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I plan to do it over open ocean in unrestricted airspace, and I will have a visual spotter in addition to myself. I would like to maximize my chance that it comes back, and minimize the loss of my time and money if it does not.
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Last edited by SirRexAlot; Jun 09, 2018 at 12:49 AM.
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Good plan.
I don't know what, "Fly within visual line-of-sight" really means. Does it qualify if a telescope is used? Is it OK as long as the plane is in a theoretical visual line-of-sight, even though you can't see it? It's a silly rule. My wife once acted as my spotter and mostly I had to tell her where to look. She had lost the plane, but I knew exactly where it was at all times, I wasn't flying over open ocean, however. |
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The rules I've seen specify unaided line of sight.
I've worked with the FAA in the full-scale world, and what's happening here is pretty typical. If they get confronted with something new, they simply don't know what sort of rules are needed, so their initial response is to heap every rule and test on it that they can dream up, in hopes that either they have "almost by accident" covered everything necessary, or else make things so impossibly difficult that this new thing will just "go away". I have seen full-scale aircraft designs fall victim to this, the cost of this unending list of testing got so high that they simply ran out of money. I agree, it's a silly rule, vague and essentially unenforcible. However, right now it's what we have to work with. SirRexAlot, one thing you will have to confront getting from 60 KPH to 60 MPH is power requirements. Power is work per unit time. Work is force times distance. In this case the force is drag, which is proportional to the square of the airspeed. That has to be multiplied by the airspeed again (the distance per unit time) to get power needed for an increase in speed. The net result is that power required is proportional to the CUBE of the airspeed. So, 60 KPH = 37.3 MPH, and 60 / 37.3 = 1.61 times faster. 1.61^3 = 4.16 You need over four times as much power to go that much faster. Note, that does not include dealing with the extra drag due to the extra weight that the much bigger powerplant will add, which could itself be considerable. The only way to get even partially around that is by reducing drag. If you wanted to do this without a significant power (and weight) increase, you would need to get your drag down to less than a fourth of what it is now. And, since you're dealing with high-speed performance, the dominant factor by far is parasite drag. Playing with things that help induced drag, such as planform and span, will not make much difference. You need to look primarily at whetted area, also the profile drag coefficient of the airfoils, and of course the drag of anything sticking out in the breeze. Also, look at whetted area of the fuselage, tail surfaces, etc. There is a reason why planes like the F-104 had such ridiculously tiny wings. Massive reductions in wing area can make a difference, but you still have to be able to land this beast, so there's some serious restrictions on that route. This sort of thing tends to quickly devolve into a "You can't get there from here" situation. |
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Last edited by Don Stackhouse; Jun 09, 2018 at 08:21 AM.
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10km at 60km/h should not be too much of a stretch. And not just because using proper units clearly makes things much easier
My little ~1.2m span correx FPV wing has done 19.1km at an average of 57km/h on just a 4S 1300mAh lipo, with quite some juice left in the tank. It was never built with long range in mind, so I put no thought at all into optimizing it. Also, our flying restrictions here are more ludicrous than yours, so that 19.1km was all line-of-sight over a small park with lots of energy wasting turns. So you should be able to hit the 10km target with just about any reasonably sleek plane. Now 100km....that is a far more ambitious target, and one of the reasons I've been following this thread. Given that my plane surprised me at how far it could fly on a small Lipo, I subsequently did a lot of test flights to see how low I could get the Wh/km...and if 100km could be possible with some Li-ion cells. Some of this may be of help to you:
Hope that helps Also, as a full size pilot, may I ask you to please think very carefully about where you fly. Over the ocean and in uncontrolled airspace does not mean that there won't be any full size planes. I assume you've picked a spot with no regular traffic (GA pilots like to fly low over the sea just off the coastline). Also please stay low. A lot of 'normal' GA traffic is below 1500ft and certainly above 1000ft, I'd be pretty pissed to run into a 'drone', whereas below 500ft, I'd have my eyes out looking for birds and similar small 'things' sharing the airspace. Finally, your spotter's primary purpose must be to look out for full size planes...even better if s/he has a coms radio to listen out for GA calls. |
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Last edited by Extreme Sports; Jun 09, 2018 at 10:42 AM.
