Calculating propeller drag

No! There isn't any simple method. A best guess approximation would be found in Skeene's Element of Yacht design. Chap 11 ( I think) is the section on propeller design. We normally find the resistance from emperical data derived in a towing tank test. Your question is a frequent topic for sailboat designers who has to calculate the drag of the non functional prop when the sailing yacht is powered by wind.

The results of such tests led to the development of feathering or folding props. From practical experience sailors have determined that a locked prop is less drag than a free wheeling one. Yes that does seems counter intuitive, but actual tests have proven the point.

Your mention of multi screw propeller boats suggest you are talking about power boats, not sail boats. Be aware that many gear boxes have the oil pump driven from the input shaft. Letting a propeller free wheel in a power boat can cause considerable damage depending on gear box brand. Securing the shaft in such cases is recommended. Use a pipe wrench or vice grip as the situation dictates. Sailboats usually have a purpose made shaft lock incorporated in the drive line.

You'd be not too far off it you considered the drag coefficient of the prop to be 1. Then, you can calculate the drag using 1/2 rho V² x prop projected area x Cd. In the English system, rho for water is 2 slugs per cubic foot. Therefore 1/2 rho is 1, as is Cd (in this estimation) so the equation boils down down to V² x prop area.

So at 5 knots (8.45 fps) each square foot of prop area would create about 70 lb of drag.

To get a closer figure you could do tow tests: lock all the props, and tow the boat, recording drag at some fixed speed. Remove one prop, and repeat the tow test. The reduction in recorded drag would be the drag of the removed prop. Assuming you don't have a tow tank, you'd need a near windless/waveless day.

Even then, this all goes haywire if you dramatically change speed: the Cd of a fully ventilating (and partially cavitating) prop dragging through the water at planing speeds will be different that the prop at displacement speeds.

Hi thanks for your reply.

If the propeller is not windmilling the drag is 0.5*rho*V^2*projected area of the propeller. When the propeller is allowed to windmill the velocity would be the net velocity vector (of prop tangential speed and water speed), and the area would be that projected into this vector? And then to calculate drag I need to take the component in the forward direction. So it would indeed turn out that the drag force is greater when the propeller windmills?

Thanks in advance for your reply.

I think it is hard to go (accurately) beyond the approximation of dragging a flat plate (equal in projected area to the prop) through the water.

A feathered prop has very low drag, even though many of these are truly feathered (with the foil shape creating no lift) only at one diameter, with points closer to center being twisted one way, and points further out twisted the other. I suspect that a very high pitch prop (as seen on very fast power boats) might have less drag when windmilling in neutral than when locked: I'd think the water might be not too disturbed in its wake. With the flat pitch prop of a sailboat, then the blade is flying through the water relatively faster, and a large portion of the wake is probably quite turbulent, accounting for the high drag. (Also, the blade is disadvantageously shaped when windmilling, and with a relatively sharp leading edge, you'd expect flow on the ordinary thrust side to be completely unattached.)

Good 3D CFD would tell you pretty accurately, and tank towing would tell you accurately, but I think your vector solution alone might not be sufficient: depending upon the prop, I think the area projected into the vector would be so much smaller than the locked prop area (projected in travel direction) that the difference in velocity would not be great enough to explain the drag incurred. But it would certainly explain much of the problem, and the disadvantageous shape (especially with cupped blades) would explain the rest.

I suppose, in simple terms, you could say that a locked three blade prop creates three very turbulent wakes, the total of which might be 80% of the area of the prop disc area. The windmilling prop might create one large turbulent wake, of perhaps 120% of the disc area.

Most of this is conjecture on my part: my experience with underwater foils has been with untwisted sections, like centerboards, rudders, and hydrofoils.

If the propeller has only 2 blades and this can be locked in line with the hull, I doubt if it will create much drag......or spinning freely would be my second choice, but as bearings etc make drag, it may revolve, but not as quickly as it should, still creating drag!! I do not see how it can rotate really free, if the watertight bearings are doing their job properly.....

A complex problem!!

Contrary to the posts made thus far, propeller drag is greater for a locked or stationery propeller than it is for a propeller allowed to freewheel.

Propeller drag is a continuum, where when a propeller is spinning fast enough it is creating thrust, as it slows it will reach an RPM where neither thrust nor drag is created. As the propeller continues to slow down more drag is created until ultimately, when it comes to a stop, it creates its maximum drag. Beyond this state, the only way to increase the drag of the propeller is to put it into reverse gear and increase the RPMs. Obviously as the RPMs increase in reverse the so-called drag increases.

If one graphed the drag continuum described above, it would be a straight line starting well below the zero drag point (in the thrust mode), this line would continue up through the zero drag point as RPMs fell until it reached the stopped or fixed propeller position. From here it would continue to rise as the propeller is put into reverse gear and RPMs are now increased.

