leonz
Super Member
I want and hope to provide a little more backround information for everyone as far as a single stage snow throwers capacity for clearing snow pack.
Using the Riest 48 inch single stage snow thrower for the example I hope to show how everyone how efficient the single stage system is.
The information I have listed below is from tractor data.
SO lets start with the Mule and for the sake of saying it an Kubota L2800 gear drive with the 1.5 liter engine is used in the example.
1. 30 gross horsepower
2. 2,800 RPM at high idle
3. constant mesh hydro transmission with 8 forward speeds 4 reverse speeds with a shuttle shift between froward and reverse.
4. 25 horse power at the power take off with the later gear driven tractor models.
5. the engine RPM is 2,430 RPM at the 540 RPM PTO Speed
The Riest folks claim that their single stage units only require 80% of the power needed to operate a 2 stage unit of the equivalent size snow blower.
SO if a 2 stage unit operates at 2,430 RPM engine speed with the 540 RPM PTO speed;
The 48 inch single stage will only require 80% percent of the engine speed needed which is 1,944 RPM to spin the snow blower rotor through the PTO at 540 rpm and the snow blower rotor will spin at 600 RPM and in their saying that the tractor will have 20 percent of its engine speed and torque in reserve with the single stage 48 inch unit.
SO using the formula for calculating the volume of a cylinder to determine the volume of snow moved through the 20 inch snow blower rotor we would reduce the volume by the volume of the auger ribbons and the four paddles.
SO using a 90 percent figure for the volume of the cylinder of a 20 inch snow blower rotor that is 48 inches wide we will begin here.
Let us assume the augers are 4 inches deep and the solid portion of the snow blower rotor is 12 inches in diameter 4" + 4" + 12" equals 20 inches in outside diameter.
Area= 0.7854 * d squared minus f squared
.7854 * 400-144= 256 square inches
SO doing some short cutting I deducted the volume of a 12 by 48 inch cylinder from a 20 inch by 48 inch cylinder to obtain the volume of the annulus which is 3,600 cubic inches (rounded lower).
3,600 cubic inches divided by 1728 cubic inches(one cubic foot) equals 2.083 cubic feet so we will say its 2 cubic feet=3,456 cubic inches for the sake of simplicity.
SO if one were to encounter a 20 inch deep snow pack the snow blower would be discharging 2 cubic feet per revolution of the snow blower rotor and at 600 revolutions per minute that would be 1200 cubic feet per minute= 36,000 pounds per minute of snow removal
A front mount Riest unit will allow the user to operate the snow thrower at an even faster speed with a mid mount PTO system if the Mid mount PTO is operating at 2,500 RPM through a reverser and different gear set on the snow thrower.
I used a volume calculation of 20 inches by 48 inches by 2,400 inches in length as a starting point
Making 2,304,000 cubic inches of snow pack divided by 1728 inches per cubic foot equals 1,334 cubic feet in volume times 30 pounds per cubic foot=40,000 pounds(rounded lower)=20 US tons
So just to be conservative driving slower would let us remove less tonnage per hour;
One mile per hour would create 88 feet per minute ground speed; a cut 48 inches wide and 20 inches high operating at 600 revolutions per minute would clear 800 cubic feet (rounded higher) per minute times 30 pounds per cubic foot equals 24,000 pounds per minute=12 tons per minute.
It would or could be less of course due to the volume of the auger ribbons and the four paddles in the snow blower rotor and any loss of available traction if ice is encountered.
I think my math is correct.
Using the Riest 48 inch single stage snow thrower for the example I hope to show how everyone how efficient the single stage system is.
The information I have listed below is from tractor data.
SO lets start with the Mule and for the sake of saying it an Kubota L2800 gear drive with the 1.5 liter engine is used in the example.
1. 30 gross horsepower
2. 2,800 RPM at high idle
3. constant mesh hydro transmission with 8 forward speeds 4 reverse speeds with a shuttle shift between froward and reverse.
4. 25 horse power at the power take off with the later gear driven tractor models.
5. the engine RPM is 2,430 RPM at the 540 RPM PTO Speed
The Riest folks claim that their single stage units only require 80% of the power needed to operate a 2 stage unit of the equivalent size snow blower.
SO if a 2 stage unit operates at 2,430 RPM engine speed with the 540 RPM PTO speed;
The 48 inch single stage will only require 80% percent of the engine speed needed which is 1,944 RPM to spin the snow blower rotor through the PTO at 540 rpm and the snow blower rotor will spin at 600 RPM and in their saying that the tractor will have 20 percent of its engine speed and torque in reserve with the single stage 48 inch unit.
SO using the formula for calculating the volume of a cylinder to determine the volume of snow moved through the 20 inch snow blower rotor we would reduce the volume by the volume of the auger ribbons and the four paddles.
SO using a 90 percent figure for the volume of the cylinder of a 20 inch snow blower rotor that is 48 inches wide we will begin here.
Let us assume the augers are 4 inches deep and the solid portion of the snow blower rotor is 12 inches in diameter 4" + 4" + 12" equals 20 inches in outside diameter.
Area= 0.7854 * d squared minus f squared
.7854 * 400-144= 256 square inches
SO doing some short cutting I deducted the volume of a 12 by 48 inch cylinder from a 20 inch by 48 inch cylinder to obtain the volume of the annulus which is 3,600 cubic inches (rounded lower).
3,600 cubic inches divided by 1728 cubic inches(one cubic foot) equals 2.083 cubic feet so we will say its 2 cubic feet=3,456 cubic inches for the sake of simplicity.
SO if one were to encounter a 20 inch deep snow pack the snow blower would be discharging 2 cubic feet per revolution of the snow blower rotor and at 600 revolutions per minute that would be 1200 cubic feet per minute= 36,000 pounds per minute of snow removal
A front mount Riest unit will allow the user to operate the snow thrower at an even faster speed with a mid mount PTO system if the Mid mount PTO is operating at 2,500 RPM through a reverser and different gear set on the snow thrower.
I used a volume calculation of 20 inches by 48 inches by 2,400 inches in length as a starting point
Making 2,304,000 cubic inches of snow pack divided by 1728 inches per cubic foot equals 1,334 cubic feet in volume times 30 pounds per cubic foot=40,000 pounds(rounded lower)=20 US tons
So just to be conservative driving slower would let us remove less tonnage per hour;
One mile per hour would create 88 feet per minute ground speed; a cut 48 inches wide and 20 inches high operating at 600 revolutions per minute would clear 800 cubic feet (rounded higher) per minute times 30 pounds per cubic foot equals 24,000 pounds per minute=12 tons per minute.
It would or could be less of course due to the volume of the auger ribbons and the four paddles in the snow blower rotor and any loss of available traction if ice is encountered.
I think my math is correct.
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