I'm a bit perplexed as to why I always hear of airplane engines in terms of horsepower but torque is never mentioned.

I'd think that torque, as a twisting force, would pair nicely with a propellor and be the key determinant of airplane performance.

I can't figure out why a 520 cubic inch engine puts out a measly 275 horsepower with great fuel. Is the tradeoff for torque or simply reliability?

HP = Torque x RPM ÷ 5252
So, if you know any two of the three, you know the third.
For example, if you know hp at rated rpm, divide the hp by the rpm and multiply it by 5252 to give you the torque.

Torque and horsepower are interrelated. You're right that torque is a twisting force, but horsepower is a unit of power. If we want to talk about anything getting done, units of force are nearly useless by themselves. Units of work tell a more complete story, since they talk about the force and the distance that force acts on, but units of power bind that work to a specific timeline.

For example, if I ask you to move a 25 lbs weight from the floor to a table it'll be easy to figure that you'll need to exert 25 lbs of force. But what really matters is the weight and the distance I need you to move it. In this case, the table is 3 feet tall. So I need you to do 75 ft-lb of work. (This is in a linear frame, so the product of the weight times the distance is a scalar (non-directional) and is a unit of work, this is not the same as torque, which is a vector (directional) describing a rotaional force)

Back to the example, so as of right now I'm only asking you to complete 75 ft-lb of work. Is there a significant difference if I require you to do that in 2 seconds vs. 30 seconds? Yes! That's what power is all about. If I need you to complete that 75 ft-lb of work in 2 seconds, I'm asking you to provide 37.5 ft-lb/s of power. To do it in 30 seconds only requires 2.5 ft-lb/s of power. That's a significant difference.

In a rotational frame (in US customary units), the units we use are ft-lb for angular force. There's not actually a unit of work and power in the rotational frame that we use here, so it's all converted back to a linear frame. We convert the angular force and the angular distance that force acts to linear ft-lb (work), and then we can use the time it takes to complete that work to compute horsepower (power).

(If you're familiar with the old "hp= (tq x rpm)/5252" equation, the 5252 is the constant that puts us in hp units (33000 ft-lb/min) divided by the conversion that brings us from angular to linear (2*pi rad/rev).)

Another note is that gearing multiplies the force exerted, but reduces the distance that force is acted through. Therefore gearing can change a power-plant's torque output but will not change how much power it produces.

Anyways, the long story short is that comparing the maximum rotational force each engine can produce doesn't give a full picture of what that engine is capable of - it's more a snapshot of a very specific value. Power, on the other hand, is a good picture of an engine's capabilities and is a more appropriate basis to compare different engines.

As an example, I can give you an engine that produces 290 ft-lb of torque and I can give you one that produces 16 ft-lb of torque. They'll both fly your 2000lb gw airplane nearly identically. Both produce ~150 hp. The first at 2,700 rpm (piston) and the second at 48,000 rpm (turbine). That's a little confusing, but if I tell you I can get you two different 150 hp engines, it's a more complete description of the engines.

Hopefully that was clear, I wrote it in between doing a bunch of stuff at work.


EDIT: cleaned up 2 typos!

I see Cam Tom is on the same track as my post, but I'll leave mine up for the same explanation with a change on the wording:

Therefore, 5252 x HP/RPM = torque.

Propellers are optimized with low RPM and large diameter. Tip speed to produce highest thrust needs to be at about .95 Mach according to testing I've seen documented on Pponk Aviation's web site. An example is like what is offered on a 185. 86" diameter prop at a maximum rated speed of 2850 RPM.

Now to drive it, if the designer has decided the airframe needs 300 hp to achieve the desired performance, and the highest RPM that he can turn the prop is 2850, then:

5252 x 300/2850 = 553 ft.lb. of torque required. To achieve that much torque requires many cubic inches. In this case, 520 does the job.

I recently installed an IO-550, which has the same HP rating as the IO-550 it replaced. Why? Lower red line RPM, 2700.

5252 x 300/2700 = 584 ft.lb. torque.

So it produces more torque at a lower RPM and achieves the same horsepower result.

There are some here who would like to debate the pros and cons of the IO-520 vs the IO-550, but please see the old thread and don't clutter this explanation.

