Belt Speed

Wayne Coe

Forum Owner - Moderator
For all of you who have wondered or had the question asked and did not know the answer I was able to obtain the use of a tachometer to measure belt speed in feet per minute. I have two Moe's grinders in my shop at the present time. Both use 1750 rpm motors with 4" drive wheels. Each has different size pulleys which is reflected in the chart.
After taking the measurements in Feet Per Minute (fpm) I converted to Inches Per Second (ips) which I think is a more understandable measurement. The formula that I used for the conversion is:
feet X 12 = total inches divided by 60 = inches per second. If someone out there understands this conversion better than I do and if I am incorrect please inform me so that I can correct this thread.


Build A Belt Grinder Belt Speed


With 1750 rpm motor with step pulleys in feet per minute

4 inch pulley
3.5 inch pulley
2.5 inch pulley
2 actually 1 7/8 pulley
4,000 fpm
2,450 fpm
1,436 fpm
906 fpm
800 ips
490 ips
287 ips
192 ips
5 inch pulley
4 inch pulley
3 inch pulley
2 inch pulley
4,750 fpm
2,475 fpm
1,400 fpm
737 fpm
950 ips
495 ips
280 ips
147 ips


I hope that if you are interested this chart will be useful.

Well the software did not acept my chart so I will make an attachment. View attachment Build A Belt Grinder Belt Speed.pdf
 
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Thanks, I found some formulas online, but they were very confusing. It doesn't get easier than this.
 
Very nice work Wayne. That is very useful to see how the pulley size changes the speed.
What size is the pulley on the motor shaft?
 
Stabber, I see that you asked a question that I never saw. The drive wheel is 4". The pulleys on the motor and the drive shaft are the same with the belt riding on the large step and the small step. Think back when you saw my grinder at IGs. I hope that this us understandable.
 
Ok, this is freaky... I just asked this question in another thread, and PM'd Wayne about it too, and then THIS gets revived! Am I in the Twilight Zone?

So what speed is "best" for major bevel grinding, Wood/Micarta grinding and Finishing work?
 
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Your initial first grinds are the fastest and then slow down as you go and same for wood 80grit rips off wood real fast but less presure is needed at high FPS
 
Ok Wayne, I know we talked about this briefly, but do you have any info on direct drive with a 6 inch drive wheel?
I'm seriously loving this grinder! I will post pics of it maybe tomorrow or so.
 
No, I don't but you can get a contact tachometer at Harbor Freight for about $20.00. With it you can check rpm and fpm.
 
Ok Wayne, I know we talked about this briefly, but do you have any info on direct drive with a 6 inch drive wheel?

Actually, direct drives are among the easiest to calculate, since pulley reduction ratios do not need to be inserted into the formula.
No worries about pitch diameters, either.
That's the plus side of it.
The negative side of direct-drives is that one is more limited on changing the belt speed if a variable-speed motor (VFD, etc.) is not used. Changing wheel diameters is certainly an option, but it not only requires the investment in different wheels, but is also inconvenient to constantly change them.

If you do not have a tachometer, you can easily calculate belt speed provided you know the diameter of the drive wheel and the motor RPM.
Have to go back to math class, but it's still very simple.

The "standard" form of measurement is SFPM, or surface feet per minute. Simple conversions will allow one to easily determine other forms of measurement if so desired (inches/minute, feet per second, metric, etc.)

In a nutshell, the SFPM formula is as follows:
SFPM = wheel RPM x wheel circumference (in inches) divided by 12

Let's break that down for explanation. Please understand that I do not know your level of experience with this subject, but others are also reading this; therefore I attempt to be thorough in my explanation.

1. In a direct drive system, the drive wheel RPM is the same as the motor RPM.
I do not know specifics on your motor, but for example's sake let's call it 1725.

2. Find the circumference of the drive wheel. The formula for circumference is: C= D x pi
In our example, we use a 6" diameter drive wheel.
Circumference= 6" x 3.1416 = 18.85"

3. For sake of convenience, we measure the wheel diameter in inches, so the circumference is also in inches. But we are trying to find the answer in feet. So we divide by 12.

Belt speed= 1725 RPM x 18.85" divided by 12 = 2710 SFPM

Plug in the numbers for a 3450 RPM motor, and you'll see that the belt speed also doubles.
With direct drives, the options for varying belt speed are limited to the following options:
a. Use a variable-speed motor.
b. Change the size of the drive wheel.

Hope this helps.
 
Dazed & Confused...

