kubota vs. kioti

[LD1]

My Cubota rig has 10% more horsepower than my brother's Kioti tractor but I doubt it can pull like the Kioti. Toque vs HP?

IT has nothig to do with torque vs horsepower.

It has EVERYTHING to do with machine size, traction and gearing.

A small 70 or 80hp tractor with a few hundred ft-lbs or torque has more pull than a 400hp/800ft-lb new diesel 1-ton. Its all about weight, geating and traction. Nothign at all to do with the motor.

 

[oldpilgrim]

IT has nothig to do with torque vs horsepower.

It has EVERYTHING to do with machine size, traction and gearing.

A small 70 or 80hp tractor with a few hundred ft-lbs or torque has more pull than a 400hp/800ft-lb new diesel 1-ton. Its all about weight, geating and traction. Nothign at all to do with the motor.

maybe I should have used a few more smiley faces in my post.

 

[LD1]

maybe I should have used a few more smiley faces in my post.

My reply would have been the same.

Too often people want to compare a tractor engine to a mower by referencing how much they can pull/tow.

And it has nothing to do with that. Think of it this way. A small 27hp kohler or honda or briggs or (pick your poison) zero turn or other mower. Now compare that to something like a 20hp tractor with a bigger engine and a mower deck. The ZTR will power through tougher thicker taller grass at a faster rate. Regardless or torque difference.

The engines (amd machine) are geared for different things. A tractor is geared for pulling. And has the weight and traction to maximize that.

 

[tomato coupe]

Okay, who wants a trip back to high school physics? Here's a problem for you to solve:

How fast can a 4400 lb (2000 kg) tractor go up a 30 degree incline powered by an engine with performance shown in the attached graph? The graph is for a an actual Kubota engine that produces 50 hp (@2700 RPM) and 110 lb-ft of torque (@1600 RPM). Let's assume the tractor has a 12 speed, geared transmission that can cover the range of 1-12 mph, and it is lossless. Let's also assume that road friction is negligible. I recommend you use metric units for the calculation, and then convert the answer to mph at the end. (It's much, much easier that way.)

To save you some time looking for your old text books, here are some hints:

1. The work performed is equal to m * g * h, where m is the mass of the tractor, g is the acceleration due to gravity (9.8 m/s), and h is the vertical height the mass is lifted. Work is measured in joules (J).
2. Power (P) is the rate at which work is performed. It is measured in W (watts), which is the same as J/s (joules per second). P = (m * g * h) / t, where t is time.
3. For a 30 degree incline, the distance traveled is double the height the mass is lifted.

and some conversion factors:

1 m/s = 2.24 mph (velocity)
1 HP = 746 W (power)
1 kg = 2.2 lb (mass)
1 N-m = .74 lb-ft (torque)

It shouldn't take more than a few minutes to come up with the answer.

Cheers

 

[sawtooth]

Okay, who wants a trip back to high school physics? Here's a problem for you to solve: How fast can a 4400 lb (2000 kg) tractor go up a 30 degree incline powered by an engine with performance shown in the attached graph? The graph is for a an actual Kubota engine that produces 50 hp (@2700 RPM) and 110 lb-ft of torque (@1600 RPM). Let's assume the tractor has a 12 speed, geared transmission that can cover the range of 1-12 mph, and it is lossless. Let's also assume that road friction is negligible. I recommend you use metric units for the calculation, and then convert the answer to mph at the end. (It's much, much easier that way.) To save you some time looking for your old text books, here are some hints: 1. The work performed is equal to m * g * h, where m is the mass of the tractor, g is the acceleration due to gravity (9.8 m/s), and h is the vertical height the mass is lifted. Work is measured in joules (J). 2. Power (P) is the rate at which work is performed. It is measured in W (watts), which is the same as J/s (joules per second). P = (m * g * h) / t, where t is time. 3. For a 30 degree incline, the distance traveled is double the height the mass is lifted. and some conversion factors: 1 m/s = 2.24 mph (velocity) 1 HP = 746 W (power) 1 kg = 2.2 lb (mass) 1 N-m = .74 lb-ft (torque) It shouldn't take more than a few minutes to come up with the answer. Cheers

I need another beer first.

 

[LD1]

I didn't use metric, and not sure if it works out the same, but here is what I did.

The fundamental definition of a HP is 33000 ft-lbs per minute. Or the ability to lift 33000# 1' in 1 minute.

Multiply that by 60 to get an hourly rate, and multiply again by the 50hp the tractor has and you get the total weight the tractor has the ability to lift 1ft in 1 hr.

Divide that by the tractors weight, and you get the vertical distance it can lift itself in 1 hr measured in ft. Multiply that by 2 for the 30 degree slope, and divide by 5280 to get miles per hour. And I get 8.5 is the speed it would maintain. And it would take 70hp for it to maintain the 12mph.

 

[tomato coupe]

I didn't use metric, and not sure if it works out the same, but here is what I did.

The fundamental definition of a HP is 33000 ft-lbs per minute. Or the ability to lift 33000# 1' in 1 minute.

Multiply that by 60 to get an hourly rate, and multiply again by the 50hp the tractor has and you get the total weight the tractor has the ability to lift 1ft in 1 hr.

Divide that by the tractors weight, and you get the vertical distance it can lift itself in 1 hr measured in ft. Multiply that by 2 for the 30 degree slope, and divide by 5280 to get miles per hour. And I get 8.5 is the speed it would maintain. And it would take 70hp for it to maintain the 12mph.

So, a "horsepower guy" says 8.5 mph. Anyone from the "torque camp" want to give their answer?

 

[LD1]

I get the same thing using the metric system. Converting the power of the tractor to 37,300 watts, it can move vertically at 1.9m/s. Multiply by 2 for distance, and by 2.24 for mph. 8.5 on the button.

 

[LD1]

In order for it to maintain top speed of 12, with its 50hp, it would have to be 20 degree grade or less. (36% grade)

 

[LD1]

So, a "horsepower guy" says 8.5 mph. Anyone from the "torque camp" want to give their answer?

Are you looking for more than 1 answer?