No other explanation makes any sense, since you cant possibly have a situation where drag all of a sudden reduces when the propeller is stopped, and then increases once again once the propeller is put into reverse gear.

darmoose

I decided to consult with a colleague who is a naval architect and who has done tow tank testing plus he did a tour of duty in a hydraulics test lab building scale models of major civil structures like dams, river diversions, etc. Water flow and resistance was key.

His answer was disappointing. The measured drag of fixed and free wheeling propellers will be greater or lesser as compared to the opposite, depending on hull shape, strut configurations prop design and position of prop relative to hull. In other words everybody is right, depending on what their personal experience was. Therefore you cannot predict or calculate what the drag is going to be with any certainty. Only emperical data from a tow tank will tell accurately.

His answer was disappointing.

I find his answer very encouraging... mainly because it squares with my own.

As I wrote above, "Good 3D CFD would tell you pretty accurately, and tank towing would tell you accurately, but I think your vector solution alone might not be sufficient:..."

Even 2D CFD does a large number of calculations, and 3D CFD is far more complex. It's only been fairly recently that 3D CFD can get quite close to real world results, partly because modelling fully turbulent flows is very difficult: what is the equation for determining the motion of a particle in fully turbulent flow?

The Aptera (three wheel vehicle) story demonstrates some of the difficulties. They claim to have determined that the coefficient of drag of their vehicle would be .06, by using CFD. The actual figure, they now say (a year later), is .11. If they used CFD, the program used would have been fairly crude, because you can certainly get closer than 50% of actual values.

I am now claiming to be a human CFD machine, because when they claimed .06 I wrote this in a CR4 thread:

• Way down the list is the Aptera, for which I've done the calculations. They are exaggerating re the 330 mpg at 65mph: doing so would require a Cd of .06 (which they have actually claimed, according to a source I dug up). .06 is the Cd for a finless torpedo, and impossible to obtain in shape that has niceties such as wheels, suspension members, etc. They may or may not be aware that they are exaggerating (CFD programs can give all sorts of incorrect numbers for all sorts of reasons -- many related to blind faith and limited knowledge of the person driving them) But their shape is very clean: could be as low as .010.

So, from 3000 miles away, without any hard dimensional data, I was able to come quite close to actual, whereas they were off by a large margin even with the benefit of accurate dimensions and 3D CFD.

Actually, my estimate was partly gut feel, and partly rule of thumb. I suspect their "CFD" might not have really been employed. I am by no means really a human CFD machine, and I would trust good CFD, run by someone aware of the limitations, to get much closer than I can get by my guesstimating.

But a good rule of thumb is this: real world testing is best, tank testing (or wind tunnel testing) is next, good CFD is next, educated guessing might be next, tabular info (as from Skenes) is next*, marketing department info (which I suspect was the Aptera case) is last. (This says nothing about cost -- sometimes, real world testing is cheapest, sometimes it is most expensive.)

* because 1. the test conditions are often not spelled out in enough detail and 2. these don't take into account the often dramatic effects of all the other variables involved. Skenes has some rule-of-thumb tables and graphs for planing resistance, if I recall. The actual calculations involved however, are discussed in a 24-page Savitsky paper produced in the mid 60's, which is still considered the gold standard if you want to take into account all the many variables, such as deadrise, longitudinal weight distribution, porpoising avoidance, etc.

Blink wrote:

I find his answer very encouraging... mainly because it squares with my own.

I meant disapppointing to all those who think computers and calculatiosn wil solve all problems.

In support of what i said above "A locked or fixed propeller creates more drag than a propeller allowed to freewheel", let me offer an experiment in logic.

We have a sailboat whose sails are up and is under a 15kt steady wind, the boats engine is turned off and the propeller is freewheeling. The sailboat is equipped with a shaft brake (friction brake) so that the shaft can be slowed down and even stopped by the brake.

We all know that the water rushing into the propeller is what makes it spin. WE also know that whatever forces are at play, they all must manifest themselves in only one way, by adding to or subtracting from the rate of spin of the propeller (RPMs)

So, as we begin to apply pressure to the shaft brake the shaft, and thusly the propeller slows. The more pressure we apply, the slower the propeller spins, even though the water continues to rush against it, because we are under sail with that 15kt wind. Eventually we apply enough pressure to stop the propeller completely.

What conclusions can we draw from this experiment? Simply, that no matter where we are on the RPM scale, in order to slow the propeller even more, we must apply more pressure, because the slower the propeller turns the more drag it creates, as evidenced by the increased force trying to turn the propeller, and the more pressure it takes to maintain that rate of spin, and the more pressure it takes to reduce the rate of spin.

Then, when we want to stop the propeller, we must again increase the pressure on the brake to overcome the force created by the water trying to turn the propeller.

