grinding wheel question

[farmENGR]

Hi all,

I am looking to find how long it takes a grinding wheel to tavel 10'' along a workpiece. The grinding wheel revolves at 60 fpm.

My grinding wheel is 16'' in OD and 1'' wide.
The workpiece diameter is 3''.
feed is 1/4'' per revolution.

I cannot seem to find the correct equation.

thanks

 

[farmENGR]

i get 28.2 minutes. I don't think that is right though.

I am using the time to machine equation for a turning operation.
Tm= (L + Ap + Ao)/ f x rpm

The rpm i find is very low. 14 rpm

 

[SnTMan]

farmENGR, assuming 60 surface ft/min at wheel OD, ignoring workpiece dia. & wheel width:

16/12 * pi = 4.19 ft circumference (in / in * ft^-1)

60/4.19 = 14.3 r/m (ft * min^-1 / ft * rev^-1)

14.3 x 0.25 = 3.58 in / min (rev * min^-1 * in * rev^-1)

120 / 3.58 = 33.5 min (in / in * min^-1)

Regards,

Mike

 

[farmENGR]

i do not understand what you have done.


can you explain?

 

[Occupant]

I have (just like SnTMan) 14.324 Rev's/min or 0.0698 min/Rev x 44 Rev's (11/0.25) = 3.07min, that is if you want the wheel entirely off the workpiece.

 

[farmENGR]

Occupant,

STNMan calculated 33 minutes and you calculated 3 minutes. Something doesn't seem right

 

[SnTMan]

farmENGR, I have done the numbers based on the geomerty and data given. The notations in parentheses are a carry-through of the units involved to insure each stage of the answer had the right units. If I learned one useful thing in college, that was it

For example the first calculation (in / in * ft^-1) is shorthand for (inches) divided by (inches per foot), giving units of foot.

1)Circumference of the grinding wheel in feet.
2)Conversion of ft/min to rev/min
3)Conversion of rev/min to in/min
4)Calculation of time for given inches travel at calculated in/min rate

I see now my calculated time is off by a factor of 12, 10 inches was given not 10 feet. Duh.

Should be 2.79 minutes. Sorry for the inconvenience. Decent agreement w/ Occupant, given the slightly different travel assumed by each.

Regards,

Mike

 

[farmENGR]

After looking at my work i was in inches and not feet for my first calculation and after seeing feet/ min i saw i made a mistake too.

I come up with

10in/12in *ft-1= 0.83 ft to tavel

Turning operation:

RPM= 12*V/pi * D = 12(60)/ 3.14 * 3 = 76.28 rev * min-1

0.06 and 0.06 are used as Ap and Ao values. Standard i think
Time to tavel =0.83ft + 0.06in + 0.6in / .025*rev-1 * 76 rev *min-1 = 0.49 minutes

I do not need the OD of the grinding wheel or the width of it. I could not calc sfpm or velocity because i did not have rpm of the wheel. I would have needed something.

what do you think sTnMan?

 

[SnTMan]

farmENGR, I think a problem properly stated is a problem half solved

Reworking my calcs with the first calculation using 3 inch instead of 16 (0.785 ft) I get 0.52 min, close enough to your 0.49 as makes no difference, at least for ONE piece.

Have no idea what the Ap, Ao terms represent. not sure I trust your feed, I get .0208 ft/rev. Also I infer that (L + Ap + Ao) is divided by (f x rpm).

You will get the idea by now I know nothing really about the process, just crunching the numbers given

Regards,

Mike

 

[SnTMan]

So here's what I interpret:

The grinding wheel is turning 60 fpm, as is the 3 inch workpiece.
The feed is 1/4 in per rev of the workpiece, not the grinding wheel.

 

[MikeHalloran]

Uh, 60 sfpm seems a reasonable speed to rotate a workpiece in a grinder so as to not 'crowd' the abrasive. That and the workpiece diameter gives you an rpm for the workpiece spindle, which with the feed/rev of the workpiece and the workpiece length gives you a cycle time.

But 60 sfpm is _way_ slow for a grinding wheel.



Mike Halloran
Pembroke Pines, FL, USA

 

[racookpe1978]

Also, there's no "realistic" way to feed anything at 1/4 inch per revolution of the grinding wheel ......

That would be feedrate for a circular saw perhaps.

 

[Jaydenn]

The feedrate will be linked to the work RPM... Not the wheel RPM.
Figure out the work RPM.

Time will be:

Distance to be traversed is 10 inches.
Feedrate is 0.250 inches per revolution.

So,

10 * 0.250 = 40.

Think about it. Every 4 revolutions, you will traverse 1 inch.

So, if you are spinning at 40 RPM you will traverse the part in 1 minute.

If your actual RPM is different, you can divide to get the time.

----

If the work speed you quoted is correct at 60SFPM and the part is 3 inches in diameter the RPM will be approx 76 RPM.
(Based on the formula (3.81 * cutting speed)/diameter)

At that RPM, it will take just over 30 seconds to traverse the part. 0.526 minutes to be exact.

That is my Guerrilla math method of solving this. Hope it helps rather than hinders.

J

NX 6.0.5.3

 

[farmENGR]

Jaydenn,

You seem to have gotten very close to my result.

Where does the 3.81 come from?

thanks,
Joe

 

[farmENGR]

Jaydenn,

You also say

"if you are spinning at 40 RPM you will traverse the part in 1 minute."

how are you coming up wth this conclussion?

 

[Jaydenn]

FarmENG,

The 3.81 is an approx. constant from the RPM calculation. The full formula is (cutting speed * 12)/(Pi * Diameter)

The 3.81 comes from 12 divided by Pi(3.14159265) when you re-write the formula.
It's a machinist's shortcut.

The "40rpm in one minute" is more intuitive.

The 1 minute comes from the base units "RPM". Revolution per minute. That how many turns you make each minute.

So, I work backwards from the 1 minute.

Your stated feedrate is 0.250" per revolution; therefore, if you turned 1 rpm, you would move 0.250" in one minute.

So, if you turned 4 rpm, you would move 1" in one minute, right?

Making the next leap, 40 rpm will move you 10" in one minute.

Make sense?

J


NX 6.0.5.3