# Why, with a constant speed propeller, does moving the propeller control forward increase drag?

Could someone explain why, let's say in a constant speed propeller, with moving propeller control forward the drag increases? It is noticeable when we push the blue knobs forward in flight that the airplane has an additional portion of drag.

• If you move the propeller control you increase thrust, therefore increasing speed and therefore increasing drag. Maybe this is an oversimplification to your question or I didn't understand it correctly. Jul 25 at 12:51
• How do you know there's an "increased portion of drag" on the airplane? I don't know about any drag-o-meter.... Jul 25 at 12:55
• What I mean is that it is feeled by the crew that with moving propeller controls forward aircraft slows down a bit Jul 25 at 14:15
• @Sanchises when you are landing the plane and you switch from cruise propeller angle to full-power angle, the instant deceleration which accompanies the engine's increase RPM is very noticeable. Jul 25 at 16:06
• So, you're really asking about what causes the deceleration of the airframe. It's quite an assumption to think this is due to an increase in drag and not due to a decrease in thrust... Jul 25 at 17:22

The prop control sets the engine RPM. Moving it forward allows the governor to flatten the blades, which results in a lower angle of attack. In fact, in certain circumstances the AoA goes so low that it actually becomes negative, meaning that the airflow is moving faster through the propeller than the propeller is moving forward. This is like running around with a pinwheel.

In our case, propellers will generally always have a difference between the effective pitch and the geometric pitch[*]. When the effective pitch is shorter then the prop is making positive thrust and when it is longer the prop is making negative thrust.

We can immediately infer that the sharp deceleration felt when the prop control is sped up is due to the effective pitch being longer than the geometric pitch. This is how we know the AoA is negative.

### Effects of Negative AoA

@MichaelHall explained it well in his answer, so I won't go into it much here. In short, the negative AoA means that instead of the high pressure area being the underside of the blade, it becomes the topside. This inverts the $$F$$ vector direction, and means that the propeller is no longer absorbing torque from the shaft, but is instead transmitting torque to the shaft.

### Where is the energy going?

The prop is spinning faster, true, but why is this slowing the plane down? After all, $$E = F d$$, and so if there's a deceleration Force, there must be somewhere that Energy is flowing.

The answer is two-fold. A small amount of energy goes to parasitic drag losses in the propeller, but the vast bulk of it goes to the engine pumping and friction losses.

Simplifying things somewhat[**], for a classic gasoline ICE the energy lost per ideal cylinder cycle is $$E = p v$$. If we have a bog-standard $$360\rm{in}^3$$ engine, such as the venerable Lycoming O-360, then, converting into standard units, and assuming a 75% vacuum behind the closed throttle plate, we have E = 101300[Pa] * 0.75 * 5.90[L]/1e3[L/m^3] = 448J per cycle. Since the Lycoming is a four stroke, a piston cycle takes two revolutions, and thusso at 2700RPM-- a typical max speed--, we have 1350 cycles/minute, or 22.5cycles/sec.

22.5 * 448 = 10kW, or approximately 13hp.

Considering that the 0-360 only makes 180hp, an engine braking factor of approximately 10% of peak thrust is certainly going to make itself felt.

While we haven't taken into account internal friction nor parasitic drag, nor deviations from an ideal Carnot engine, we can still see that the primary power sink is the engine pumping air against a closed throttle plate.

This engine-braking effect is very familiar to motorcycle riders and stick-shift drivers, and the same principles apply. Interestingly, this does not apply to diesels, as they do not have a throttle plate. They have far lower engine braking forces and diesel trucks have to go to some length to create engine braking force, which is a useful backup to traditional friction brakes.

[*] Read about the fascinating and ingenious method to achieve true 0-thrust, required in order to empirically test a powered plane's glide ratio: https://engineering.purdue.edu/~andrisan/Courses/AAE490A_S2010/Buffer/AIAA-46372-872.pdf
[**] We assume no friction losses and minimal heat transfer.

• Sorry, but I could give you only one upvote. Great answer! Jul 26 at 3:15
• Agreed, big improvement over version 1.0, and I like it better than my own... (although I think mine has some merit as a math-free alternative!) Jul 26 at 15:36
• @MichaelHall Team effort! My answer steers readers to your answer for an explanation of why the AoA goes negative. Jul 26 at 16:49

Let’s take the engine out of the equation to simplify things. In fact, let’s take the prop out too, and consider a very elementary comparison – the old “hand out the window” example...

