> -----Original Message----- > From: Vivian Meazza > Sent: 04 May 2004 7:38 pm > To: 'FlightGear developers discussions' > Subject: RE: [Flightgear-devel] Spitfire Propeller vs. YASim > > Richard Bytheway wrote > > > Sent: 04 May 2004 10:42 > > To: FlightGear developers discussions > > Subject: RE: [Flightgear-devel] Spitfire Propeller vs. YASim > > > > > > > I had already shown by some pretty simple math that at 2850 > > > rmp the tips of > > > a 1.65m radius propeller would be supersonic and therefore highly > > > improbable, but we now know that the data of hp, gear ratio, > > > rpm etc all tie > > > together. > > > > > > Thanks > > > > > > Vivian Meazza > > > > > > > I have a memory from years back of being told that the reason > > the Spitfire had such a distinctive sound was that the > > propellor tips _were_ supersonic. Maybe it was just heresay. > > > > Richard > > > > I think it is possible that the propeller tips went supersonic in the > corners of the flight envelope of some of the later versions. > However, the > math seems to show that in most circumstances they were not. It seems > unlikely that this could explain the distinctive sound when > heard from the > ground. > > Here are some calculations on propeller rpm. > > The propeller the tip speed should be as high as possible > with the only > limitation being that the tip should not get into the region > of aerodynamic > compressibility. Typically a figure of Mach 0.85 is used as > the magic number > that should not be exceeded. (This makes some allowance for the speed > increase as the air passes over the aerofoil curved surface > and the increase > in air velocity caused by the propeller operation.) > > If we take 8000 ft as the operating altitude then Mach 1 = > 1085 ft/sec > (approx) > > Assuming that the forward velocity of the aircraft is 300 mph > = 440 ft/sec > > Then the maximum rotational velocity may be calculated by Pythagoras: > > Max Rotational Velocity = ((M *1085)^2 - (V)^2)^0.5 > > where M is the designed Mach Number (0.85) and V is the > aircraft forward > velocity > > = ((0.85*1085)^2 -(440)^2)^0.5 = 810.52 > ft/sec > > RPM at Max rotational velocity is given by: > > RPM = Max rotational velocity*60/(PI * D) > > Where D is the propeller diameter (ft) > > = 810.52*60/(PI * 10.75) = 1439.98 rpm > > At 3000 rpm the propeller rpm is 1431 rpm, but the Merlin > only did this when > the throttle was through the gate, and the Boost Control > Valve Cutout was > operated. This was allowed for 5 minutes. > > We can calculate the Max Rotational Velocity @ 1431 rpm > > Max rotational velocity (PI * D) = (RPM/60) * (PI * D) > > = (1431/60) * > (PI * 10.75) > > = 805 ft/sec > > We can also calculate the Mach Number (M) of the tip by > rearranging and > substituting > > M = ((805^2+440^2)^0.5)/1085 > > = 0.8459 > > I hope that all the maths are correct. > > I think all this shows that under normal operating > conditions, and observing > the normal operating limit of 2850 rpm, it is unlikely that > the propeller > tips would exceed M1. > > Regards > > Vivian >
Very clear, thanks, Richard _______________________________________________ Flightgear-devel mailing list [EMAIL PROTECTED] http://mail.flightgear.org/mailman/listinfo/flightgear-devel