I'm terrible at math:confused2::eek: so hopefully you can help me out and tell me if changing the v belt pulley wheels can get my old Koval knives 2X72 belt grinder's belt speed to run "any where near the 400 rpm mark" for final sharpening and felt stropping (I like the edges to stay cool on my blades) ?:biggrin:

And if it can be done what size pulleys would I use and where can I buy them?

If it cant be done then how slow can I get it to run?

NOTE:
motor
1HP spinning 1750rpm shaft speed.

Thank you a million times in advance for any help you can provide for me on the above issue!!:60::happy::42:
 
I'm terrible at math:confused2::eek: so hopefully you can help me out and tell me if changing the v belt pulley wheels can get my old Koval knives 2X72 belt grinder's belt speed to run "any where near the 400 rpm mark" for final sharpening and felt stropping (I like the edges to stay cool on my blades) ?:biggrin:

And if it can be done what size pulleys would I use and where can I buy them?

If it cant be done then how slow can I get it to run?

NOTE:
motor
1HP spinning 1750rpm shaft speed.

Thank you a million times in advance for any help you can provide for me on the above issue!!:60::happy::42:



Yes, it can be done.

A few questions:
1. I am not familiar with the machine you mention. Is it the type with a motor transmitting power to one driveshaft? In other words, is this a 2-sheave setup, where one sheave is on the motor and another is on the driveshaft which holds the drive wheel?
2. You mention @400 RPM. Is this the rotational speed you wish to accomplish in the final drive, or do you mean to accomplish @400 SFPM on the belt?
3. If your goal is SFPM, what is the diameter of your drive wheel (the wheel that powers the sanding belt)?
4. Belt specs will affect the final outcome. By far, the most common for this type of application is the A/4L size. Is this what you're using?

More to come.
 
Still Dazed & Confused

Thank You for your response!

Grinder

KOVAL 2X72" Belt Grinder I got it in my teens the mid 80's from a knife maker and supplier from Ohio by the last name KOVAL HIS CO. HAS SINCE BEEN ACQUIRED by another knife maker supply Co. that no longer seems to carry the machine any longer and my machine is inaccessible to me at the moment, (I'm even unsure of new owner's

name at the moment but it might be Jantz or something?):31:
DSCF3582.jpg

but anywho here's a pic of a similar type of setup (I found the pic on the net an saved it to my PC) and in case it does not show for some reason I will also describe how mine is driven.

The motor is separate from the grinder body and has a V belt type wheel mounted to its output shaft and a belt goes from there to a shaft that also has a V B wheel mounted to 1 side and on the other side of this shaft is mounted either an 8" or 10" serrated rubber coated contact wheel, I am unsure of either of the 2 shaft sizes involved here. (Back in my day this grinder was a very popular starter grinder with the poeboys that wanted to get into knife making on the cheap even though motor+machine went @ $600.00 ish)

RPM VS SFPM
answer I am unsure, but as long as I can achieve my goal of slow enough speed sharpening polishing of edges in order to not have too much risk of over heating the edge while mastering the skill W/O having to water dip the blades about every 3 seconds or so then I will be a happy camper.

I hope this helps you enough for you to be able to help advise me.

Thank you very much in advance!:biggrin:
 
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On your machine, the enemy of slow speed is the contact wheel. The 10" wheel turns out to be your drive wheel. There is a lot of math, but the simple explaination is that using say a 2" drive pulley on the motor - and a 5" driven pulley on the shaft for the contact wheel, you are going to get down to about 700 RPM. That sounds pretty good until you realize that 700 roations of that 10" wheel moves over 1800 SFM of belt. That's OK for sharpening but you have to dip to keep it cool if you're doing much work on an edge.
 
Posting a photo of your machine really helps clear things up.

Knifemaker.ca is correct. That large contact wheel (which also functions as the drive wheel) will make massive speed reduction more difficult.

If you're wanting to slow your sanding belt's surface speed down to about 400 SFPM, while still using a 10" drive wheel, you're gonna need to have a drive shaft speed of 153 RPM. With 1750 RPM at the motor, that's a speed reduction ratio of over 11:1, which is not good for a 2-sheave setup.

Consider going to a much smaller drive wheel as the first and most important step to attaining your reduction in surface speed.
Your goal is to lower the speed reduction ratio required between the motor shaft and the drive shaft. Large ratios are not efficient. The most evident problem is reduced wrap angle (belt contact) on the smaller sheave. Slipping of the v-belt leads to rapid wear of both the belt and the groove in the sheave. It also wears out your bearings very quickly; since you're over-tightening the v-belt to compensate for slippage, you're asking the bearings to withstand a higher load.
In a nutshell, all components of the system wear more rapidly if large reductions are obtained through the use of only two shafts.


