What is the Mean Aerodynamic Chord of a disk/platter? Whatever that is, I would think that the CG of a flying saucer should be no further aft than 25% or even 20% of the MAC. I am interested because I have a bucket list project for a CL flying saucer, sort of modified, based on a sketch from a 1952 Air Trails.
Keith
THERE's the AT quote! So Keith, are you thinking of the "Clough saucer" or another? Re, your question...
I just remembered your mentioning the MAC - 'woke up thinking about that. Several years ago, I derived the MAC's of several wing plan forms (straight and swept versions of straight-tapered, elliptical, and parabolic), as well as their positions and some comparisons. No one was interested, when I offered to send a copy of the file, and since there seemed such animosity toward things mathematical. I just quit contributing that sort of thing.
The interesting thing about elliptical forms is that they are easier to analyze than usual wing forms with straight leading and trailing edges and taper. Since circles are just a simpler ellipses, it's not difficult to find their MAC's. I tried to draw the diagram, but my eyesight sort of wrecked that effort. So I used Microsoft Office Tools to draw a circular wing, which is pictured below. For the circle, you can get the MAC by just multiplying the radius by 16 and dividing that answer by Three pi (3 x pi). You can find the quarter chord point of the MAC (assumed approx. aerodynamic center) by dividing that answer by four. For a circle, the distance outboard on the wing is the same as that answer). Then you can find the point on the root chord corresponding to that point by subtracting that answer from the radius. That will tell you how far back from the root leading edge the a.c. is. SO...
MAC length = 16R/(3 pi) = about 1.7 R
MAC/4 = 4R/(3 pi) = about .424 R
a.c. distance back from root leading edge: x' = R - 4R/(3 Pi) = about .576 R
Distance of MAC outward from Root = 4R/(3 pi) = about .424 R (same as MAC/4, for a circle)
I suppose a conservative version of the a.c. could be found by just going back .424 x R from the actual leading edge.
The diagram shows something else that some internet sites get wrong. Unlike straight-tapered wings, the MAC is not the same length as the local chord at its location; it is shorter. If you put the MAC at the location where it matches the local chord, you can get a ridiculous answer. Also, you would have to have a very large taper ratio to get a straight tapered wing to have the a.c. as far inboard as an elliptical half-wing (Taper Ratio T = .376). I put the top of the summary page for elliptical wings in a separate picture below. You can see that the equations simplify to what we have above:
MAC = approximately .85 x Root Chord. For a circle, that's about 1.7 x Radius.
d = .424 x Radius, for a circle
The third equation is big, because it includes sweep. For a circle it is just x' = R - .4244R = (1 - .4244) x R = about .576 x Radius
So you can get any of the three things you might want to know about the circle by just multiplying the Radius by some number. "Simpler" Wings take a bit more work.
One comment on the whole things about MAC's. These are really mean geometric chords (calculated without regard to actual air flow and pressures). They are just based on the idea of lift being the same at each point on a wing's surface, which is not true. We know by experience that the lift of a chord-length sliver of wing at some point out for a symmetrical wing section is about .24 to .25 of the way back from its leading edge. So we make up for that first assumption by considering the aero center of the wing to be at the quarter-chord point of the MAC or, better, the MGC. That gets us pretty close, but tip losses actually move the a.c. and thus the MAC inboard a bit. This works well as a benchmark for determining longitudinal balance though and we choose c.g.'s relative to this point. The convenient thing though is that you can find the point where these imaginary, equal lifts are centered simply by cutting out a cardboard scale planform of your wing and finding where it balances, because the lift and thus weight increments are directly associated with weight.
Well, if I've unsettled anyone by posting equations, 'sorry. As always, it's a FWIW.
SK