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Propeller pitch seems to be a widely misunderstood parameter.
Let's use an automotive analogy. For an airplane, the propeller is the equivalent of your car's transmission, drivetrain and wheels. The propeller pitch is the equivalent of what gear your car is in. The diameter and blade area are roughly analogous to the diameter and width of your tires. The gear (pitch) you need to be in depends on what you're trying to do. If you want to accelerate away from a stop light (or you want lots of static and low speed thrust for takeoff), you need to be in first gear (flat pitch). But for cruising on the highway (or in high-speed cruise in your airplane), first gear is the LAST thing you need. You should be in high gear (or coarse pitch). As far as rules of thumb such as "square" combinations of diameter and pitch (i.e.: both values equal to each other, such as an 8-8 prop), That's a purely human construct, with no real basis in science. In many cases, the best setup for most efficient cruise tends to be "over-square", i.e.: with a value for pitch that HIGHER than the diameter. In the full-scale propeller business, we don't generally use "pitch" in the aero calculations, we use a "non-dimensional" version of it called "Advance Ratio", or "J": J = Va / nD where Va is freestream airspeed in feet per second n is rotation speed in revolutions per SECOND (NOT RPM) and D is diameter in feet https://en.wikipedia.org/wiki/Advance_ratio As that link indicates, the Advance Ratio is the ratio of forward speed to tip speed. Another handy parameter that non-dimensionally indicates how hard the prop s working is "Power Coefficient", or "Cp": Cp = P / rho * n^3 * D^5 Where P is power and rho is the air density. Note, from the Cp formula we can see that power is related to RPM CUBED, but it's related to the FIFTH POWER of diameter. Yes, power is extremely sensitive to RPM, but it's even more sensitive to diameter. We can plot efficiency vs Cp and J to create a "Cp-J map". I've attached an example of one, taken from a report Hamilton Standard made for the Army many years ago. They actually took a whole bunch of full-scale 2, 3 and 4-blade propellers, ran them in a large wind tunnel, then plotted the results in a book about an inch thick. Yes, at model Reynolds numbers the efficiencies won't be quite as good, but the combination of pitch and diameter that give the optimum results will still be fairly close. The plot looks like a contour map of a mountain, with the peak efficiencies at the top of the mountain. Note that the highest efficiencies tend to occur at surprisingly high values of "J", i.e.: at very high pitches. However, there is a down side. Keeping RPM low for a given airspeed also tends to result in a very narrow range of airspeeds where that propeller "works". This is not a problem if you have a variable-pitch prop, but for a fixed pitch prop (like the vast majority of model airplane props), it basically gives you a one-speed airplane. If you have enough pitch for great efficiency at high speed, the poor thing is often stalled at takeoff and climb speeds. If you have the pitch flat enough for good takeoff performance, it's like trying to run on the highway in first gear when you're in cruise. Going to a higher-than optimum RPM and a lower than optimum value for J on a fixed pitch prop for a model airplane costs significant efficiency over the whole operating range, but it gives you a prop that can perform "good enough" over most of the plane's operating envelope. Also, the RPMs that props like tend to be much lower than motors like, so the RPM and pitch you end up with tends to be a compromise between the two. |
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PS: Found one of them - http://m-selig.ae.illinois.edu/pubs/...Propellers.pdf PS2: It did occur to me that a catapult launch would probably be a good thing for a long range plane anyway - a disproportionate amount of energy seems to be consumed for the takeoff! |
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OK, now that you're still reeling from that last "drink from a fire hose", let's see if we can simplify it a bit.