So, direct drive aircraft piston engines are configured for high torque, low RPM. Plug the numbers into the formula for a late model car, and you'll see small displacement producing moderate torque at high RPM to achieve a high horsepower number. Want more power for your airplane without more displacement? Put a gear box between your engine and prop and turn the engine faster while keeping the propeller within its sweet spot RPM. Examples are a Cessna 175 or 421. Plenty of reading available on the trade offs involved if you're curious.

So far the simplicity of a high torque low RPM engine has been the favourite, as suggested by the popularity evident in the current fleet.

Still thinking about engine upgrades for your float plane? There is no replacement for displacement.

CamTom12 wrote:Torque and horsepower are interrelated. You're right that torque is a twisting force, but horsepower is a unit of power. If we want to talk about anything getting done, units of force are nearly useless by themselves. Units of work tell a more complete story, since they talk about the force and the distance that force acts on, but units of power bind that work to a specific timeline.

For example, if I ask you to move a 25 lbs weight from the floor to a table it'll be easy to figure that you'll need to exert 25 lbs of force. But what really matters is the weight and the distance I need you to move it. In this case, the table is 3 feet tall. So I need you to do 75 ft-lb of work. (This is in a linear frame, so the product of the weight times the distance is a scalar (non-directional) and is a unit of work, this is not the same as torque, which is a vector (directional) describing a rotaional force)

Back to the example, so as of right now I'm only asking you to complete 75 ft-lb of work. Is there a significant difference if I require you to do that in 2 seconds vs. 30 seconds? Yes! That's what power is all about. If I need you to complete that 75 ft-lb of work in 2 seconds, I'm asking you to provide 37.5 ft-lb/s of power. To do it in 30 seconds only requires 2.5 ft-lb/s of power. That's a significant difference.

In a rotational frame (in US customary units), the units we use are ft-lb for angular force. There's not actually a unit of work and power in the rotational frame that we use here, so it's all converted back to a linear frame. We convert the angular force and the angular distance that force acts to linear ft-lb (work), and then we can use the time it takes to complete that work to compute horsepower (power).

(If you're familiar with the old "hp= (tq x rpm)/5252" equation, the 5252 is the constant that puts us in hp units (33000 ft-lb/min) divided by the conversion that brings us from angular to linear (2*pi rad/rev).)

Another note is that gearing multiplies the force exerted, but reduces the distance that force is acted through. Therefore gearing can change a power-plant's torque output but will not change how much power it produces.

Anyways, the long story short is that comparing the maximum rotational force each engine can produce doesn't give a full picture of what that engine is capable of - it's more a snapshot of a very specific value. Power, on the other hand, is a good picture of an engine's capabilities and is a more appropriate basis to compare different engines.

As an example, I can give you an engine that produces 290 ft-lb of torque and I can give you one that produces 16 ft-lb of torque. They'll both fly your 2000lb gw airplane nearly identically. Both produce ~150 hp. The first at 2,700 rpm (piston) and the second at 48,000 rpm (turbine). That's a little confusing, but if I tell you I can get you two different 150 hp engines, it's a more complete description of the engines.

Hopefully that was clear, I wrote it in between doing a bunch of stuff at work.

That was an excellent refresher CamTom. It's been a while since I sat in a physics classroom!

I enjoy a good HP vs Tq discussion.

In a STOL situation isn't Tq what we really care about? Tq is what is going to get our prop up to desired rpm quickly thus giving us a better holeshot. Doesn't it take a turbine a second to spool up?

whee wrote:I enjoy a good HP vs Tq discussion.

In a STOL situation isn't Tq what we really care about? Tq is what is going to get our prop up to desired rpm quickly thus giving us a better holeshot. Doesn't it take a turbine a second to spool up?

It does take most turbines a second to spool from a low idle, and this is usually due to a narrow powerband. This is usually avoided by using a high idle speed that significantly reduces spool up to operating RPM. On gas-coupled turbines, another source of lag is the design of the engine. The gas producer does exactly that: produces hot gases to spin a turbine that is connected to the power user (that turbine is called the "power turbine"). When you demand the power, the gas producer has to produce the gases by injecting more fuel, heating the air, driving its own turbine faster (which is shaft coupled to the compressor), compressing more air than it was, injecting more fuel, etc. This then provides a greater volume of hotter gases for the power turbine to use, increasing the power output of the engine. You can see where a lag could form!

