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Old 04-16-2019, 07:25 PM   #33 (permalink)
Vman455
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Quote:
Originally Posted by freebeard View Post
Okay, so 3/4ths the tire height minus the ground clearance, over 1/2 the tire height, I guess. Thanks.
It's easier to think of it as the simple ratio of body cover:wheel diameter, I think.

Quote:
Originally Posted by freebeard View Post
Except the text is talking about the wheel well radius, not the tire radius called out in the illustration. According to this a close fitting cycle fender would have more drag than a looser fitting one?
I don't think a cycle wheel in a fender is analogous. For one thing, it doesn't have a vehicle body around it. Second, as Hucho notes, in zero-yaw conditions the windstream at the front wheels is approaching at an angle of ~15 degrees, where no such shear would be present for the bike wheel.

The last few sentences of that paragraph I take to mean "...increased [wheel] radius and reduced [wheel housing] clearance," based on what's written before it.

We should also note that Hucho, citing an earlier study by Cogotti, indicates a completely linear relationship between wheel housing volume reduction and wheel drag. Interesting.

Quote:
Originally Posted by freebeard View Post
A tight-fitting Beemer-style lip would isolate the wheelwell more than an open 4x4 wheelwell — that's different than the cavity volume.
But that is directly related to the cavity volume, if we're holding other conditions constant, like the body cover.
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