With endless discussions about what is the most aero front wheel and how they handle in cross winds, can’t somone figure out how much material there is on the 808 vs. the Hed 3 coming from a direct cross wind. I know there are other factors at work here, but I am curious, let’s eliminate all other factors. We know the rim depth of the 808, we know the circumference of a 700c wheel, you could then figure out the circumferance of the inner edge of the rim, multiply the difference between the two circumferances by the rim depth. Also, we know the rim depth of the H3 and the width of the spokes. Someone, please, do the math.
…divide by 4, carry the 1…answer is…2.7356891315748
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I am a math major and actually this would be more of a physics question. Actually, knowing what little I know about aerodynamics, the actual amount of material would have very little to do with the effect of a crosswind on a wheel. The shape of the wheel would have more to do with it. While not everyone agrees on this example, look at aero tubing (I know Gerard can back me up with some proofs since we know is the dominant force of aero tubing in the market). An Aero downtube, like on the Soloist, would probably have more material (i.e. surface area) than a standard round tube, but we know that it would do better in the wind vs. round tubing. Also, you have to remember the bike is moving down the road and the wheel is whirrling about and that totally changes the problem and makes it much more complex than simply a surface area. Just my thought though…
Brent
EXPOSED SURFACE AREA FOR 1 SIDE OF EACH WHEEL
OK, for the 808…
(pi x radius1(squared)) - (pi x radius2(squared))
Where:
radius 1 = center to outer edge of 808
radius 2 = center to inner edgo of 808
Now for the Hed H3…
(pi x radius1(squared)) - (pi x radius2(squared))
Where:
radius 1 = center to outer edge of H3
radius 2 = innermost portion of cutout to outer edge of cutout
The three cutouts on the H2 really are just a much smaller inner circle, split into 3. It’s like putting gaps in a pie chart - they eventually fit together into a fairly perfect circle. So forget that the H3 is a tri spoke, but think of it as two circles, much like the 808.
I’n not a math major, so don’t hack me apart if I’m wrong. I’m a physiology guy, with and inquiring mind…but I think I’m on the right track here.
OK, you’re right, but I don’t care about any of that. Just out of pure curiousity I want to know which one has more exposed surface area from a direct crosswind.
H3 does.
I am curious, let’s eliminate all other factors.
You kind of lost me there…
See, now we are getting somewhere…
In Reply To I am curious, let’s eliminate all other factors.
You kind of lost me there…
How did I lose you? Was it because you quoted only half of my sentence? Read the first half, I am admitting there are other factors at work in terms of how each wheel would handle in the crosswind, but I am not requesting anything scientific, just trying to get the answer to something I am curious about…so, therefore, let’s eliminate all the other factors at work. Did I really lose you with that? I don’t think it was so complicated, my sentence structure may not have been perfect, but I think you could figure out what I meant…or are you just living up to your name 
If you’re eliminating all the important factors, what’s the point of knowing side area?
For the third time now…because I am CURIOUS…simple as that, nothing more to it.
Then how come you don’t do the math?
Because I am not smart enough…and it’s the off-season, nothing too challenging for the next couple of weeks 
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If you have photoshop then all you need to do is get a side view photo of each wheel, convert it so the wheel surface and spokes are black and everything else is white, then adobe photoshop can count the number of pixels. Once you have that, you need to have a known area within the same frame and count the pixels of that. Then you simply do a proportion of known area / # pixels in known area * # of pixels in wheels.
Okay. Remember 700’s are EXACTLY 28in. in diameter, not exactly 700mm., at least according to what I’ve read. So that is exactly 711.2mm. We’ll assume that as the outer diameter. The Zipp 808 rim is 81mm deep.
Zipp 808: ((711.2/2)^2 - (549.2/2)^2)*PI = 0.1603 m^2, ignoring surface area of spokes & hub.
Hed3: 72mm255mm3 (approximate rectangular area of “spokes,” using my trusty tape measure) + ((711.2/2)^2 - ((711.2-55*2)/2)^2)*PI (55mm is the rim depth according to my same trusty tape measure) = 0.1685 m^2
H3 > Zipp 808… What can I say, I’m bored too…
Since what you are getting at is the effective lateral force on the wheel in a crosswind, you can look here: http://www.zipp.com/tech/documents/ANoteonRimWidth_002.pdf
Page 5 has a graph of actual side force on the wheel relative to wind angle for the two wheels being discussed as well as numerous others.
The thing that you won’t get just figuring the side areas of the two wheels is that the curvature of the surfaces has as much or sometimes more of an effect on side load than the pure surface area, so that a wheel with high surface area, and also high curvature like the 808 can have less drag as some wind angles than a wheel with more surface area but less curvature like a 3 spoke.
josh
technically, you need to know the curvature of the rim and compute a triple integral…but I am really not that bored to do this and your approximation will have to do…
I’m not measuring rim curvature for you, even if you were…
Oh damn, I forgot about spokes, this could be close huh?
i always knew princeton alumnis were slackers…