Torque is immensely important in all aspects of the engine, but its an incomplete picture. If you think about saying an engine doesn't have enough torque at low RPM, you're describing the horsepower (talking about torque and rpm put together is effectively the same as just talking about horsepower). A more meaningful discussion would be the shape of the power curve. Riding a 2-stoke dirtbike, you'll experience a very "peaky," or "narrow" powerband. Riding a big cube 4-stroke dirtbike, you'll experience a very "flat" powerband. Even though both may have the same power output, one might be more effective for racing and another might be better for trail riding.

In reality, you could also say that an engine has a flat torque curve, but that gets away from talking about what really matters in an engine, and that's the power output.

I think a flat powerband is very useful in backcountry flying - specifically with a fixed pitch prop. If my static RPM is 2300, I'd like to develop as close to my maximum rated power at that RPM - all the way up to my maximum of 2700 RPM. In reality we're talking about 400 RPM, which in higher revving automotive applications is easy. I'm not familiar enough with aircraft engine cam design, head flow, typical dyno charts, etc. to say if that's common in my O-320. Anyways, due to those extra 400 RPM, I'd probably see more thrust from the propeller at 2700 RPM than 2300 RPM, but if I had a flat powerband I wouldn't be losing thrust output at 2300 RPM due to feeding less than my rated horsepower to the prop. Fixed pitch props being a compromise, I think that's the best we can hope for in that scenario.

With a constant speed prop, I'd argue the minimum pitch stop setting and speed of the prop governor matter more and the powerband could easily afford to be more narrow. That's probably a different discussion though

Torque is static force on a moment arm.

Tim

I think a flat powerband is very useful in backcountry flying - specifically with a fixed pitch prop. If my static RPM is 2300, I'd like to develop as close to my maximum rated power at that RPM - all the way up to my maximum of 2700 RPM. In reality we're talking about 400 RPM, which in higher revving automotive applications is easy. I'm not familiar enough with aircraft engine cam design, head flow, typical dyno charts, etc. to say if that's common in my O-320. Anyways, due to those extra 400 RPM, I'd probably see more thrust from the propeller at 2700 RPM than 2300 RPM, but if I had a flat powerband I wouldn't be losing thrust output at 2300 RPM due to feeding less than my rated horsepower to the prop. Fixed pitch props being a compromise, I think that's the best we can hope for in that scenario.

You nailed it!
This is spot on.

People get excited about big torque numbers.. but without enough RPM to go with it, actual power output isn't very impressive. But, as mentioned above, in an airplane, especially one with a fixed pitch prop.. that flat powerband (flat HP band.... the torque curve will actually peak early and fall off) is helpful!

I can't figure out why a 520 cubic inch engine puts out a measly 275 horsepower with great fuel. Is the tradeoff for torque or simply reliability?

Think this was basically covered in above replies.. but I'll throw out my way of explaning it as well. It's a combination of things. Long stroke engines have a harder time revving out, breathing and making power up high (high piston speed means the piston is prone to outrun the fire.. timing and other factors have to be spot on to make power with that high piston speed).... low RPM/long stroke engines are naturally going to want to make more torque than HP numbers wise. That 520 makes a big impressive torque number, but can't rev high enough to make an impressive HP number. Though the whole power curve is pretty "meaty" The propeller is the reason they spin so slow, so they were designed around that RPM limitation. And as mentioned, reliability plays a role. Others such as Rotax use a gear box to then allow the use of a shorter stroke, higher revving engine which is advantageous in my opinion.. but that is a whole other discussion.

Very good technical discussions, I can provide a very relevant example of what is being discussed.

A typical small block chevy motor will develop 147 to 160 hp on the low end and can go up to a few thousand hp on the extreme high end.

I run a 350 chev on my airboat, it has headers, hot ignition, big carb, big cam, and so on, in a car with a different cam it would make around 300 hp. The big problem is that i cannot exceed about 3000 rpm so lots of that potential power in not available. In working with the folks at crane cams they came up with a design that works with all the other components on the engine and within the rpm limits to get the best power possible at less than 3000 rpm.

My airplane used an 0-435-1 Lycoming rated at 190 hp, again 435 cu in. to only make 190 hp but it does it well at 2500 rpm. When run with a controllable pitch prop it worked very well.

Nearly all airboats running V8 engines now run gear reductions so that the engine can run free and the prop can produce max thrust, these are compromises that must be made so that a particular engine can do a specific job. I could easily get more power out of my engine but then you have to factor in the extra expense and weight of a reduction drive.

Like they say, "you can move a freight train with a washing machine motor, all it takes is lots of gears" but you ain't gonna go very fast.

