Increasing blade size and number of teeth??

**camoz** · 26th November 2012, 11:17 AM

I have possibly a silly question, but in my defence I have asked two saw sharpening places and got two different answers.

I am in the process of restoring an old Stenner ABM 18" table saw (https://www.woodworkforums.com/f163/s...ration-157563/), and as christmas is coming up I was hoping santa might provide me with some new saw blades. My question is when you increase the size of the blade, what is the relationship to the teeth in regards to cleanness of cut?

Now I know this will probably raise all sorts of questions, like what are you cutting, what pattern of teeth are you using, etc. etc., but in a hope to simplify the issue, on my 12" saw I tend to use a 60 & 96 tooth blades (for most of what I do). So my initial thought was teeth per inch, work out the circumference and get the closest teeth per inch in the 18" blade. Then I remembered that the speed of the teeth on the 18" will be moving a lot quicker than the 12" due to the size difference of the circle, and then I realised that my 12" saw blade rotates at 3600 rpm and the 18" is connected directly to the motor which only rotates at 2860 rpm. After that I got a headache and had to lay down

.

I am not sure if I am thinking this out too much, but at over $300 for a blade, I would hate to take a guess and find I am not happy with the number of teeth I chose.

Any help would be greatly appreciated!

Thanks,

Cameron

P.S. If this is turns out to be a dump question, please be gentle

.

**BobL** · 26th November 2012, 11:51 AM

The linear teeth speeds for the 18" blade = 13500 fpm while for the 12" = 11300 fpm so if you want equivalence on 18", as a 96 tooth on the 12", you will need a 120 teeth on the 18" blade.

**Jim Carroll** · 26th November 2012, 01:08 PM

Another thing to consider is the weight of the blade.

Was the original blade just a HSS blade with teeth ground etc or did it have a TCT blade.

If just a normal blade then the power to weight ratio would have been worked out and the motor matched.

If it was a normal blade and you want to put on a TCT blade then you may need to up the anti on the motor as there is a lot more drag with a TCT blade, there is a lot more load when cutting as you have basically a square tooth cutting where before there was a slicing type blade.

**camoz** · 26th November 2012, 04:01 PM

Thanks guys for the fast responses.

BobL is there a formula that you used to work out the fpm (just out of curiosity)? Also just to check, to make sure I have the math right (you do 96/11300*13500), so the 60 teeth equivalent would be 72 teeth in the 18"

Jim, (I had not thought about that) the blade that was on it when I got it was a TCT blade, but it is a circa 1957 table saw, so not sure about originally. I think I should be OK as the blade that is on it looks to have done a lot of work, and even if I wanted to change the motor it wouldn't be viable, as it is a direct drive system (the motor shaft is the arbor, and the blade bolts directly to the shaft). I definitely don't think it will hurt the motor (I have had the motor apart and it is built like a tank, and weighs like one too

). Obviously with this design although the blade is 18" the maximum depth of cut is a lot less than half (the motor is a 5HP Brooks motor). I did originally consider going with a smaller blade, but wanted to try to get the maximum depth of cut due to this design (the advantage is no belt or pulley to add vibration into the equation).

P1020829.jpg motor in 4.jpg motor in 5.jpg

Cheers,

Cameron

**BobL** · 26th November 2012, 04:57 PM

Originally Posted by camoz

Thanks guys for the fast responses.

BobL is there a formula that you used to work out the fpm (just out of curiosity)? Also just to check, to make sure I have the math right (you do 96/11300*13500), so the 60 teeth equivalent would be 72 teeth in the 18"

All that matters is the # teeth per unit time passing any specific cutting point, so an easier way to think of it (rather than using the FPM argument) is as follows

If both blades were doing the same RPM, then to maintain the same # teeth per unit time, 96 teeth blade on a bigger (or smaller blade) would do the same job.

The main difference here is the RPM of the smaller blade is 3600/2860 = 1.258 of the smaller one, so to maintain the # teeth per unit time one needs to increase the number of teeth
96 X 1.258 = 120.

For a 60 tooth blade its 60 1.258 etc

**camoz** · 26th November 2012, 05:13 PM

Originally Posted by BobL

All that matters is the # teeth per unit time passing any specific cutting point, so an easier way to think of it (rather than using the FPM argument) is as follows

If both blades were doing the same RPM, then to maintain the same # teeth per unit time, 96 teeth blade on a bigger (or smaller blade) would do the same job.

The main difference here is the RPM of the smaller blade is 3600/2860 = 1.258 of the smaller one, so to maintain the # teeth per unit time one needs to increase the number of teeth
96 X 1.258 = 120.

For a 60 tooth blade its 60 1.258 etc

Thanks for the clarification.

Cheers,

Camo

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