my thoughts on propeller pitch

[5-in-50]

The bike analogy is fine, but doesn't account for Constant Speed Units, which aren't present in CoD anyway.

[Skoshi Tiger]

Originally Posted by 5-in-50

The bike analogy is fine, but doesn't account for Constant Speed Units, which aren't present in CoD anyway.

Rotol Hurricane and MKII Spitfire have CSU's

Cheers!

[JG53Frankyboy]

Originally Posted by Skoshi Tiger

Rotol Hurricane and MKII Spitfire have CSU's

Cheers!

& Ju87, Ju88(what is wrong), He111, G.50, Br.20

[Lololopoulos]

Thanks everyone for ur opinion.
I have one question though, so in the video is it mentioned that how fast the airplane will be going depends on 2 factors:
1. pitch or how much air the propeller grabs
2. the rpm or how fast the propeller rotates.
theoretically more pitch and more rpm combined will make the airplane go faster

This is what I don't understand, in the constant speed propeller take off run, doesn't a fine pitch counteracts the high rpm? intuitively for me, if the pitch is fine then it doesn't really matter much how fast the propeller spins anymore because the prop doesn't get much air anyways??

Can someone point me in the right direction?

[adonys]

they are in a symbiotic (!) relation : the more air the propeller bites, the slower till be the RPM.

think about it in this way: if you have the engine without any propeller, and working at maximum power, the shaft gets to a certain RPM. it will NEVER went higher than that by itself (unless helped by an external force).

now, when you fit the propeller on max fine (biting the smallest amount of air), some engine force will be spent on pushing that air, so the force rotating the shaft will be smaller, therefore the RPM will go down. The more air bites, the slower it will rotate.

the exception is when you are diving, because while diving, the finest propeller setting will expose the max propeller surface against the airflow going up (as your aircraft is going down), and that will apply the biggest external force on the engine's shaft, making it exceed its maximum self obtained RPM, and breaking the engine.. that's why in dives you have to set your propeller on max coarse, to offer its smallest profile to the airflow going up, in order to not over-rev your engine.

[bw_wolverine]

Originally Posted by adonys

they are in a symbiotic (!) relation : the more air the propeller bites, the slower till be the RPM.

think about it in this way: if you have the engine without any propeller, and working at maximum power, the shaft gets to a certain RPM. it will NEVER went higher than that by itself (unless helped by an external force).

now, when you fit the propeller on max fine (biting the smallest amount of air), some engine force will be spent on pushing that air, so the force rotating the shaft will be smaller, therefore the RPM will go down. The more air bites, the slower it will rotate.

the exception is when you are diving, because while diving, the finest propeller setting will expose the max propeller surface against the airflow going up (as your aircraft is going down), and that will apply the biggest external force on the engine's shaft, making it exceed its maximum self obtained RPM, and breaking the engine.. that's why in dives you have to set your propeller on max coarse, to offer its smallest profile to the airflow going up, in order to not over-rev your engine.

A very good explanation. I think this is something a lot of people miss and end up over-revving their engine as a result.

[CaptainDoggles]

Originally Posted by Lololopoulos

Thanks everyone for ur opinion.
I have one question though, so in the video is it mentioned that how fast the airplane will be going depends on 2 factors:
1. pitch or how much air the propeller grabs
2. the rpm or how fast the propeller rotates.
theoretically more pitch and more rpm combined will make the airplane go faster

This is what I don't understand, in the constant speed propeller take off run, doesn't a fine pitch counteracts the high rpm? intuitively for me, if the pitch is fine then it doesn't really matter much how fast the propeller spins anymore because the prop doesn't get much air anyways??

Can someone point me in the right direction?

Propellers and the mathematics behind them are quite complex, but we can do a crude analysis.

An approximation of the thrust produced by a propeller can be made by computing:

Thrust = [efficiency factor * power] / speed

For takeoffs, you want to accelerate to liftoff speed in as short a distance as possible, which means you want as much thrust as you can get.

Looking at the above equation, there are only 3 terms. Speed is very small as you begin the takeoff roll and the efficiency factor is determined by the design of the prop.

That leaves engine power as the only remaining variable we can manipulate. Since the engine develops more horsepower at high RPM than it does at low RPM, you want to run the engine at high RPM so that you can produce the most thrust.

Going a little deeper: Power is computed by multiplying shaft torque by rotational speed. So to increase power you want the prop to spin faster.

Therefore you want to go to fine pitch.

------

I really dislike the "car analogy" that many make, but if it helps: In a car you accelerate from a standstill in lowest gear. This is akin to putting the prop at fine pitch.

[Art-J]

Ehm... Sorry, Cheesehawk, but I'm sure Cpt.Doggles' formula is right . It's a simplified, basic formula for prop thrust, obtained from the definition of prop efficiency, which can be found in any book about props... or on the same page You linked to, just a couple of lines below the formula You quoted. Note the formula being used for thrust estimation in the numerical example at the bottom of the same page.

Cheers - Art

[Lololopoulos]

Originally Posted by CaptainDoggles

Propellers and the mathematics behind them are quite complex, but we can do a crude analysis.

An approximation of the thrust produced by a propeller can be made by computing:

Thrust = [efficiency factor * power] / speed

For takeoffs, you want to accelerate to liftoff speed in as short a distance as possible, which means you want as much thrust as you can get.

Looking at the above equation, there are only 3 terms. Speed is very small as you begin the takeoff roll and the efficiency factor is determined by the design of the prop.

That leaves engine power as the only remaining variable we can manipulate. Since the engine develops more horsepower at high RPM than it does at low RPM, you want to run the engine at high RPM so that you can produce the most thrust.

Going a little deeper: Power is computed by multiplying shaft torque by rotational speed. So to increase power you want the prop to spin faster.

Therefore you want to go to fine pitch.

------

I really dislike the "car analogy" that many make, but if it helps: In a car you accelerate from a standstill in lowest gear. This is akin to putting the prop at fine pitch.

thanks a lot!!
when it comes to things that are unintuitive, looking at a math formula is very very helpful, because it gives u an idea of what is important and what is irrelevant. Good job with the condensed version of the formula and explanation!!

[CaptainDoggles]

Originally Posted by cheesehawk

Sorry Doggles, I am sure your formula is off.

No need to apologize, cheesehawk. The formula I posted is perfectly valid (if a bit simplistic). You'll notice I used the phrases "a crude analysis" and "an approximation". In real life we must also account for components of thrust not in line with the direction of motion, rotation of the fluid within the slipstream, etc. The formula you posted doesn't account for these either. It all depends on how much effort we are willing to invest and how precise we want our approximation to be.

The concept of delta-v added to the flow is not easily discerned from in-cockpit instrumentation. Horsepower on the other hand is at least related to engine RPM, and we have a tachometer in the cockpit