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Theory of rotational drag of wheels
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There has been a lot of discussion already about this topic in several threads, and there is no consensus about the effect the rotational drag has on the total aerodynamic drag which a cyclist encounters.

I want to explain here my understanding of the problem at the moment.

I devide a wheel in four sections as such:



Let's assume that the wind conditions in the upper, front and rear sections are the same in the windtunnel as on the road (I treat the lower section further below).

To further simplify the problem, assume that there is only a horizontal component of drag (Du) in the upper section. In the front and rear sections there is a horizontal (Dfh and Drh) and a vertical component (Dfv and Drv) of drag.

We consider the drag caused by the wind in the upper section and the drag caused by the vertical movement of the front and the rear sections as rotational drag. This drag can be measured by measuring the torque of the motor of the roller which moves the wheel in the windtunnel. The drag of the (horizontal) wind in the front and rear sections (Dfh and Drh) are not measured by the torque of this motor.

In a windtunnel the horizontal forces are being measured by the horizontal scales with which the bicycle is fixed. The vertical forces not though.

If now (what is done often), the torque of the driving motor is not measured, but only the drag measured by the horizontal scales, the vertical drag in the front and rear sections of the wheels (Dfv and Drv) are not measured. The total drag measured is thus smaller than the drag in the real world is.

If however the torque of the driving motor is taken into condideration, and the power of it is simply added to the power measured by the horizontal scales, a figure is determined which is too high, because the drag in the upper section is measured double, once by the horizontal scales and again by the torque of the motor.


QUANTIFICATION OF THE ROTATIONAL DRAG

The total drag in reality is thus the drag measured by the scales and a part of the drag measured by the torque of the motor. To quantify this part is not easy to determine, but I give it a try:

The speed of the wheel at the very top is twice as high as the speed of the bicycle on the road (which latter corresponds to the speed of the wind in the windtunnel) let's therefore assume that the speed in the upper section is twice as that of the bicycle on the road.
To simplify further, assume that the vertical components of the speed in the front and rear sections is the speed of the bicycle (which is of course only true at the very front and rear of the wheel).

So the vertical components of the speed in the front and rear sections relatively to the air are the speed of the bicycle. Also the horizontal components of the speed in the front and rear sections relatively to the air are the speed of the bicycle. The speed of the upper section realtively to the air is twice the speed of the bicycle.

To simplify assume that the horizontal and vertical drag in both the front and rear sections are the same (in reality, with spoke wheels, probably the vertical drag will be greater than the horizontal drag, because in the vertical movement the spokes is crossways relative to the air). The drag in the upper section is though four times greater, because drag increases to the square of the speed.

If we set the horizontal drag Dfh in the front section "D", we have the following components of drag to the wheel:
Horizontal drag in the front section (Dfh) = D
Vertical drag in the front section (Dfv) = D
Horizontal drag in the rear section (Drh) = D
Vertical drag in the rear section (Drv) = D
Drag in the upper section (Du) = 4*D
The total drag of the wheel is thus 8*D.

If we only measure the drag with the horizontal scales, we only measure 6*D which is 75% of the total drag.

If we add the torque drag to the drag measured by the horizontal scales, we measure 6*D + 6*D is 150% of the total drag.

If we want to get to 100%, we would have to add a third of the torque drag to the drag measured by the horizontal scales.


DRAG IN THE LOWER SECTION OF THE WHEEL

Upto now we have neglected the drag in the lower section. There should indeed not be a lot of drag in this section in the real world, since the lower part of the wheel has little speed (the contact point of the wheel on the road has zero speed).

To simplify again, let's assume that we want to emulate the situation with no wind on the road. In the windtunnel, we must have then an airstream from the front which has everywhere the same speed. This is though not the case in windtunnels. The pictures I have seen show that the bicycle is fixed on a plate. Fluid mechanics teach us, that the speed of an airstream at a wall or a plate is zero. So the airstream in our windtunnel at the lower section of the wheel will be lower than the airstream higher up (you see that in the drawing). The roller will move the wheel though with a speed of the airstream higher up. This means, that the lower section of the wheel moves faster backward than the airstream does at this position, giving the wheel a forward thrust.
I call this forward thrust the "Mississippi-effect" (Dm) because of the paddlewheels of Mississippiboats.
The Mississippi-effect of a spokewheel should be greater than that of a disc.
If drag is measured only with the horizontal scales, the measure-result is thus flawed in that the drag measured is smaller than the drag in the real world.
A torque motor would measure this drag which is not there in the real world.


CONCLUSION
The theory of drag of a wheel is a very very very very very complicated problem.
Last edited by: longtrousers: May 12, 17 3:58
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Re: Theory of rotational drag of wheels [longtrousers] [ In reply to ]
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Very good summary. As an automotive engineering student with backgrounds in aerodynamics, I can just say well done. There is nothing to add.

And the conclusion that it's a very very very complicated problem should shut down a few discussions wether the 4500$ wheel set or the 5000$ wheelset will make ones 5hour split 2 seconds faster!

There needs to be a lot more engineering, research ,calculating and testing for those marginal gains (in comparison to a descent fit)


Thanks for taking your time and explain everything in detail

sent from my iPhone
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Re: Theory of rotational drag of wheels [Stromlinie] [ In reply to ]
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Why not measure the power it takes to spin the wheel with no wind? The drag should be the same in all four quadrants. Then crank up the wind speed and get values at different wind speeds to get a curve.

Great fun.
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Re: Theory of rotational drag of wheels [nglitz] [ In reply to ]
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nglitz wrote:
Why not measure the power it takes to spin the wheel with no wind? .

