Belt drive alternative

JPG · Post by **JPG** » Mon Jun 20, 2011 5:17 pm

paulmcohen wrote:This picture while called a speed reducer does not change the speed. For it to be a speed reducer you would need for the pulley sizes to be different. 2/7 * 7/2 = 1. What am I missing?

Saying what Dusty said, but slightly differently. The 'math' is 7/2 * 7/2 = 49/4 = 12.25/1

Put another way. speed reduction/increase = driven pulley/driving pulley.

From the headstock(driver) 2" to the jackshaft(driven) 7" the ratio is 7/2.

From the jackshaft(driver) 2" to the bandsaw(driven) 7" the second ratio is 7/2'

The effective overall ratio is the product of the individual ratios.

holsgo · Post by **holsgo** » Mon Jun 20, 2011 7:59 pm

This is a very cool concept since I've been trying to figure Prmindartmouth's previous work in this area. This is too easy. I think I'm going to try it.

paulmcohen · Post by **paulmcohen** » Tue Jun 21, 2011 2:39 am

Thanks all, I realize what I was doing wrong.

JPG · Post by **JPG** » Tue Jun 21, 2011 8:20 am

paulmcohen wrote:Thanks all, I realize what I was doing wrong.

De Nada!!!!!!;)

dusty · Post by **dusty** » Tue Jun 21, 2011 8:26 am

[quote="JPG40504"]De Nada!!!!!!]
Yeh. NO conozco la canción, así que un poco más de contexto sería útil. A secas me parece que es "absolutamente nada.

holsgo · Post by **holsgo** » Tue Aug 09, 2011 9:19 pm

Ok. I'm going to take this on in the next couple weeks. This pulley thing is making my head hurt. So, the. 2 on the headstock to the 7 underneath I get. It's the 2 back to the. 7 on the bandsaw that doesn't compute. Any clarity Jpg? I've tried to understand but it just seem to come back to there being no speed reduction. No, this is not my area of expertise.

JPG · Post by **JPG** » Tue Aug 09, 2011 11:25 pm

holsgo wrote:Ok. I'm going to take this on in the next couple weeks. This pulley thing is making my head hurt. So, the. 2 on the headstock to the 7 underneath I get. It's the 2 back to the. 7 on the bandsaw that doesn't compute. Any clarity Jpg? I've tried to understand but it just seem to come back to there being no speed reduction. No, this is not my area of expertise.

Forget pulley diameters for a moment.

Think in terms of belt speed and pulley circumference.

Also for simplicity, approximate the pulley circumferences as 6" and 21".

Also look at this from the driven end(band saw drive shaft).

For every revolution of the 7" bandsaw drive shaft pulley, the belt will move 21".

For that to happen, the 2" jack shaft pulley to which that belt is attached will turn 3 1/2 revolutions. 21" / 6" = 3 1/2.

When the jack shaft turns 3 1/2 revolutions, the belt on the 7" pulley on the jack shaft will travel 73 1/2". 3 1/2 x 21" = 73 1/2 ".

When that belt travels over the 2" headstock pulley the pulley will rotate 73 1/2" / 6" = 12.25 rpm.

So the ratio is 12.25 : 1, orrr from the headstock to the bandsaw the ratio is 12.25 : 1.

Now finally since the circumferences used were reduced by the same ratio(3/3.14...)[used 3x2 and 3x7 instead of 3.14x2 and 3.14x7] the final ratios are the same. It is the constant (pi)[3.14...) that allows figuring the ratio by merely using the pulley diameters rather than the circumference(belt motion).

bs shaft = 1 rpm-------belt = 21"

Jack shaft = 3.5 rpm---belt = 21" , 73.5"

headstock = 12.5 rpm-------belt = 73.5"

holsgo · Post by **holsgo** » Wed Aug 10, 2011 6:08 am

OK. I'm following you. I will try and let this soak in while at work today. Don't know why this is throwing me off. I used a couple on-line calculators and maybe that's where my head hurt. It seems that motor pulley is small, driven pulley is large then that's speed reduction but to go back in reverse would simply send the speed back to the original. There's no calculator out there for this kind of arrangement so you are it!

dusty · Post by **dusty** » Wed Aug 10, 2011 8:20 am

holsgo wrote:OK. I'm following you. I will try and let this soak in while at work today. Don't know why this is throwing me off. I used a couple on-line calculators and maybe that's where my head hurt. It seems that motor pulley is small, driven pulley is large then that's speed reduction but to go back in reverse would simply send the speed back to the original. There's no calculator out there for this kind of arrangement so you are it!

Remember that this arrangement produces a 7:2 speed reduction followed by another 7:2 speed reduction.

Analyze the actions of each pulley pair individually and I believe the fog will clear.

pennview · Post by **pennview** » Wed Aug 10, 2011 8:41 am

I don't know if this makes it any clearer or not regarding how a compound pulley or compound gears work, but I'll give it a try.

Say you have two pulleys, one with a circumference of 2" (pulley A) and a second one with a circumference of 10" (pulley B).

Pulley A is the drive pulley and Pulley B is the driven pulley.

It takes five revolutions of A to turn B one revolution, or B is turning at one fifth the speed of A.

Now add a 2" pulley (Pulley C) to the axle (jackshaft) that B is connected to. You now have a compound pulley -- B plus C. And, since B and C turn together, it also takes five revolutions of A to turn C one revolution.

Now, connect C to a second 10" pulley (Pulley D) and for each five revolutions of C, Pulley D turns once.

So, it takes 5 revolutions of A to turn D one fifth of a revolution, and 25 revolutions of A to turn D once. So in this example, you have a speed reduction of 25 to 1. With a 3600 RPM motor, you have pulley D turning at 144 RPM.

Mathematically, using the circumferences of the pulleys, you have (A/B) * (C/D), or (A divided by B) times (C divided by D). For the example given, it's (2/10) * (2/10) = 1/25.