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I'm still pondering and researching and scribbling in my spare time with the intent of building a small crawler. I'm kind of a buy-once cry-once measure twice cut once individual (only because of multiple, multiple past mistakes), so I'm trying to make sure that what I build is going to do what I want.
Basically, I'm looking at this ZTR setup: https://www.surpluscenter.com/item.asp?item=13-1489&catname=powerTrans
As a way of driving and controlling the drive sprockets for my track frames. SC specs that it is capable of delivering 115 ft/lb of torque continuous, or 150 intermittent.
So what does that mean (assuming I have the engine power necessary to develop that at the speed I want to apply it)?
115 ft/lb of torque, driving a 12" sprocket. Sprocket circumference is 37.7" or 3.14'. Torque arm length is 6" or 0.5'.
115 ft/lb / 0.5 = 230 pounds of linear force x 2 (pump/motors) = 460 lbs of linear force forward?
150 RPM * 3.14 feet = 471 ft/min * 60 = 28,260 ft/hr / 5280 = 5.35 MPH?
5 MPH I'm happy with as a top speed for a crawler that would for the most part be used to move snow. But 460 lb of continuous linear force sounds light to me? Even at intermittent 150 ft/lbs of torque, linear force comes out to only 600 lbs? That sounds a little better, and would be closer to the point of the tracks spinning on the ground rather than pushing the snow in front of it.
My thinking is that if the crawler + me = >600lbs it would never lose traction before it ran out of power. Would that be accurate? If I want to effectively push snow, do I need to look for a bigger pump/motor combo with higher torque output?
Basically, I'm looking at this ZTR setup: https://www.surpluscenter.com/item.asp?item=13-1489&catname=powerTrans
As a way of driving and controlling the drive sprockets for my track frames. SC specs that it is capable of delivering 115 ft/lb of torque continuous, or 150 intermittent.
So what does that mean (assuming I have the engine power necessary to develop that at the speed I want to apply it)?
115 ft/lb of torque, driving a 12" sprocket. Sprocket circumference is 37.7" or 3.14'. Torque arm length is 6" or 0.5'.
115 ft/lb / 0.5 = 230 pounds of linear force x 2 (pump/motors) = 460 lbs of linear force forward?
150 RPM * 3.14 feet = 471 ft/min * 60 = 28,260 ft/hr / 5280 = 5.35 MPH?
5 MPH I'm happy with as a top speed for a crawler that would for the most part be used to move snow. But 460 lb of continuous linear force sounds light to me? Even at intermittent 150 ft/lbs of torque, linear force comes out to only 600 lbs? That sounds a little better, and would be closer to the point of the tracks spinning on the ground rather than pushing the snow in front of it.
My thinking is that if the crawler + me = >600lbs it would never lose traction before it ran out of power. Would that be accurate? If I want to effectively push snow, do I need to look for a bigger pump/motor combo with higher torque output?