What does Physics prove ?
#27
For every car, for example, there's a more-or-less well defined threshold speed, over which the turbulent component starts to dominate (the exact value depends on the aerodynamic properties of a specific car). Under that threshold the car's fuel consumption (MPG) depends very little on the actual speed. Over that threshold, as v^2 component begins to become more prominent, the car's MPG is dropping noticeably as the speed increases. It is often assumed that the threshold value is 55 mph, while in reality the number makes little sense, since for each car model it is different (sometimes significantly).
A typical bike with a cyclist on it is not a very efficient aerodynamic shape, so it is not a surprise that the relationship will largely depend on v^2 even for low speeds. A trailer can be built in a much more efficient aerodynamic shape, thus extending the range of speeds in which the linear air resistance component dominates over the squared component. This is splitting hairs, of course, especially if the trailer is pulled by a bike (which immediately negates any aerodynamic benefits).
In any case, to call the relationship "exponential" (as previous poster did) is certainly a major error.
Last edited by AndreyT; 12-01-11 at 04:09 PM.
#28
Strictly speaking, air drag increases linearly with speed as long as the air flow remains laminar. As the speed increases, the flow eventually detaches and becomes turbulent, and the air drag becomes proportional to the square of speed. In real life it is usually somewhere in between, and for faster speeds it moves closer to v^2.
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A typical bike with a cyclist on it is not a very efficient aerodynamic shape, so it is not a surprise that the relationship will largely depend on v^2 even for low speeds.
<snip>
A typical bike with a cyclist on it is not a very efficient aerodynamic shape, so it is not a surprise that the relationship will largely depend on v^2 even for low speeds.
#29
Senior Member
Joined: Sep 2007
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From: Gaseous Cloud around Uranus
Have fun "cipherin".....
where
FD is the force of drag, which is by definition the force component in the direction of the flow velocity,[1] ρ is the mass density of the fluid, [2] v is the velocity of the object relative to the fluid, A is the reference area, and CD is the drag coefficient — a dimensionless constant related to the object's geometry and taking into account both skin friction and form drag.
Short of having a wind tunnel and some way of measuring resistance(like a spring scale....LOL!)I'm not sure what good this will do.Your going to have a few problems coming up with some numbers.
Look mom...I have a wind tunnel and a spring scale.....I'm ready to do some math.....should I get panniers or a trailer....let's find out.....
whereFD is the force of drag, which is by definition the force component in the direction of the flow velocity,[1] ρ is the mass density of the fluid, [2] v is the velocity of the object relative to the fluid, A is the reference area, and CD is the drag coefficient — a dimensionless constant related to the object's geometry and taking into account both skin friction and form drag.
Short of having a wind tunnel and some way of measuring resistance(like a spring scale....LOL!)I'm not sure what good this will do.Your going to have a few problems coming up with some numbers.
Look mom...I have a wind tunnel and a spring scale.....I'm ready to do some math.....should I get panniers or a trailer....let's find out.....
Last edited by Booger1; 12-02-11 at 02:42 PM.
#30
If things were that simple... 
Cd in the above formula can only be assumed constant under certain set of circumstances. In general though Cd is a function of so called "Reynolds number" (https://en.wikipedia.org/wiki/Reynolds_number_)
"At a low Reynolds number, the flow around the object does not transition to turbulent but remains laminar, even up to the point at which it separates from the surface of the object. At very low Reynolds numbers, without flow separation, the drag force is proportional to v instead of v^2; for a sphere this is known as Stokes law."

Cd in the above formula can only be assumed constant under certain set of circumstances. In general though Cd is a function of so called "Reynolds number" (https://en.wikipedia.org/wiki/Reynolds_number_)
"At a low Reynolds number, the flow around the object does not transition to turbulent but remains laminar, even up to the point at which it separates from the surface of the object. At very low Reynolds numbers, without flow separation, the drag force is proportional to v instead of v^2; for a sphere this is known as Stokes law."
#31
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Joined: Apr 2006
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From: Albuquerque, NM
Cd in the above formula can only be assumed constant under certain set of circumstances. In general though Cd is a function of so called "Reynolds number" (https://en.wikipedia.org/wiki/Reynolds_number_)
#32
Senior Member

Joined: Apr 2006
Posts: 4,609
Likes: 507
From: Albuquerque, NM
I guess you haven't beed following what people have been doing with power meters (do a search on Coggan & regression or Chung method). It's possible to measure CdA with high levels of accuracy and precision; better than the power meters themselves.
#33
Same weight, the difference in rolling resistance will be trivial. If any, because the rolling resistance is proportional to weight (and speed) so while more wheels can give more resistance each of those wheels has less resistance in proportion to the lesser weight on each. The mechanical losses in the extra sets of bearings is almost nothing.
At the speeds mentioned, 6-12 mph, air resistance is not significant enough to be concerned about any delta. So the power requirements will be very close either way.
Handling and balance may be affected, which may make the rider less efficient on one setup or the other. My "physics" guess is that this is where the only real difference in wattage is found. Unless you want to consider higher speeds.
At the speeds mentioned, 6-12 mph, air resistance is not significant enough to be concerned about any delta. So the power requirements will be very close either way.
Handling and balance may be affected, which may make the rider less efficient on one setup or the other. My "physics" guess is that this is where the only real difference in wattage is found. Unless you want to consider higher speeds.