That figure seems to be neglegably small and not very indicative for the situation with wind, as I understood from some tests. Probably the air starts circling around such that hardly any drag remains.
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Re: Theory of rotational drag of wheels [longtrousers] [ In reply to ]
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longtrousers wrote:
...
CONCLUSION
The theory of drag of a wheel is a very very very very very complicated problem.

No, it's not. I tried to lay it out here, with free body diagrams, to no avail.

Short version:
1 understand how the wheel moves
2 measure where the resultant aero drag vector acts, and thus the aero drag torque
3 calculate the horizontal hub force required to overcome the aero drag torque
4 normalize to CdA through dynamic pressure

High school level really... not the aero part but the measured and calculated forces, vectors and torques.

Main problem IMO is the notion of translational AND rotational.
There is ONLY rotational, but not around the hub...
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Re: Theory of rotational drag of wheels [longtrousers] [ In reply to ]
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longtrousers wrote:
nglitz wrote:
Why not measure the power it takes to spin the wheel with no wind? .


That figure seems to be neglegably small and not very indicative for the situation with wind, as I understood from some tests. Probably the air starts circling around such that hardly any drag remains.


A more complete diagram (not "discretized into four sections, but described as continuous functions) would show that the relative speed of the wheel/tires/spokes is U + omega*r, which looks like a curtate cycloid (https://en.wikipedia.org/wiki/Cycloid). It's not a linear of a calculation in real life, but it gives you a general sense of how the magnitude differs from a statically rotating wheel.

The act of getting the air closer to the relative velocity of the body is actually the manifestation of drag, i.e. the air has taken energy from the wheel. Although the wheel is imparting shear stresses tangentially, the local air will gain approximately linear momentum and you should feel a radial flow component from the tires/wheels.

I like to think of wakes and energy losses not just from the body frame of reference, but in an inertial frame, where a wake has actually gained momentum relative to ambient air. It's interesting to see how significant the wake is for a pack of crit riders from the sidelines, or to feel the duration of the wake leaving the riders linearly if you stand at the end of a straight just as they turn a corner.
Last edited by: codygo: May 14, 17 5:42
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Re: Theory of rotational drag of wheels [Nicko] [ In reply to ]
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Nicko wrote:
longtrousers wrote:
...
CONCLUSION
The theory of drag of a wheel is a very very very very very complicated problem.

No, it's not. I tried to lay it out here, with free body diagrams, to no avail.

Short version:
1 understand how the wheel moves
2 measure where the resultant aero drag vector acts, and thus the aero drag torque
3 calculate the horizontal hub force required to overcome the aero drag torque
4 normalize to CdA through dynamic pressure

High school level really... not the aero part but the measured and calculated forces, vectors and torques.

Main problem IMO is the notion of translational AND rotational.
There is ONLY rotational, but not around the hub...


Maybe I should have written "it is a very very very complicated problem at least to me", because to me it is and it can well be that my current understanding is incorrect.

At least, as you can see if you read my thoughts in the first post, you can conclude that I understand that the wheel rotates on the road around its contact patch (i write that the lowest point of the wheel has has zero speed and the upper double speed of the bike). Of course in a windtunnel it is different, here the wheel rotates around its hub.

And I assume that in a windtunnel the horizontal power with which the bike (and its platform, and all of the system which is fixed to it) is fixed in the windtunnel is measured. Isn't that what is often done in a windtunnel, that only one horizontal force is measured? (I saw you write about a six axis load cell, but I must confess I do not know how to understand that).
Last edited by: longtrousers: May 14, 17 9:15
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Re: Theory of rotational drag of wheels [longtrousers] [ In reply to ]
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longtrousers wrote:
... I understand that the wheel rotates on the road around its contact patch (i write that the lowest point of the wheel has has zero speed and the upper double speed of the bike). Of course in a windtunnel it is different, here the wheel rotates around its hub.
...
You're almost there.
Now picture a point on the perimeter of the tire, in the wind tunnel. If the test is conducted properly, this point on the tire will experience NO WIND the moment in time it is at the bottom, right? This is exactly what happens in real life, when a point of the tire is in contact with the road (if there is no wind), agree?
So in the wind tunnel, the wheel DOES rotate around an imaginary contact patch, not around the hub.

In the aerodynamic modeling of a bicycle through still air, the solid objects move through the molecules of air that initially are "at rest", with the road acting as a "wall" for the pressure waves and induced swirling. If the solid object is a wheel, nothing moves through the air in a linear trajectory. Hence there is nothing "translational"...

In the wind tunnel, the frame of reference is the moving air, not the static hub. The moving air is then "still".
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Re: Theory of rotational drag of wheels [Nicko] [ In reply to ]
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Nicko wrote:
longtrousers wrote:
... I understand that the wheel rotates on the road around its contact patch (i write that the lowest point of the wheel has has zero speed and the upper double speed of the bike). Of course in a windtunnel it is different, here the wheel rotates around its hub.
...
You're almost there.
Now picture a point on the perimeter of the tire, in the wind tunnel. If the test is conducted properly, this point on the tire will experience NO WIND the moment in time it is at the bottom, right? This is exactly what happens in real life, when a point of the tire is in contact with the road (if there is no wind), agree?
So in the wind tunnel, the wheel DOES rotate around an imaginary contact patch, not around the hub.

In the aerodynamic modeling of a bicycle through still air, the solid objects move through the molecules of air that initially are "at rest", with the road acting as a "wall" for the pressure waves and induced swirling. If the solid object is a wheel, nothing moves through the air in a linear trajectory. Hence there is nothing "translational"...

In the wind tunnel, the frame of reference is the moving air, not the static hub. The moving air is then "still".

I agree with this (nearly) totally: and my first post is in line with this.
(I write "(nearly)", see what I call the "Mississippi-effekt". )
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