Even with all the right aero tools, for model props we don't have a lot of good data to plug into all the formulas. That means that we have to take our best guess on a starting point, and then do some flight testing to fine-tune things from there. However, there are some things we can do to cut down on the guess work. Getting the right pitch as a starting point is a good initial strategy. We usually have some idea of about what airspeed we're expecting to fly (if you don't, then we can cover that in another discussion), and we can usually get a reasonable estimate of what RPM we want to fly at. If nothing else, the published specs for a given electric motor usually give some indication of what the most efficient RPM is for a given motor. If we have an expected airspeed and RPM, and if we can make a reasonable guess at the prop efficiency, we can estimate an approximate pitch. For a really good model prop (such as the Friedenthaler/Aeronaut CAM props), at takeoff and climb we're probably looking at around 60% efficiency, while the other brands fall somewhere below that, perhaps as low as 25%. At cruise the efficiencies will quite possibly be a bit better, perhaps as much as 70% or so. A prop makes thrust by taking air from in front of it at the free stream velocity and accelerating it out the back at a higher speed. Half the acceleration occurs in front of the prop, and the other half behind. One way to calculate prop efficiency is the ratio between the free stream and the propwash velocities. So, if we have a prop that's only 60% efficient (0.60), that means the velocity in the slipstream several prop diameters behind the prop will be 1/0.60, or 1.66 times the airplane's airspeed. The speed at the prop disk itself will be around the average of those two, or 1.33 times the plane's speed OK, I can see you're all starting to get "glassy-eyed". Sorry. Let's try an example. Let's say we're trying to fly at 60 KPH, which is 37.3 MPH. (SirRexAlot, did I get your attention?) Convert that to feet per second (to do that, multiply by 22 and then divide by 15). That gives us 54.7 feet per second. Since we want to know our pitch in inches, multiply by 12 to get inches per second, which in this case is 656 inches per second. Now, we have to include half the acceleration of the air due to the prop. Let's say we have an Aeronaut prop, well-sized and fitted to the airplane, and we're in cruise at a moderate power setting (note, if we were at full throttle, the efficiency would be less). Let's guess that our efficiency is somewhere around 70%. 1/0.70 is 1.43, and the average between that and 1 is (1.43+1.0) / 2 = 1.21. The airspeed actually at the prop disk is about 1.21 times the plane's airspeed: 54.7 x 1.21 =66.4 feet per second, or 797 inches per second Let's say our motor is happy at 5000 RPM. Divide that by 60 and we get 83.3 RPS (Revolutions Per Second). Now, divide the airspeed at the prop disk (797 IPS) by the rotation speed (83.3 RPS): 797 / 83.3 = 9.57 inches per revolution. That means we need a prop with somewhere around 9.5 inches pitch. Note, that's an approximation, subject to all the uncertainties of all the estimated numbers you used for input. HOWEVER, it's likely to be far closer than the number you would probably come up with using the old "TLAR" method. Now that you have a reasonably good starting point for pitch, you can estimate what sort of diameter will load the motor properly at that pitch. Yes, it's a bit more work than just grabbing a handful of props off the rack at your LHS, but in my experience this method results in finding the right prop with a lot less time and expense than the old "try everything until you find the right one" method. |
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Last edited by Don Stackhouse; Jun 09, 2018 at 01:19 PM.
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But I think you have something else in mind...and, since it takes over 3 months to get props from my 'LHS' (in China), I'm very interested in finding a less random approach to finding the best prop! |
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Every one of your posts gets my full and undivided.