So how is one to deal with a large reduction ratio? Assuming we're still talking about the use of v-belts and sheaves as the method of power transmission, here are a couple of options.

1. The best method is multiple reduction. With this method, you have more than two shafts. For example, the motor shaft speed is reduced by an intermediate shaft. Then the intermediate shaft transmits the power at yet another reduction to the final drive shaft.
This allows you to achieve the desired final reduction in "small steps" rather than one giant leap.

2. Some situations require large ratios while still using only two sheaves. An example would be an application that operates under space constraints (no room for intermediate shafting). Or possibly the reason is because the budget is very tight and the expense of the extra components (extra set of bearings, two additional sheaves, another shaft) makes it cost-prohibitive. Whatever the reason, you have decided to still obtain a large reduction yet still use only two sheaves (driver on the motor shaft, driven on the drive shaft).
There is a method that still allows this, but it is very compromising. The use of an outside idler pulley will increase the amount of contact the v-belt has on the sheaves. Increasing the contact the belt has with the sheave reduces the increased slippage inherent with large ratios, but does so at the expense of belt life. The addition of the idler adds another point around which the belt must flex.
I have used the idler approach a couple of times, and I can say from experience that if this is the method you plan to use, you had best have a lot of spare belts nearby.


But all of the above-mentioned compromises can be minimized if you lower the ratio (the relationship of RPM at the motor shaft compared to the RPM at the drive shaft).
Since you're wanting to obtain lower surface speeds on the sanding belt, you're best approach (mentioned above) is to reduce the diameter of the final drive wheel.
Assuming a sanding belt speed of 400 SFPM is your goal, here is an example for your machine:
A 4" drive wheel requires 382 RPM on the drive shaft, which equates to a ratio of 4.58:1 Now we're getting into an area where using two sheaves (pulleys) could be effective.

Were I "in your shoes", the first thing I would do is replace the large 10" contact wheel with a 4" drive wheel.
Doing so means that you no longer have the use of that large contact surface, unless you like going through the inconvenience of constantly switching out wheels.
The tension adjustment on your machine may need to be altered due to the fact that you're still using a 72" sanding belt. For your machine, the range of adjustment is based upon the height of the vertical column. Some situations may require replacing the original steel tube with a longer piece.

Take a close look at your machine and decide if you want to eliminate the large contact wheel and replace it with a smaller-diameter wheel. Only then can we decide the diameter of the sheaves (pulleys) you need.
 
RE: Knifemaker,Ca 's response

On your machine, the enemy of slow speed is the contact wheel. The 10" wheel turns out to be your drive wheel. There is a lot of math, but the simple explaination is that using say a 2" drive pulley on the motor - and a 5" driven pulley on the shaft for the contact wheel, you are going to get down to about 700 RPM. That sounds pretty good until you realize that 700 roations of that 10" wheel moves over 1800 SFM of belt. That's OK for sharpening but you have to dip to keep it cool if you're doing much work on an edge.




Knifemaker,CA

Thank you very much for the info, and most importantly for doing that nasty evil math for me lol! :35:

:les:
So if a 2" Drive Sheave mounted on the motor's own shaft and a 5" Driven Sheave which would be mounted to the shaft that spins the (assumed) 10" contact wheel gets me 700RPM of said 10" Contact wheel which nets me a bit more than 1800 SFM of belt speed, then what SFM would say a 2" or if I could find a 1.5" or 1 & 7 eighths D P and a 6" D P get me and maybe I could even find a bit smaller of a contact wheel as well (as $ allows) to go along with the other 2" an 6" DP's?


Thank you very much again for the above answers as well as any that are forthcoming !:biggrin:
 
Yes, it can be done.

A few questions:
1. I am not familiar with the machine you mention. Is it the type with a motor transmitting power to one driveshaft? In other words, is this a 2-sheave setup,=yes where one sheave is on the motor and another is on the driveshaft which holds the drive wheel?
2. You mention @400 RPM. Is this the rotational speed you wish to accomplish in the final drive, or do you mean to accomplish @400 SFPM on the belt? I'm still unsure?:les:
3. If your goal is SFPM, what is the diameter of your drive wheel (the wheel that powers the sanding belt)?
4. Belt specs will affect the final outcome. By far, the most common for this type of application is the A/4L size. Is this what you're using? at this time I do not know the designation of the belt that is mounted on my machine other than I remember that a old car type (think old style fan belt) V belt was mounted on the machine but I would be willing to run whatever length of belt necessary if that is what you mean? :biggrin:

More to come.

