Motor torque question for a tiller
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Baby Grand
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Assuming 5 hp/ft ... 20 HP @ 540 Rpm, and assuming your forward speed is enough that it's starting to drag down the RPMs, because I don't know if your soil is light or heavy:
T = HP * 5252 / RPM
T = 195 foot pounds
x12 = 2334 inch pounds
T = HP * 5252 / RPM
T = 195 foot pounds
x12 = 2334 inch pounds
Baby Grand
Elite Member
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- Tractor
- Kubotas: L3240GST B2320HST B5100D & G5200H
5 hp/ft is just an approximate guide for sizing implements that I learned from others, here on TBN.
Power required to run a single tine will vary widely by differences in tine design, reverse or forward rotation, PTO speed, forward tractor speed, operating, # of tines/ft, operating depth and soil type.
To determine hp/tiller tine you really have to back into that one: divide hp by the # of tines.
You can check that approximation by examining a few different tillers and dividing the minimum recommended hp by the # of tines.
Within a group of similar tiller types you should see a fairly constant value.
Power required to run a single tine will vary widely by differences in tine design, reverse or forward rotation, PTO speed, forward tractor speed, operating, # of tines/ft, operating depth and soil type.
To determine hp/tiller tine you really have to back into that one: divide hp by the # of tines.
You can check that approximation by examining a few different tillers and dividing the minimum recommended hp by the # of tines.
Within a group of similar tiller types you should see a fairly constant value.
Thanks.
If I go with two tines per foot that would be 2.5hp per tine, so:
T = HP * 5252 / RPM
T = 24.3149 foot pounds
T x12 = 291.7778 inch pounds
24.3 ft pounds does not seem to be much torque when comparing it to the amount of force used to tighten a bolt. I don't mean to sound uneducated but is my comparison of the required torque to a torque wrench force incorrect? Sorry for the rookie question that's probably been asked many times before.
If I go with two tines per foot that would be 2.5hp per tine, so:
T = HP * 5252 / RPM
T = 24.3149 foot pounds
T x12 = 291.7778 inch pounds
24.3 ft pounds does not seem to be much torque when comparing it to the amount of force used to tighten a bolt. I don't mean to sound uneducated but is my comparison of the required torque to a torque wrench force incorrect? Sorry for the rookie question that's probably been asked many times before.
The reason I ask is my tractor has max 18hp at the PTO and the highest torque I can run using two hydraulic motors is around 450 inch pounds per motor, which based on the reply above is about 37 ft pounds. Because each of these motors will be used to directly drive an 12" dia. in-row cultivator blade 37ft pounds does not seem like enough torque when using a torque wrench as a tangible comparison.
The largest PTO driven pump I can use can be found here:
http://www.northerntool.com/images/downloads/manuals/1050.pdf
And the motors I have chosen to run in parallel using a flow divider valve:
1.93 cu in CHAR-LYNN 129-0004 HYD MOTOR
At 18hp I do not have enough power to drive a pump at > 1800 psi in order to drive any higher torque motors in parallel. I find it odd that 18 hp is not enough to power two < 5hp motors. That is only < 55% efficiency.
5hp x 5252/ 660rpm = 39ft lb = 468 inch lb
I hope have misunderstood something, any guidance or correction would be appreciated.
The largest PTO driven pump I can use can be found here:
http://www.northerntool.com/images/downloads/manuals/1050.pdf
And the motors I have chosen to run in parallel using a flow divider valve:
1.93 cu in CHAR-LYNN 129-0004 HYD MOTOR
At 18hp I do not have enough power to drive a pump at > 1800 psi in order to drive any higher torque motors in parallel. I find it odd that 18 hp is not enough to power two < 5hp motors. That is only < 55% efficiency.
5hp x 5252/ 660rpm = 39ft lb = 468 inch lb
I hope have misunderstood something, any guidance or correction would be appreciated.
bigdeano
Platinum Member
I put a hydraulic powered tiller on my garden tractor. From the published figures for the pump and motor I used, 19 hp input results in 10.2 hp output. The tiller is 42" wide. It works, but will stall if I go over 6" deep, and the ground has already been loosened up with the 18" plow on my big tractor. Every time I use the tiller I wish I'd used mechanical drive instead of hydraulic. You're only going to be 24" wide and not going very deep, so you'll probably be ok.
Forgot to mention, I'm turning the tines at about 250 rpm if I remember correctly.
Forgot to mention, I'm turning the tines at about 250 rpm if I remember correctly.
J_J
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Your 18 HP can only provide about 11.4 GPM.
Your motors run at 1750. so you will need a relief valve.
That is 5.6 GPM through each motor.
5.6GPM at 1750 psi for the motors, and 1.93 cu in = 538 in lbs.
Each motor can develop about 5.61 HP X 2 = 11.22 HP
The pump you selected was a 40 GPM pump and at 2250 psi, it would take about 62 HP
I did not take into account the efficiency of hyd pump HP to the HP develop by hyd motors.
Your motors run at 1750. so you will need a relief valve.
That is 5.6 GPM through each motor.
5.6GPM at 1750 psi for the motors, and 1.93 cu in = 538 in lbs.
Each motor can develop about 5.61 HP X 2 = 11.22 HP
The pump you selected was a 40 GPM pump and at 2250 psi, it would take about 62 HP
I did not take into account the efficiency of hyd pump HP to the HP develop by hyd motors.
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