Put your hand out the window when you are driving and feel the drag it creates. Now rotate your hand. Can you feel how there is little drag when your hand forms a “knife edge” versus when your palm is flat, and perpendicular to the airflow? (I know you know this and I'm not trying to insult anyone's intelligence, but I find it useful sometimes to pull way back to the basics to make a point...)

This is the same principle at work when you vary the pitch of the prop. When you move the prop control aft towards course pitch you are decreasing the surface area of the prop that is facing the airflow, decreasing drag. (remember, we are considering a free spinning prop here, in a gliding descent, because I took the engine power out of the example for now…) With a full feathering prop it is possible to “knife edge” the prop blades, creating very little drag.

Now consider the opposite, that we move the prop control full forward. This has the effect of flattening the blades to a fine pitch. If we go far enough they form a slow rotating disc, with a lot of surface area exposed to the oncoming wind, creating lots of parasite drag. (just like the palm of your hand.)

Let’s add the engine back in - During take off and climb we select a fine pitch, but it isn’t to create drag, it is to get the most power at high RPM and slower airspeeds when we need it. Just like shifting your car or bike to a lower gear when going up hill. Normally in cruise or descent we pull the prop lever back to select a coarser pitch for high speed lower RPM efficiency, just like you might use overdrive on the freeway.

When the prop is under load from the engine it is producing power. The blades have a positive angle of attack, and they are taking “bites of air” and accelerating them rearward to create thrust. However, in a descent with the power at idle, the reverse might be true. The angle of attack can be negative as the props extract energy from the airstream to rotate instead of adding energy. This contributes to the decelerative effect of the drag.

Basically by varying the blade angle and engine power setting we can use the prop to either pull ourselves forward, or to resist forward motion.

So, when might we use the prop to create more drag and resist forward motion? Well, on a descent at flight idle when trying to slow down to landing gear and flap extension speed. Without speed brakes it can sometimes be a challenge to slow down, but if we move the props to a fine pitch they flatten out and become a very efficient drag device.

I know, it feels counterintuitive to shove the props up to slow down, but trust me it works quite well. It's similar to the idea of downshifting your car on a steep decline instead of riding your brakes. (and it sounds from your question like you have experienced this effect firsthand...)

• I have my doubts that this explanation is correct. The angle of attack for a freely spinning propeller blade is going to be very well low. True, additional propeller parasitic drag is created by virtue of the prop spinning faster, but I suspect these losses are negligible in face of the pumping losses from engine braking at several thousand RPM. Jul 25 at 17:45
• @Kenn Sebesta, I am open to suggestions to improve this answer, but a prop under power will have a positive AOA as it "bites" the air and pulls the aircraft forward, while a free spinning blade will have a negative AOA from the force of the airflow making it turn. Excellent point about the fact you can't ignore the effect of the engine turning because the prop really isn't spinning free, but my gut tells me that the primary contribution to deceleration is the increased drag from the flat pitch. Jul 25 at 17:50
• And might the answer be different between a turboprop, (the basis of my answer) and a piston engine? (the basis of your answer) Because a turboprop is closer to a free spinning blade, while a piston engine has the resistance of compression, which puts it much closer to the example of downshifting a car or motorcycle. Jul 25 at 18:03
• @Kenn Sebesta: I respect Peter's IQ and scholarly acumen, but I am a simple unfrozen caveman pilot and prefer common language explanations based on relatable observation over complex equations. ;) However, your remark about "pumping air" leads me to believe there might be a better answer than simple drag. That the prop is extracting energy from the airflow rather than adding energy to create thrust. I'm just not sure where to go with that idea.... Jul 25 at 18:42
• @PeterKämpf, Michael Hall, I revised my answer and worked out the math. The engine braking for an O-360 is an astounding 13hp! The prop spinning faster isn't what's causing the deceleration on its own, it's the negative AoA (which M. Hall explains well) which is being driven by the piston engine braking. Without the braking, the AoA would likely be scarcely negative. I'm not sure the magnitude is the same for a turbine, though, since turbines and diesels share the common trait of not having a throttle body. Jul 26 at 0:55