Thank you very much Rob for your response!
 
RE: Rob 's response

Posting a photo of your machine really helps clear things up.

Knifemaker.ca is correct. That large contact wheel (which also functions as the drive wheel) will make massive speed reduction more difficult.

If you're wanting to slow your sanding belt's surface speed down to about 400 SFPM, while still using a 10" drive wheel, you're gonna need to have a drive shaft speed of 153 RPM. With 1750 RPM at the motor, that's a speed reduction ratio of over 11:1, which is not good for a 2-sheave setup.

Consider going to a much smaller drive wheel as the first and most important step to attaining your reduction in surface speed.
Your goal is to lower the speed reduction ratio required between the motor shaft and the drive shaft. Large ratios are not efficient. The most evident problem is reduced wrap angle (belt contact) on the smaller sheave. Slipping of the v-belt leads to rapid wear of both the belt and the groove in the sheave. It also wears out your bearings very quickly; since you're over-tightening the v-belt to compensate for slippage, you're asking the bearings to withstand a higher load.
In a nutshell, all components of the system wear more rapidly if large reductions are obtained through the use of only two shafts.


So how is one to deal with a large reduction ratio? Assuming we're still talking about the use of v-belts and sheaves as the method of power transmission, here are a couple of options.

1. The best method is multiple reduction. With this method, you have more than two shafts. For example, the motor shaft speed is reduced by an intermediate shaft. Then the intermediate shaft transmits the power at yet another reduction to the final drive shaft.
This allows you to achieve the desired final reduction in "small steps" rather than one giant leap.

2. Some situations require large ratios while still using only two sheaves. An example would be an application that operates under space constraints (no room for intermediate shafting). Or possibly the reason is because the budget is very tight and the expense of the extra components (extra set of bearings, two additional sheaves, another shaft) makes it cost-prohibitive. Whatever the reason, you have decided to still obtain a large reduction yet still use only two sheaves (driver on the motor shaft, driven on the drive shaft).
There is a method that still allows this, but it is very compromising. The use of an outside idler pulley will increase the amount of contact the v-belt has on the sheaves. Increasing the contact the belt has with the sheave reduces the increased slippage inherent with large ratios, but does so at the expense of belt life. The addition of the idler adds another point around which the belt must flex.
I have used the idler approach a couple of times, and I can say from experience that if this is the method you plan to use, you had best have a lot of spare belts nearby.


But all of the above-mentioned compromises can be minimized if you lower the ratio (the relationship of RPM at the motor shaft compared to the RPM at the drive shaft).
Since you're wanting to obtain lower surface speeds on the sanding belt, you're best approach (mentioned above) is to reduce the diameter of the final drive wheel.
Assuming a sanding belt speed of 400 SFPM is your goal, here is an example for your machine:
A 4" drive wheel requires 382 RPM on the drive shaft, which equates to a ratio of 4.58:1 Now we're getting into an area where using two sheaves (pulleys) could be effective.

Were I "in your shoes", the first thing I would do is replace the large 10" contact wheel with a 4" drive wheel.
Doing so means that you no longer have the use of that large contact surface, unless you like going through the inconvenience of constantly switching out wheels.
The tension adjustment on your machine may need to be altered due to the fact that you're still using a 72" sanding belt. For your machine, the range of adjustment is based upon the height of the vertical column. Some situations may require replacing the original steel tube with a longer piece.

Take a close look at your machine and decide if you want to eliminate the large contact wheel and replace it with a smaller-diameter wheel. Only then can we decide the diameter of the sheaves (pulleys) you need.

Sorry rob I must have missed this response to my ?

Okay I might just be starting to get a little of this thanks to everyone's gracious attempts to extract my head from the land of the lost! lol :3:

My funds are extremely limited right now to just buying the 2 pulley sheaves, and some abrasive belts, and I also saw a 6" sheave for sale somewhere on the WWW, but in the future in order to get the very slow/cool sharpening speed I want (400SFPM?) I may opt to also get a 2" to 4" Contact wheel providing that I can get my machine to accommodate them W/O more hassle (complicated mods) than I can deal W/ at that time.

Thank you very kindly for your response!:biggrin:
 
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