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Unread Today, 11:31 AM   #16
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Re: Repair or replace table saw


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Originally Posted by SeniorSitizen View Post
The formula is: drive sheave diameter X motor RPM divided by driven sheave diameter.

Example IIR correctly: 1750 RPM X 3" motor sheave divided by arbor sheave diameter of 2" = 2,625 blade RPM.

It is the circumference you play with and your speed would be 3500
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Unread Today, 11:36 AM   #17
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Re: Repair or replace table saw


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Originally Posted by Nealtw View Post
It is the circumference you play with and your speed would be 3500
I'd rather play with diameter.
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Unread Today, 11:51 AM   #18
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Re: Repair or replace table saw


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Originally Posted by SeniorSitizen View Post
I'd rather play with diameter.
but the key is 2" is 1/2 the size of a 3"
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Unread Today, 02:58 PM   #19
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Re: Repair or replace table saw


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Originally Posted by Nealtw View Post
but the key is 2" is 1/2 the size of a 3"
If the above is correct, I'm certain we attended different arithmetic schools.
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Unread Today, 03:16 PM   #20
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Re: Repair or replace table saw


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Originally Posted by SeniorSitizen View Post
If the above is correct, I'm certain we attended different arithmetic schools.

Yes, not sure what I did wrong but yes. I spent 20 years in an industry where we had to show all the long math, for belt speed. Hopefully my math skills were better then.
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Unread Today, 04:19 PM   #21
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Re: Repair or replace table saw


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Originally Posted by moosehaed View Post
Not sure if direct drive is the correct term. I was referring to non belt drive saws.
I was looking at something like this or similar.

A basic low price table saw with a stand. https://www.lowes.com/pd/DELTA-ShopM...Saw/1000795558

Sent from my Pixel 2 XL using Tapatalk
That is essentially a circular saw mounted to an inexpensive plastic box with an aluminum table top.

You can get a 1/2 hp motor for under $100 on Amazon.
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Unread Today, 04:36 PM   #22
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Re: Repair or replace table saw


The formula, I originally posted, was for driven RPM if we have both sheave sizes and a known drive RPM.

I had to re-teach myself how to determine what size sheave to attain a certain RPM of the driven if only RPM is known of the drive sheave and diameter. So I went back to pencil / paper and with difficulty re-taught myself how to solve ratio and proportion problems.

To soften the pressure on my deteriorating memory I picked an easy problem I could solve in my head.So in this instance if we want the blade to rotate 3,500 RPM we would need a 1.5" sheave in the saw arbor.
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