The other day there was some discussion on the main board about the cutting quality of various mower decks. It was suggested by some that the Crapsman deck cut very well. So what makes a mowing deck better than another - is it cutting blade design, blade tip speed or what?

I pondering this (as it continues to rain) with my coffee this morning & I decided to try to determine the cutting blade tip speed on my 42" IHC mowing deck. Now those of you way smarter than me please review my (likely) mathematical errors, and if so, PLEASE point it out to me.

Ok, here goes --

Things I assumed:

My 12HP Kohler will be run at the recommended full throttle crankshaft speed of 3600 rpms.

The PTO is directly connected to the crank. The mechanical PTO pulley (driver) is 4 3/8” (4.375") in diameter. So it is also turning at 3600 rpms.

The (driven) pulley on the top, center of the deck is 6 ¼” (6.25”) in diameter.

So: 4.375/ 6.25 = 0.70

3600 x 0.70 = 2520 rpms

Now, the sheave below the top center deck pulley which is being driven is 7-1/8” (7.125”) in diameter. It is turning at the same rpm as the sheave above it, 2520 rpms.

The center pulley drives the two outer blade pulleys, which are only 3” in diameter.

So, the driver pulley is 7.125” @ 2520 rpms.

7.125 / 3.0 = 2.375

2520 x 2.375 = 5980 rpms

Assuming my above math is correct, it's interesting to note that the two outer blades turn at nearly twice the speed of the center blade......

Blade tip speeds:

The center spindle is driven 2520 rpms. The center blade is 22” long, so this is equivalent to a pulley 22” pulley. The circumference is the distance around the imaginary 22" pulley. So, Circumference (C) = Diameter (D) x Pi (3.1416)

C = 22" x 3.1416

C = 69.11”

So, the center 22” blade tip is making a 69.11” trip with every rpm =

69.11" x 2520 RPM = 174,157 inches per minute.

174,157 / 12"/ft =

14,513 feet per minute (FPM).

The two outer blades are 10.75” long or equivalent to a 10.75” pulley.

10.75” x 3.1416 = 33.77” circumference

33.77” x 5980 rpms = 201,944 inches per minute

201,944 / 12 =

16,829 feet per minute (FPM).

So, again, assuming my above math is correct, then we've determined that my 42" IHC deck cuts grass with spindle speeds of 2600 to 6000 rpms and produce blade tip speeds of 14K to 17K FPM.

First off, is my math correct?

Secondly, I KNOW modern mowing decks have a higher blade tip speed. Does anyone know just how fast do they go?

Does anyone know what the blade tip speed of a Crapsman deck is? If not, if someone would provide some engine rpm and sheave diameter measurements here, we'd be able to calculate them for discussions sake.....

Thanks for any and all input, opinions, & discussion!

Ryan Wilke