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Model Forum / Radio Controlled / Helicopters / February 2006How much G force on the blades?
Hi all, I have a question that might tickle those grey cells! Do you know (or are you) a good maths person? A helicopter rotor blade, with radius r is doing n rpm. Calculating the speed at the tip is easy - done that. It's high! (A 70 cm blade , on a 16 cm hub, at 1900 rpm is doing about 347 M.P.H. in a hover by my calculations. Now I am trying to calculate the G force (centifugal) at the tip. I thought I could find an equation on the Internet. I found several, but all wrong ones! Can you come up with a good equation? The outward force declines towards the centre of course, in a linear way. So a slightly harder sum is to calculate the total load on the bolt. Assuming the mass of the blade to be all at it's C.G is [probably] OK for this purpose. Any good at maths?! Best wishes Malcolm | I have a question that might tickle those grey cells! Do you know (or are | you) a good maths person? [quoted text clipped - 5 lines] | (A 70 cm blade , on a 16 cm hub, at 1900 rpm is doing about 347 M.P.H. in | a hover by my calculations. Of course, you're talking only about the tip of the blade. Assuming that the 16 cm hub is 8 cm on either side of the center, 347 mph is correct. | Now I am trying to calculate the G force (centifugal) at the | | I thought I could find an equation on the Internet. I found several, but | all wrong ones! Can you come up with a good equation? centifugal acceleration = v^2/r. Multiply by mass to get centrifugal force. http://phun.physics.virginia.edu/topics/centrifugal.html gives that formula. If my calculations are correct, this works out to 3150 g's. Seems awfully high, but my math seems correct. | The outward force declines towards the centre of course, in a linear | way. So a slightly harder sum is to calculate the total load on the | bolt. Assuming the mass of the blade to be all at it's C.G is | [probably] OK for this purpose. I think that would generally be accurate. In any event, if the mass distribution is even, you should be able to replace the long blade with a single point mass at half the radius. (Which is basically what you said.) In any event, this sort of calculation is easiest done if you convert values to the metric system and stick with that.  SignatureDoug McLaren, dougmc@frenzy.com "I think there is a world market for maybe five computers." - IBM Chairman Thomas Watson, 1943 > centifugal acceleration = v^2/r. Multiply by mass to get centrifugal > force. [quoted text clipped - 4 lines] > If my calculations are correct, this works out to 3150 g's. Seems > awfully high, but my math seems correct. I didn't work the numbers, but acceleration / G gives you the number of g's. Are you working consistently with the same units? Feet, and ft/sec? Meters, and m/sec? | > If my calculations are correct, this works out to 3150 g's. Seems | > awfully high, but my math seems correct. | | I didn't work the numbers, but acceleration / G gives you the number | of g's. Are you working consistently with the same units? Feet, and | ft/sec? Meters, and m/sec? Not exactly. I know it was my advice to do so, but here's what I really did. I'll be specific with things so it can sort of serve as an introduction to using the units program to do all the heavy lifting for you : Known factors : r = 78 cm. rotational velocity = 1900 rpm. r(halfway down the blade) = (70cm /2) + 8 cm, or 42 cm. centifugal acceleration = v^2/r. F = m * a calculations : % units 2084 units, 71 prefixes, 32 nonlinear units You have: 78 cm * 1900 rpm You want: radians * mph * 347.16061 radians are usually a dimensionless unit, but the program doesn't realize this, so I have to specify it ... You have: 347.16061 mph * 347.16061 mph / 78 cm You want: m/s^2 * 30878.703 That was just the application of the v^2/r formula. You have: 347.16061 mph * 347.16061 mph / 78 cm You want: force * 3150.8881 (`force' is a built-in constant for 9.8 m/s^2.) So, I really do think that 3151 g's is correct. Now, that's at the tip of the blade, so the acceleration at halfway down the blade (note that it's not half of the radius, because there's the hub (who's mass I ignore)) will be -- You have: ( 42 / 78 ) * 30878.703 m/s^2 You want: m/s^2 * 16626.994 and I don't know how much these blades will weigh, but let's guess four ounces each. So the force pulling the hub apart would be - You have: 2 * 4 oz * 16626.994 m/s^2 You want: pounds force * 847.74077 The two factor is for two blades, of course. Also note that I'm using that 9.8 m/s^2 `force' constant again. To the units program, `pounds' is a unit of mass rather than force (when in practice it's used in both ways, and people rarely state exactly which variation they're using, but the program needs to be specific, and so it is), but by throwing in the `force' constant, it's easily converted to a unit of force. Definately, we want an answer in pounds-force rather than pounds-mass, because pounds-mass would be nonsense. 848 pounds of force pulling that hub apart. Quite a bit, no? :) In any event, the `units' program is incredibly powerful, in large part because dimensional analysis is so powerful. Possibly relevant URLs : http://www.gnu.org/software/units/units.html http://en.wikipedia.org/wiki/Dimensional_analysis http://en.wikipedia.org/wiki/Pound_%28mass%29 http://en.wikipedia.org/wiki/Centrifugal_force http://phun.physics.virginia.edu/topics/centrifugal.html The last two are similar, but the last one is probably better for the layperson. Note that the `centrifugal force' doesn't even really exist, but it still feels quite real, and so that hub on your helicopter still needs to be *very* strong. And if the tip of that helicopter hits you at 350 mph, even if it only weighs 4 ounces total ... it's gonna hurt! There was a page of lots of pictures of R/C helicopter induced wounds, but I can't seem to