Well, you'd never 'complete the loop'.

Define loop as 360 degrees of turning. It won't be a circle, it'll be a sort of ellipsoid thing.

I'll post the answer for doing this at constant speed as well as constant power.

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Ed O'Malley
www.VeloVetta.com
Founder of VeloVetta Cycling Shoes
Instagram • Facebook

Seems like you wouldn't go in a loop. If you're rotating the bike at a constant rate and pedaling at a constant power, your average position will get pushed by the wind.

see my answer above. when I say loop, I do not mean circle.

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Ed O'Malley
www.VeloVetta.com
Founder of VeloVetta Cycling Shoes
Instagram • Facebook

I know. But it seems like the shape would be something like this:



Possibly with some drift to the side as well... I haven't sorted that out.

Well if your solution is like your CFD,

it will be wrong.

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HELLO HAMBINI FANS!!!

I don't want to give too much out about the shape of the path as that may be a hint. But you are right that it will not be a circle.

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Ed O'Malley
www.VeloVetta.com
Founder of VeloVetta Cycling Shoes
Instagram • Facebook

The issue isn't whether the path is circular, it's whether it's a closed loop.

The problem statement seems like a trick question.

Are you measuring yaw on the frame or the cockpit/front wheel?

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"They're made of latex, not nitroglycerin"

If you're rotating at a constant rate, isn't everything else irrelevant? It's basically the same as if you were sitting on a turntable that was rotating (0s - 0°, 1s - 1°, ...). Now this is a bit contrived since on a path where with a consistent input power you'd travel different distances each second based on the effective forward force (your input plus/minus the wind's force) so you'd need to do this on a large flat paved surface (e.g., parking lot) to follow the path these parameters would force on you.

Constant speed would end up being a circle with constant angular rotation and the power would vary quiet a bit.

I might also be totally wrong 🙂

...if you stop the trace after 360 seconds, you'll be facing north, but be positioned somewhere south of your original position. Like a capitol "C".

For the sake of the thought experiment, the wording was misleading. OK, if that revealed some hint, instead of saying "close the loop", it should've said once you "are again facing due north".

Well it's not like a turn table because there is a wind component from your forward motion as well as from the actual wind, which obviously you understand but just to make sure I'm clear...

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Ed O'Malley
www.VeloVetta.com
Founder of VeloVetta Cycling Shoes
Instagram • Facebook

Measured from the frame.

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Ed O'Malley
www.VeloVetta.com
Founder of VeloVetta Cycling Shoes
Instagram • Facebook

It's not meant to be a trick question. I was just surprised by the result and I'm wondering if I'm the only one.

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Ed O'Malley
www.VeloVetta.com
Founder of VeloVetta Cycling Shoes
Instagram • Facebook

Sorry for being confusing. Yes, when you are facing due north is a better way to describe it.

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Ed O'Malley
www.VeloVetta.com
Founder of VeloVetta Cycling Shoes
Instagram • Facebook

stop entertaining this guy. Anyone who asks a question then stays they'll give the answer later just wants attention and to flaunt how smart they think they are

I'm gonna stop being cryptic and just post! So here are the results:







I was surprised to see this. Even though this is a completely manufactured situation (chosen because I could easily model it in excel) I was expecting something that at least resembled the normal distributions that everyone who's measured this in the real world has seen (Flo, Swissside, Mavic, etc). There's a good summary here:


https://www.slowtwitch.com/...Yaw_Angles_5844.html

The reason for it is that the rate of change of yaw angle is lowest during the periods of high yaw - that is the amount yaw angle changes with a 1 degree change of heading:



(oops... the right vertical axis there should not be speed, it should be yaw rate of change dYaw/dtheta)

Now an 8mph wind is a bit on the high side. If I cut that down to 4mph, those outer buckets go down to [7 - 9].

I just thought it was interesting that this is so different from real-world measurements. I do not really understand why yet. Thoughts?

Disclosures - this model is simplified using only aero drag and rolling resistance. Inertia, bearing drag and drivetrain friction are ignored.

I tried to put the spreadsheet in google sheets to share but it's having trouble with some of the math....

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Ed O'Malley
www.VeloVetta.com
Founder of VeloVetta Cycling Shoes
Instagram • Facebook

I think part of it is that a steady 8mph at ground/bike level is a fairly intense wind. Standard wind measurements are, what, 20 feet above ground?

And I think another thing is that, compared to riding a perfectly circular course, with the parameters in your model that lead to a skewed ellipsoid path (I’d need to chart this out to visualize better) it is going to naturally have more time/distance at higher yaw angles.