Once more I'm humbled by the depth of expertise on this site. Thanks all. I learned something today.

I was already convinced to PPonk, moreso now.

Where is your new plane now? Has it made it closer to its new home?

Power is power, and for those who understand the metrics of electrical energy, it is very easy to make an analogy to help simplify the relationship between torque, speed and HP. Let’s call them Amps, Volts and watts respectively.
A typical 10,000 volt neon sign transformer sounds pretty powerful, but can barely melt the edge of a razor blade. Its all “speed” volts, but little torque or amps.
A soldering gun could be seen as being on the opposite end. Instead of stepping up the 110v to 10,000 it steps it down to merely a volt or so, but can quickly heat the conductive copper tip to temps that easily melt lead, it’s all “torque” but little speed. “Volts” . Both show perceived high power, but doing the ohms law equation would reveal they dissipate about the same power (wattage), about enough to run a desk fan.
Wattage or HP is what gets “work “ done. Volts and Amps is the yin and the yang, one cannot exist without the other. Same for speed - rpm (time) and torque without either, there is nothing created.
Engines are designed for specific tasks, gear boxes and transmissions shift the ratio of speed and torque to optimize the efficient production of power from one end, and absorption of power at the other.. To climb a hill a truck will shift ratio to the wheels to higher torque at the expense of speed. But the HP remains the same. Time is a critical factor of HP.

Theoretically one could lift the Empire state building with a small clock motor, given an incredibly high gear ratio that provided vast torque, but it could take 1000 years.

Analogies using water flow, such as Pressure = volts- speed-time. Volume = amps-torque. Pressure x Volume = Watts, HP,

Now for those who are really technically challenged, let make one out of the effectiveness of guard dogs in regard to their viciousness. Let’s look at the elements, Bark and bite. Bark being the speed (volts) and bite being the torque or amperage. Dogs that just make a lot of noise and bluff are usually harmless and have no power. They usually scare easy, so are weak. Those that would just come up to you and clamp down on a part of your anatomy will not be intimidating enough, but will be painful. So the amount of Brk X Bte= could be expressed as viciousness. I think this one needs a little refining, perhaps differentiating between Chihuahua’s and pitbulls would be better.

Pinecone wrote:Where is your new plane now? Has it made it closer to its new home?

Nope, Still sitting in Calgary (Springbank). Ferry captain has been watching the weather and thought today might be the day, but not yet.

I'm not a patient guy at the best of times... this is killing me.

Well it's a good thing he didn't come further north. It's been brutal up here!

Great discussion, I'm really enjoying this as I'm in college right now.

HP=Torque*rpm/5252

Can't say it any simpler than that.

JimC wrote:HP = Torque x RPM ÷ 5252
So, if you know any two of the three, you know the third.
For example, if you know hp at rated rpm, divide the hp by the rpm and multiply it by 5252 to give you the torque.

On a gear reduction engine do you use engine RPM or Prop RPM to calculate torque?

tcj wrote:

JimC wrote:HP = Torque x RPM ÷ 5252
So, if you know any two of the three, you know the third.
For example, if you know hp at rated rpm, divide the hp by the rpm and multiply it by 5252 to give you the torque.

On a gear reduction engine do you use engine RPM or Prop RPM to calculate torque?

It depends... where do you want to know the torque? Before or after the gear reduction?

Yeah, and if you want it at the prop, you also have to apply the gear box losses (the power required to drive the gear box). Don't forget to use the prop map and airspeed to convert the shaft power to thrust power.

Personally, as long as I know rpm, I've never cared whether I was given torque or horsepower as a known. One specifies the other.

CamTom12 wrote:

tcj wrote:

JimC wrote:HP = Torque x RPM ÷ 5252
So, if you know any two of the three, you know the third.
For example, if you know hp at rated rpm, divide the hp by the rpm and multiply it by 5252 to give you the torque.

On a gear reduction engine do you use engine RPM or Prop RPM to calculate torque?

It depends... where do you want to know the torque? Before or after the gear reduction?

I was trying to calculate prop torque using engine HP and prop RPM and couldn't get a believable result. I am thinking I can't mix engine and prop numbers in the formula. So I guess I need to know prop HP to calculate prop torque?

For what its worth. Engine HP is 46 at 6000 RPM and 41 ft/lb of torque. The gear box reduction is 3:1 so prop RPM is 2000.