View Full Version : CSU Question

Big Kev
30th Jun 2006, 08:46
Hi All,

Just a quick question as I'm studying for my CSU endorsement. I have read in the books that the pitch must always be set to fully fine for take-off and just prior to landing to enable the engine to deliver maximum thrust but wouldn't the engine deliver max thrust at a lower RPM with the pitch blades in a more coarse position.

If they were in a more coarse position for takeoff sure, the engine wouldn't rotate as fast but the bit of air that the propeller takes would be greater than if it were at a finner pitch spinning faster. That bigger bite of air would result in a bigger production of thrust shouldn't it?

So why is it common practice to go pitch fully fine on takeoff when at a coarser setting more thrust is being developed?

Thanks all to those who reply.

Piltdown Man
30th Jun 2006, 12:32
You are not setting a pitch with the Blue knob but an RPM. Horsepower = Torque x RPM. To Takeoff you need lots of grunt (ie. Horspower and for landing the option of producing the same, but quickly) Therefore you select max RPM for these two events. In addition, having the prop at max. RPM will give you some useful drag on the approach. Try gliding with the engine at idle and then, without increasing throttle, move to Min. RPM and see what happens.


30th Jun 2006, 19:09
Prop blade and angle of attack like a wing; no forward speed,or slow, as in takeoff (or landing) at coarse pitch the prop blade will "stall" at high rpm whereas at fine pitch its a of a can cope.At higher airframe forward speed and rearward airflow at coarse pitch the prop blade a of a becomes good.

Simplistic but tue!

2nd Jul 2006, 10:01
At take-off and in a go-around we require lots of power and thrust at low airspeed. For best effect we need both the prop and the engine to be operating efficiently.

Most prop blades are cambered aerofoils. This means that they are most efficient when their angle of attack is about 4 degrees. But at any given blade angle, the angle of attack increases as aispeed decreases. So a coarse pitch prop would have a very large angle of attack during take-off and go-around. This would make the prop very inefficient.

What we need is fine pitch to keep the angle of attack at about 4 degrees. To do this we select a high RPM, which causes the CSU to move the blades towards fine pitch.

Engine power output = Torque x 2pi x RPM. This means that for any given power output, if we reduce RPM we must increase the torque. But exerting a high torque requires the engine to produce high cylinder pressures. These high cylinder pressures tend to cause detonation. To demonstrate the effect of asking an engine to produce high torque at low RPM, try driving your car up a steep hill at low speed in fifth gear.

So in order to produce a large amount of power efficiently the engine needs to run at a high RPM.

So by selecting a high RPM we ensure that the prop is operating at an efficient (low) angle of attack and the engine is operating at an efficient (high) RPM.

Big Kev
3rd Jul 2006, 13:17
Thanks to all that replied.

I understand that if you start of with say 25" and 2500RPM and then reduce to 25" and 2000RPM then the amount of power being developed is less.

But why should the thrust being produced by the propeller be decreased as well?

If we were to stand behind the propeller at 25"/2500RPM and measure the thrust and then do the same at 25"/2000RPM, shouldn't we feel the same?

I ask this because even though at 25"/2000RPM less power is being produced, the blade will slice a larger chunk of air because it is in a coarser position so even though the rate of air acceleration being produced by the prop is lower at the 25"/2000RPM combination, the larger chunks of air that the prop takes should mean that thrust pretty much stays the same - right??

Thanks to all that reply.....


Old Smokey
4th Jul 2006, 06:22
Some very important points are being missed here.

Even if all other things are equal (which they are not), it must be remembered that Thrust, not Power, is required by the aircraft. Power is produced by the engine to drive the propeller and produce the thrust, and the useful thrust is the fore and aft component of the propeller reaction, the lateral component of propeller reaction is torque.

Seeing as some sample figures have been included in the last post, I’ll use them, i.e. 2000 R.P.M. and 2500 R.P.M. To provide a sample propeller of 4 feet radius, the mean propeller position (in terms of thrust production) is at 4 ft X 70.7% = Radius 2.828 ft.
Thus, at 2000 R.P.M. and 2500 R.P.M., propeller rotational speed at the mean position is 35,537.7 ft/min and 44,422.1 ft/min respectively.

Assuming that the propeller accelerates the incoming air flow by 60 knots, i.e. 6076.1 ft/min (about normal), the vector sum of the relative airflows over the propeller for the 2 engine speeds whilst the aircraft is stationary are –

2000 R.P.M. = 36053.4 ft/min : 2500 R.P.M. = 44835.7 ft/min

As the lift force generated by the propeller is directly proportional to the speed squared, the 2500 R.P.M. propeller is generating 1.55 times that of the 2000 R.P.M. propeller, or put inversely, if the 2500 R.P.M. propeller is producing 1 unit of reaction, the 2000 R.P.M. propeller is only producing 0.647 of a unit.

Now, assuming that both propellers are at an angle of attack of 4°, the blade angle of the propellers are –

2000 R.P.M. = 13.7° : 2500 R.P.M. = 11.8°

As the useful thrust equals the reaction force multiplied by the Cosine of the blade angle, the Thrust comparison becomes –

2000 R.P.M. = 0.647 Units X Cosine 13.7° = 0.629 Units : 2500 R.P.M. = 1 Unit X Cosine 11.8° = 0.979 Units

Thus, at zero speed and maximum power set in both cases, the 2500 R.P.M. propeller is producing 1.56 times the thrust of the 2000 R.P.M. propeller, or inversely, the slower propeller is producing only 0.641 times that of the faster propeller.

The “big chunks of air” theory, if put into effect, will lead to the loss of big chunks of performance.:*


Old Smokey

4th Jul 2006, 23:31
You can even think in terms of a wing. If a plane stalls in 1g straight&level flight at 100 Kts, it can sustain 2g at 141 Kts. (twice the lift, in other words)
Exactly the same principle applies to a prop, where the airspeed of the prop airloil is proportional to rpm.

I once took off (intentionally) in high pitch, because the prop seals were leaking, and the plane was light, and I thought if I kept the oil control valve for the prop closed I could keep the windscreen cleaner. I had plenty of runway & no obstacles, which was good because performance was terrible, but the prop leaked anyway. :ouch:

5th Jul 2006, 14:34
Prop blade and angle of attack like a wing; no forward speed,or slow, as in takeoff (or landing) at coarse pitch the prop blade will "stall" at high rpm whereas at fine pitch its a of a can cope.At higher airframe forward speed and rearward airflow at coarse pitch the prop blade a of a becomes good.

Simplistic but tue!

Old smokey is technically right. It must be..... lots of writing and though went into that answer. Personally I couldn't remember the first line after I had finished reading the last one.

Anyway...Sailor had the right Idea.....KISS....:D

Keep It Simple Stupid.

When it's clear any concise you'll remember it. Bravo Sailor. :ok:

I don't mean to offend Old Smokey, all respect. Please forgive.