Never is there a case where we can reduce the brake pressure to either slow or stop the propeller. Is this not proof positive that the slower the propeller turns the more drag is created, and that a stopped propeller creates more drag than a propeller turning at any RPM?

Darmoose

Is this not proof positive that the slower the propeller turns the more drag is created, and that a stopped propeller creates more drag than a propeller turning at any RPM?

Yes, it is not proof positive.

But your logic is pretty sound. However, as you slow the prop, the lift* of the prop blade will increase as the local angle of attack with the water increases from 0 degrees to about 10 degrees. (If the propeller were driving rather than freewheeling this last figure would be higher, because the foil is operating as intended -- it could be as much as 16 degrees.) At around 10 degrees angle of attack, the prop blade will stall, and additional angle of attack (additional brake applied) will reduce lift, thus reducing shaft torque.

The situation is the same in a wind mill, with increases in angle of attack increasing torque, until the angle of attack goes beyond the stall angle of attack, at which point torque falls off.

However, shaft torque does not necessarily equate to overall drag on the boat. On a sailboat especially, flow around the surface of the boat can be nearly laminar, except in the region of the prop and prop shaft. If the prop is rotating, then the laminar flow can be disturbed over a large area of the hull and rudder. (Although we'd like to think the the propeller just propels water rearward, it also flings it outward). These disturbances happen if the propeller is locked , but over a smaller area (there being little "centrifugal" force at work). Often the zone of disturbed water is smaller with a locked prop, but the freewheeling prop does a more widespread, but less dramatic disturbing. Which effect predominates (intense but small area vs large area less intense) can only be determined on a boat-by-boat, prop-by-prop basis.

* The blade lift is what generates thrust.

darmoose wrote: Is this not proof positive that the slower the propeller turns the more drag is created, and that a stopped propeller creates more drag than a propeller turning at any RPM?

REPLY

NO! Not according to what real life and tank tests show.

In the real world the prop is connected to a shaft which in turn passes through a shaft log and a gearbox. Inside the gear box something spins. Sometimes it an oil pump which also adds resistance. The shaft packign gland also has resistance.

Which is why real life tests and realistic towing tank tests sometimes show that a spinning prop will have more resistance than a locked prop.

Your exercise in logic apparently assumes nothing is attached to the inboard end of the prop shaft. That being the case, the drag woudl pull the prop shaft out of the hull.

Furthermore; not all marine gear boxes can tolerate being driven backwards. Meaning the prop is the input shaft. Letting the shaft freewheel may well damage the gearbox. In which case the whole argument is moot, since you must lock the shaft to prevent gearbox damage. Even if this means more drag.

KEN and ELNAV,

First to elnav, all considerations of transmission and gearbox are taken out of the question, this is simply whether a stopped prop or a freewheeling prop creates more drag.

Ken, i guess i dont understand how angle of attack changes on a prop with fixed blades, or how lift changes the results of our experiment with the shaft brake?

Wouldnt the factors you speak of, if they truly changed the answer to my question, cause us at some point on the RPM spectrum to have to reduce the pressure on the brake (rather than increase the pressure) in order to slow or stop the propeller?

Can you explain angle of attack and lift a little more, as well as explain why, if what you say is true, we still have to increase the pressure on the brake to overcome drag in order to slow or stop the prop? It seems to me that the fact that we always have to increase the brake pressure to slow or stop the prop means that all the factors you are speaking of combined dont change the overall results?

Darmoose

A propeller does not simply screw its way through the water. Each blade is a hydrofoil which works on the same principal as an airplane wing, a sailboat's keel or rudder, or an airplane's propeller. All have similar airfoil or hydrofoil cross sections.

While an airfoil or rudder is straight from root to tip, a propeller (water or air) is twisted. That twist allows for the fact that near the tip, the blade travels a long distance through the water in one revolution, whereas near the root, the travel distance is much less. For any thrust to be developed, the blades must develop what is, by convention, called lift. Lift, by definition, is perpendicular to the apparent flow. Drag is perpendicular to lift (and therefore aligned with apparent flow.) The apparent flow for a propeller is, at any point on the blade (let's say half way to the tip) a helix from the perspective of someone looking down into the water. From the perspective of a bacteria stuck on the leading edge, the flow is almost straight into the leading edge. The reason I say almost is because there is an angle of attack required to make meaningful lift. That angle of attack, for a propeller operating in forward, can range from about 0 degrees to about 15. It changes more or less directly with load.

So, if the propeller is freewheeling, and is held in place on a frictionless shaft (which is not the case in the real world, of course) with no brake applied, the angle of attack of the blade with its local chunk of water will be near zero. It won't be zero, because to rotate, the blade must overcome it's foil drag (roughly aligned with the helix mentioned). But at low load, the drag is low. (Note that this drag is entirely different than the drag our original poster asked about, which is the drag aligned with the centerline of the boat, and which has many contributors.) So at low load the angle of attack is low.