I'm shocked that going from 60 kph to 60 mph will take 4x the power. I knew it must be some kind of exponent from my own testing, but didn't suspect it went hockey stick that quickly. My own experiments back this up as I play with the numbers; I thought it was inaccuracy in my flight controller's ammeter and GPS combined with my amateur build skill. My last 2 testing flights. I did my best to keep amp draw and altitude steady throughout the flight, but there were turns and avoiding trees and such, plus the 5 amp flight was in a 15-20 kph wind. 3 amps: 33.5 km / 44 min = .76 km/min = 45 kph 5 amps: 22.5 km / 26 min = .86 km/min = 52 kph 52 - 45 = 7 kph difference. 7 / 45 = 15.5% increase in speed for a 66% increase in power. 1.15^3 = 1.52 and with my mickey mouse equipment that's remarkably close. |
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I do have a 5" prop with a 5" pitch, so let's see what kind of revs that would take: 797 / 5 = 159 rev per second, or 9564 RPM. And we're right in the wheelhouse of what a little quad motor can do for just a few amps. To get some numbers from more precise equipment, I picked a 2205 2300kv motor from MiniQuadTestBench. That's the size motor I'm running. https://docs.google.com/spreadsheets...gid=1127877024 GemFan 5x4.6BN prop spins 9641 RPM at 2.58 amps. To double the speed, we have to double the rpm right? Roughly doubling that RPM to 20941 takes 14.73 amps according to chart. I can get 2.58 amps for 20 minutes with a pretty small battery. Getting 14.73 amps is going to take a much bigger more expensive lipo, raise my stall speed, etc.. |
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(plane airspeed) ^ 2 / [ (plane airspeed - wind speed) * (plane airspeed + wind speed) ] So consider a plane that flies 40 kph airspeed and a 10 kph wind speed. If you were flying somewhere 40 km away; in calm air the outbound will be 1 hr and the return will be 1 hr or 120 mins. With 10 kmh tail wind, you ground speed at 50 kph outbound. 50/40 = .8 of an hour or 48 mins. With 10 kph headwind, you ground speed at 30 kph inbound. 30/40 = 1.333 hrs or 80 mins. 80 + 48 = 128 mins total; 8 extra minutes. 8/120 = 0.0666 which is only 1/15 of total extra flight time for windspeed one quarter of your airpseed. Formula way: 40^2 / [ (40-10) * (40+10) ] = 1600 / (30 * 50) = 16 / 15 and that's the same 1/15 extra flight time. A plane that flies 40 kph airspeed and a 20 kph wind speed. 40^2 / [ (40-20) * (40+20) ] = 1600 / (20 * 60) = 1.333 or one third extra time for windspeed that is half your airspeed. 40 kph airspeed and a 30 kph wind speed. 40^2 / [ (40-30) * (40+30) ] = 1600 / (10 * 70) = 2.285 or an extra 1.285 time for windspeed that is three quarters your airspeed. It's windy a lot where I live. Even if I have to fly in wind it won't effect me as much. |
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My last two testing runs above were with a 5" prop with 4.5" pitch. I also have a 4" prop with 4.5" pitch. I'll run that next time and post the results. 3 amps on 3s with 5" prop = 45 kph 5 amps on 3s with 5" prop = 52 kph I just double checked, the motor is a 2205 2600kv if that matters. What speed should I expect at the same amp draw on a 4" prop? |
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That airspeed thing really messed with me. I had to go back and make a spreadsheet to fully understand it. It's more of a math trick than purely an airplane thing.
Code:
ratio "headwind multiplier" 10% 1.01 20% 1.04 30% 1.10 40% 1.19 50% 1.33 60% 1.56 70% 1.96 80% 2.78 90% 5.26 99% 50.25 ex. It takes you 30 minutes to get out and back to a lighthouse. Your cruise airspeed is 40 kph, and there is a 8 kph headwind, so (8/40=) 20% headwind. Multiply your 30 minutes times 1.04 to get 31.2 minutes to make the trip in this wind. Even if you had a 16 kph (40%) headwind, your half hour trip would take less than six extra minutes. An easy one to remember is a headwind exactly 1/2 your cruise speed will take an extra 1/3 of time; your 30 min trip will take 40 min. I can't imagine flying in more relative wind than that is safe or fun. It gets absurd when you think of a Cessna with a cruise speed of 120 knots flying into a 108 knot headwind then turning around and flying back with that tailwind, all to make a 1 hour trip in five and a half hours. But that's the math. |
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Good to see someone else putting some hard thought into his flying
I'm sure you will agree that calculating the flight time effect of wind is just the first step. Since your lipo 'fuel tank' has limited capacity you can't bargain on getting more flight time, so the real effect is to reduce your range. I fortunately live in a place where I can bargain on zero wind most early mornings so I have not bothered to do the calculations. I just noticed when I was flying at the coast over the Xmas break that my flight distances dropped when the wind blew. Part of this was probably from the pure maths of it, as you have demonstrated. But I also probably throttled up a bit since it is harder to judge airspeed on a windy day. And you want a bit of a speed buffer to deal with turbulence. So I think the performance penalties you have calculated are the very minimum you will see. |
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