...divide by 4, carry the 1...answer is....2.7356891315748

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.

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Portside Athletics Blog

H3 does.

You kind of lost me there...

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"Yeah, no one likes a smartass, but we all like stars" - Thom Yorke


smartasscoach.tri-oeiras.com

See, now we are getting somewhere...

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Portside Athletics Blog

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 ;-)

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Portside Athletics Blog

If you're eliminating all the important factors, what's the point of knowing side area?

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"Yeah, no one likes a smartass, but we all like stars" - Thom Yorke


smartasscoach.tri-oeiras.com

For the third time now...because I am CURIOUS...simple as that, nothing more to it.

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Portside Athletics Blog

Then how come you don't do the math?

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"Non est ad astra mollis e terris via." - Seneca | rappstar.com | FB - Rappstar Racing | IG - @jordanrapp

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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Portside Athletics Blog

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: 72mm*255mm*3 (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...

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"Non est ad astra mollis e terris via." - Seneca | rappstar.com | FB - Rappstar Racing | IG - @jordanrapp

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/...teonRimWidth_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

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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...

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"Non est ad astra mollis e terris via." - Seneca | rappstar.com | FB - Rappstar Racing | IG - @jordanrapp

Oh damn, I forgot about spokes, this could be close huh?

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Portside Athletics Blog

i always knew princeton alumnis were slackers... ;-)

You win the less surface area debate as well, though!

EDIT: Although not with the spokes.

What is the actual surface area of the rim?

And really, my math is not entirely accurate since, as zee Frenchman pointed out, this is really a problem that requires an interval...

How has this thread captured our attention so much...

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"Non est ad astra mollis e terris via." - Seneca | rappstar.com | FB - Rappstar Racing | IG - @jordanrapp

According to Sapim, the surface area of each spoke is ~ 2.3mm * 260mm = 5.98 * 10^-4 m^2 (remember the hub takes up some room).

So with 18 spokes on the front, that gives you an extra 0.0108 m^2, 0.0144 m^2 in the rear.

So the Zipp 808 is now higher!

Okay, I'm really bored.

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"Non est ad astra mollis e terris via." - Seneca | rappstar.com | FB - Rappstar Racing | IG - @jordanrapp

You may be bored but you answered my question...thanks!!!

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Portside Athletics Blog

So are you actually gonna buy some wheels now? My vote is for the 808... Well, 999, but same difference.

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"Non est ad astra mollis e terris via." - Seneca | rappstar.com | FB - Rappstar Racing | IG - @jordanrapp

yes, that's the right track. But just to be as precise as we can to answer the question (c'mon, smartass, the guy just wants to know, what's wrong with that?), we need to add back in the area of the three blades on the H3--which would be simply:

3 x length of blade * width of blade.

the radius of a 700c wheel is 35cm, give or take. I suppose somebody could measure from center to outer edge of each of those rims.

the depth of the 808 rims is 81mm, call it 8cm, so we can go ahead and solve that one: pi*34cm^2 - pi*27cm^2 = 1557 square cm.

in the process, btw, we also found the area of a full disc, which is just pi*35cm = 3847 square cm.

For the H3, again, we want to start with the area of a full disc, subtract the area of the smaller circle of nothingness inside the rim, then add back in the area of the three blades. Note that because HED makes the intersections a little swoopy, we're going to come up a tiny bit short, but I don't think it will be too significant. I think the depth of the rim is 51mm, so we'll call it 5cm (correct me if I'm wrong...or just use the right number and figure it out). I don't know what the width of those spokes is, but my carefullest estimate from looking at the picture is that the width of each blade is about 2/9ths of the radius, or a little under 8cm. I'll call it 8, and maybe this high-side estimate will cancel out my neglecting the little swoopy bits.

pi*35cm^2 - pi*30cm^2 + 3*(30cm*8cm) = 1741 square cm.

So roughly:

full disc: 3850 square cm

Zipp 808: 1560 square cm

H3: 1740 square cm

So the H3 will catch a bit more crosswind. As smartass will no doubt point out, though, that doesn't mean it will 'feel' like more, b/c the push of the wind near the center doesn't put as much torque on the wheel as the push near the edge. So my guess is that the H3 actually feels more directionally stable in a straight crosswind, even though it has a bigger area. But for what it's worth, I gave an honest stab at answering!