh...Think of the money I could make just by running a few dozen gearboxes backwards in this plant. Uhm...how long do you all suppose I could make the bearings in my spiral conveyors last at 100,000+ rpm?

Sorry JAB...just stretching your drive chain abit for fun. No offence meant.

Hi Darryl,

Can I redescribe it again? If it is not correct please feedback.

Motor(n₀=1000 rpm)'s shaft is attached sprocket with d₀=4.2 inches diameter, and this sprocket is connected to the gear box(50:1) by other sprocket with d₁=8.7 inch diameter. And what speed(n_f) is the output shaft of gearbox?

n_f = 1000 * (4.2/8.7) * (1/50) ~ 10 rpm.

Common formular is:

n₀ / n₁ = d₁ / d₀

n₂ / n₁ = d₁ / d₂

.........

Good formula. But did you copy this formula from above????

Hi man,

This formula was too old, I learned it 16 years ago.

You can copy it.

thank you, I will copy it. By the way if the formula was too old, it can not be used any more.

Hi guest,

Sorry, I means that formula is basic in mechanical. The basics are always useful forever.

Dear,

If motor is running at 1000RPM, the drive sprocket would also run at 1000RPM. Considering that the sprocket diameters mentioned by you are the pitch diameters of sprockets then the formula would be:

1000 x ratio of 4.2/8.7 this to be further divided by gear ratio 50 = 9.65 rpm

It doesn't matter which comes first, the sprockets or the gear box, the final rpm is still the same, 9.65 .....

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When I ponder, I sits and thinks, sometimes I just sits.

1000/50*(4.2/8.7)

Well repeated