Then when you start to apply the brake, the angle of attack increases, as the prop begins to generate meaningful energy (in this case serving as a cabin heater). As you increase force on the brake, the angle of attack increases, because the blade i no longer following the no-load helix: it is running too slow for for the rate of travel through the water. (If you imagine a fully stopped blade, you can see that the angle of attack would be nearly perpendicular to the blade... however the blade is angled perhaps 30 degrees at mid span, so the actual angle of attack could be 60 degrees. You can see then that the angle of attack will increase as the speed of the prop gets more an more "out of synch" with the speed of the boat... in other words as you apply more and more brake, slowing the prop.

As you apply load to the blade, eddies spill off the end of the blades, moving from the high pressure side of the blade to the low pressure side. These are caused in the same way that wingtip vortices are caused on an airliner. However their motion is more complex, because the propeller blade is moving in a helix, and because centrifugal force also comes into play.

Eventually, as we apply more brake, increasing the angle of attack, the prop blade stalls at some point in its length out from the hub -- typically out near the tip, because the blade twist is not correct for this usage. With a little more load, the entire blade will stall and the generated torque will fall off, and the blade will come to a full stop. The person controlling the brake would have no way of knowing that torque was beginning to fall off -- he would simply assume that he applied enough brake to stop the prop. Then, with the prop stopped, the operator will have to release the brake somewhat more than expected to get the prop spinning again.

Jumping to a different but related topic, the same sort of thing happens in an engine dynamometer. A typical car engine generates less torque at 1500 rpm than at 3600. So as you load the engine down to 3600 from 5000, you apply increasing pressure to the brake. If you continue to apply more pressure as the engine gets below its torque peak you'll cause the engine to stall.

Another different but similar effect occurs in flying. as you increase back pressure on the stick, the wing's angle of attack increases, and the force required on the stick also increases. However as you increase the angle of attack to beyond the stall angle of attack, lift decreases, the nose drops, and stick force reduces. You have to actively reduce stick force even more to get the plane flying again.

Wouldn't the factors you speak of, if they truly changed the answer to my question, cause us at some point on the RPM spectrum to have to reduce the pressure on the brake (rather than increase the pressure) in order to slow or stop the propeller?

If you re-read this, I think you will see that the logic is a little odd. Regardless of the shape of the torque vs angle of attack curve, there would never be a case where you would have to reduce brake force to slow the prop. There could be (and is) the case that the applied brake force (once the prop is stopped) is too much for the prop to start moving again if the release is a small increment. For example, if you applied force through a knob, you might find that a knob setting of 10 would be required to stop the prop, but that as you released the brake to a setting of 9, the prop would not start, and perhaps you would have to go all the way down to 5 before it would start.

There are all sorts of situations like this in the real world. When you learn to fly, your instructor will demonstrate how the full power of a plane capable of going 150 mph is insufficient to cause it to accelerate at all from a partial stall at 70 mph. He will also demonstrate how pulling back on the elevator can make the plane go down, not up.

So... even a freewheeling prop on a frictionless shaft takes energy to overcome its own drag. The prop's speed through the water along its helix is much faster than the boat's speed through the water. Drag increases with the square of speed, and induced drag (drag from creating lift) increases even more rapidly. So a freewheeling prop can suck up a lot of energy: it is not really free wheeling. Is it better to have your prop slogging through three feet of water for every foot forward or to have it working like a barn door, but at a much lower speed. The higher speed gives you nine times the drag, (on a small frontal area) but the lower speed gives you several times the frontal area. So for some boats and props it's a wash. For others, there is a favored configuration.

BTW, here is a link re windmill design which touches on some of this, at least in the sense that the theories are the same:

http://digital.library.okstate.edu/OAS/oas_pdf/v56/p121_124.pdf

See the link below to a downloadable pdf of a recent research paper. It includes results from towing tank tests and water tunnel tests for fixed and freewheeling blades. It also gives a predictive method for estimating the drag for freewheeling props of most common patterns if the props' key dimensions are known and frictional torque on the shaft can be measured. Categorically, spinning free produces lower drag and, furthermore, the lower the torque, the less the resultant drag.

eprints.cdlr.strath.ac.uk/5670/

In my case, when I stop the shaft, say at 7 knts speed, the slippage of the shaft begins again when speeed increase to 7.5, which means that the prop didn't reach the point of stall.

You are most probably right, but when we have a power failure on the autogyro,we try is to give maximum revs to the prop in order to go down slowly (increase the drag). Here because the blade surface is so small, the drag is obtained by increasing the speed of the blade and its own drag, creating a sort of circular wing with the prop to reduce speed down. The % of blade surface to total propeller area as well as the pitch will give drag figures at every speed for every prop. Theoretically at 0 pitch and 100% area you will have a solid disc and maximum drag.

Proplematic