find it now. SignatureDoug McLaren, dougmc@frenzy.com "Warning, cape does not enable user to fly" -Found on a Batman cape. > really did. I'll be specific with things so it can sort of serve as > an introduction to using the units program to do all the heavy lifting [quoted text clipped - 7 lines] > centifugal acceleration = v^2/r. > F = m * a ... > You have: 347.16061 mph * 347.16061 mph / 78 cm > You want: m/s^2 > * 30878.703 I get: r = 43 cm/2.54/12 = 1.411 ft. Every rev, the CG travels the circumference of a circle of radius r: v = (1900rpm/60)*(2*pi*1.411) = 280.7 ft/sec (= 191.4 mph). v^2/r acceleration = 55.84(10^3) ft/sec^2. G in imperial units is 32.2 ft/sec^2. Dividing ft/sec^2 by ft/sec^2 gives a scalar, dimensionless value: the number of g's. Multiplied by weight of blade, not mass, gives the force acting on the blade root. (Or, you can multiply acceleration by mass in slugs. Same difference, same numbers, slightly different ordering, same result.) For your 4 oz. blade, (4/16)*55840/32.2 = .25lb * 1735 = 433.7 lb. In the end, not far off your 848/2 lbs. The difference is easily accounted for in rounding of the value for G (32.2 ft is 9.815m), and possibly the different radius: 8 + (70/2) = 43, not 42. Lord knows why you would want radian/sec or mph. I think it served only to obfuscate, rather than illuminate. Give me an old HP RPN calculator for simple stuff like this any day. | Multiplied by weight of blade, not mass, gives the force acting on | the blade root. Technically, it's mass that you want there, not weight. The only thing that gravity affects in these calculations is the definition of a `g'. (And the rpm's required to hover, but I'm just treating 1900 rpms as a constant.) | For your 4 oz. blade, (4/16)*55840/32.2 = .25lb * 1735 = 433.7 lb. | | In the end, not far off your 848/2 lbs. The difference is easily accounted | for in rounding of the value for G (32.2 ft is 9.815m), and possibly the | different radius: 8 + (70/2) = 43, not 42. And that's why I used the `units' program for everything ... I didn't have any paper, and the one (should be trivial) thing I did in my head ... I did wrong. :) | Lord knows why you would want radian/sec or mph. I think it served only to | obfuscate, rather than illuminate. I calculated (and used) tip speed because Malcolm did too. As for radians/s, that was a hoop that the units program was forcing me to jump through, because it considered radians to be the unit of angle, but for my purposes it was dimensionless ... | Give me an old HP RPN calculator for simple stuff like this any day. Some paper helps too. It's wierd though ... it's difficult to find scratch paper anymore -- I have to pull some out of the printer, and find a pen. My kids (ages 2 and 4) will scribble over any paper they find with any pens they find, and unfortunately sometimes walls and clothes and other things as well, so we don't leave pens out. Some things, computers make harder :) SignatureDoug McLaren, dougmc@frenzy.com `No matter how good you are at it, I don't think you should ever list "pimping" on your resume.' - Ken's Deep Thoughts > | Give me an old HP RPN calculator for simple stuff like this any day. > [quoted text clipped - 5 lines] > > Some things, computers make harder :) MathCAD is somewhat similar in how it worked with dimensions. I last looked at it more than 5 years ago, maybe as much as ten now. Maybe it's grown some; or maybe it just fell into obscurity. I didn't use it for much, but it had a somewhat unique feel during interactive discovery. Just scribble down the information and relationships you know as you go, and sooner or later, you end up with a value in the dimensions you were working toward. Same as a crayon and napkin, I suppose, but it took care of the mundane calculations and unit conversions. Dimensional analysis is invaluable in solving some types of problems. ... in the ASP version that I've posted at www.swashplate.com I used the speed of the blade's CG rather than the tip speed to get the G force accell= m * v^2 / r where r is the radius only as far as the CG. Seems to be close. I assumed all the mass is at the CG, and that worries me a bit. > | Multiplied by weight of blade, not mass, gives the force acting on > | the blade root. [quoted text clipped - 34 lines] > > Some things, computers make harder :) CORRECTION! a = v^2 / r ; where v= speed of cg, r = radius to cg f=m*a ; thank you sir Isaac so: f = m * v^2 / r ; and by gosh I think it's right :-) > ... in the ASP version that I've posted at www.swashplate.com I used the > speed of the blade's CG rather than the tip speed to get the G force [quoted text clipped - 43 lines] >> >> Some things, computers make harder :) Thanks for the correction........I nearly had to go and get my Machineries out.... :-) > CORRECTION! > [quoted text clipped - 51 lines] >>> >>> Some things, computers make harder :) YesI worked in all metric units (but converted tip speed to MPH) >> centifugal acceleration = v^2/r. Multiply by mass to get centrifugal >> force. [quoted text clipped - 8 lines] > g's. Are you working consistently with the same units? Feet, and ft/sec? > Meters, and m/sec? Thanks! Your calculations are the same as mine which is reassuring! (see my ASP implementation at www.swashplate.com ) Malcolm > | I have a question that might tickle those grey cells! Do you know (or > are [quoted text clipped - 40 lines] > In any event, this sort of calculation is easiest done if you convert > values to the metric system and stick with that. Thanks! Your calculations are the same as mine which is reassuring! (see my ASP implementation at www.swashplate.com ) Malcolm > | I have a question that might tickle those grey cells! Do you know (or > are [quoted text clipped - 40 lines] > In any event, this sort of calculation is easiest done if you convert > values to the metric system and stick with that. |
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