What am I missing? - Wing area not right???

[hardlock]

I need to find the area of an eliptical wing but have run into a major snag.

Take a 1 sq foot area based on a 12x12" square. If you add up the sides you get 4x12=48" right?

Now working backward with a circle having a 48" circum. you divide the 48 by 3.1416 getting 15.28 and divide that by 2 for a radius of about 7.64".

Now calc the area of a circle with this radius by taking 7.64 x 7.64 = 58.37.

58.37 x 3.1416 = 183.37 sq inches!!!!! That's a long way from 144 when the same circum. is in the shape of a square. What am I missing? :o

[Sparky Paul]

Your procedure is in error.
The relation between the perimeter of a square and the circumfernce of a circle must result in different areas for the enclosed area, with the circle being larger.
Draw this out to see what you have.
Your circle's diameter is 15.28 inches, which is 3.28 inches greater than the length of the square. The area has to be different, and in this case larger.
What you are trying to accomplish with this is not stated.

[hardlock]

Thanks Sparky!

"I need to find the area of an eliptical wing" is what I'm trying to accomplish.

Throw logic out the window I guess. I was thinking if the perimeter/circumfernce were equal, so would be the area. Yes, the circle will be larger but with the corners of the square then cut off, thought they should equal out.

So a circle wing of the same outer frame material length will weigh the same as if laid out square but with will give bonus wing area?

This is a case where you do get something for nothing! Still can't wrap my logical brain around it but can't complain. Quess I'll have to do it the hard way by counting squares on graph paper. :)

[Sparky Paul]

For a regular ellipse, area is pi*a*b, where a is the root chord, b is the span.
Rather than count squares, it's more accurate to compare weights!
Find what an amount in rectangular form of something like foam board or cardboard weighs, cut your ellipse from that something, and weigh it.
Ratio the weights, to get the multiplier for the area difference.
100 sq. in. weighs 10 oz, the ellipse weighs 7 ounces, the multiplier is .7, the ellipse has 70 sq.in. of area.

[hardlock]

This is for one of those bow the CF rod and cover with cling wrap wings. The ellipse formula is just what I needed, thanks!

Considering the covering weighs next to nothing and I even save a little in the corner fittings needed for a rectangle, It looks like I will indeed get some "free" wing area. Cool! :)

(after some figuring)

Wait a minute now - The semi span is 11.5 with the root being 7 inches. That would be 3.1416*7*11.5? That must be the formula for the entire wing? I come up with 252.9 sq in. where-as figuring based on a part rectangle (7x8) with a half circle on the end (3.5 R) only got me about 75 sq in per side for a total area of 175 sq in.

Figuring it backward with the perimeter of the semi wing layout at 34" and running the circle area calc gives a semi span area of about 92 with the full wing = to 184 sq in..... Even as a circle that's still less than the elipse method. Guess I'm still lost. :(

(after some digging)

It appears that the formula is based on semi span and semi chord dimentions so it should have been about half for 126 sq in. This sounds more realistic as the same planeform outside rectangle would only be 23 x 7 = 161 and there's lots of area lost for an ellipse. :)

[slipstick]

Yes Paul omitted the critical "/ 4". If the planform is a true ellipse of span x and maximum chord y, the area is (PI * x * y) / 4. Of course a bent carbon rod is not going to form a true ellipse anyway so it's a very approximate calculation.

Steve

[Ollie]

I went into to my shop. I made a loop of 48" long, 0.03 Dia. carbon rod. I squashed the circle to a center chord of 6". The span was ~20". The fat bow-tie shape hold by the squashed loop. The chord between center and tip was about 7-3/8". My estimate of the fat bow-tie area was ~180 square inches.

I tried a 0.07 Dia. by 48" rod with the tie-bow. It had the same wing plan shape and the same area as the 0.03 Dia. rod.

Forget the ellipse math for this job. Be practical.

[Andy W]

As a side note, I've calculated areas of complex wing shapes by putting a plan view into a paint program, cutting out all but the wing, and filling it in black. I then bring up a color distribution graph (paint shop pro has this) and that gives me an easy way to calculate wing area.
..a

[Sparky Paul]

Yes Paul omitted the critical "/ 4". If the planform is a true ellipse of span x and maximum chord y, the area is (PI * x * y) / 4. Of course a bent carbon rod is not going to form a true ellipse anyway so it's a very approximate calculation.

Steve
.
Yeah, me and Archimedes leave out the good parts... :)
http://en.wikipedia.org/wiki/Ellipse#Area

[vintage1]

You can lay tghe bow over a square grid of graph paper and count the squares. Any part squares are either ignored, if less than half a square, or counted as one, it greater than half a square.

Its usually good for 95% accuracy.

[davidfee]

The equation A=pi*a*b works if you use the semispan and 1/2 chord. If you use span and chord you need A=(pi*a*b)/4. (for a true ellipse)

[slipstick]

.
Yeah, me and Archimedes leave out the good parts... :)
http://en.wikipedia.org/wiki/Ellipse#Area

Which slightly higher up the page says "The constant a equals the length of the semimajor axis; the constant b equals the length of the semiminor axis". In English this of course equates to half the span and half the widest chord ;).

That "wikipedia" is a laugh a minute, innit ? That's about the only time I've seen pi referred to as Archimedes' Constant in about the last 50 years (still at least they got the apostrophe in the right place ;)).

Steve

[Ollie]

hardlock,
"This is for one of those bow the CF rod and cover with cling wrap wings. The ellipse formula is just what I needed, thanks! "

Using carbon rod can made with an ellipse shape, you must use lots of ribs to make the outline fit an ellipse.

If you want an outline with out ribs, then you must make a fat bow-tie shape for a rod outline. That is because the rod bends it's own way under the bending stress of one rod diameter along the length.

Try it.

[hardlock]

Thanks all - great info! Since I'm using separate wing halfs for this with a root rod and played with the rods, the planform isn't the typical bow-tie shape but not an ellipse either. Closer to half circle tips with rectangle but has a slight taper also.

Either way it goes, I've learned a lot here that will help me optimize for most wing area within a given span. :)

[vintage1]

most wing area within a given span on a single wing is and infinitely large chord square planform ;)

[jay flay tway]

Not sure if this is helpful but to get a very accurate wing area figure you could get a hold of a planimeter (I think that's what they're called)--which is used for finding he area of landforms on maps--from a survey or cartographer supply shop (?) and trace your wing planform on a piece of paper.
The device is handheld and has a rolling wheel and somehow measures the area.