
Aerodynamics and power
In the latest issue of Velo News, Lennard Zinn makes the following statement:
"Aerodynamic drag increases exponentially with speed. It doesn't take twice as much power to go twice as fast relative to the air; it takes for or more times as much power."
While both parts of this statement are correct, the second is wildly inaccurate. Aero DRAG is proportional to the square of the speed, but the POWER required is drag times speed, so it is cubic. So saying "four or more" when you should have said "more than eight" is inexcusable to me. How is it possible in this day and age that someone like Zinn could get this so wrong? I expect this kind of nonsense from Bicycling magazine, but not Velo News and not Lennard Zinn.
Note: "more than eight" is because doubling speed means 8 times more power to overcome aero forces and double the power to overcome rolling resistance (tires, bearings, chain losses, etc.).

Well, at least he said "or more". Most people couldn't give you a definition of "exponentially" better than that, anyway....
"L'enfer, c'est les autres"

Originally Posted by Kerry Irons
Lennard Zinn makes the following statement:
"Aerodynamic drag increases exponentially with speed."
This is not correct either. As you say, drag increases quadratically with speed. A powerlaw function is not exponential.
But as cycling analysis has become more sciencebased, Lennard has fallen further and further behind. Does anyone else remember where he explained why "throwing" your bike at the finish does no good, because you can't accelerate the center of mass by shifting weight.

Zinn is not totally wrong, it does increase exponentially with speed, the exponent is not 2 though.
Lets see the math.

Originally Posted by duriel
Zinn is not totally wrong, it does increase exponentially with speed, the exponent is not 2 though.
"This is not correct either. As you say, drag increases quadratically with speed. A powerlaw function is not exponential. "

Originally Posted by asgelle
A powerlaw function is not exponential.
Wasn't reallly addressing this issue, I don't think any of the discussion was concerning the P/D relation, more about the original OP concerned about the square.

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Last edited by \"none\"; 03212020 at 05:02 AM.

Originally Posted by Kerry Irons
In the latest issue of Velo News, Lennard Zinn makes the following statement: ....
Stop complaining and send Zinn an email correction. Bet he posts it.
He's not afraid of admitting an error.

Originally Posted by Peter P.
Stop complaining and send Zinn an email correction. Bet he posts it.
He's not afraid of admitting an error.
Haven't been able to find an email address. Do you just respond to VeloNews editorial or is there something more specific?

Originally Posted by asgelle
This is not correct either. As you say, drag increases quadratically with speed. A powerlaw function is not exponential.
Here I am, decades after graduating engineering school, and absent looking it up I could not have told you the difference between a power law function and an exponential. I had to look it up.
That said, throughout my professional career the engineers I worked with would uniformly have said that "x cubed" represented an exponentially increasing function. Apparently we all got it wrong at least in the common usage. In my own mind, the power to overcome aero drag "increases exponentially" with speed. In this case, the exponent is 3.

Originally Posted by Kerry Irons
Here I am, decades after graduating engineering school, and absent looking it up I could not have told you the difference between a power law function and an exponential. I had to look it up.
I harp on it because it isn't a distinction without a difference. Exponential growth (y(x)=a^x) will always be much faster than any power law (y(x)=x^a) for a large enough value of x. More significantly, the two functions represent different underlying physics.

... so what is the equation for power, with aero drag & mech loses. I don't think most people understand any of this, more now than ever.

Originally Posted by duriel
... so what is the equation for power, with aero drag & mech loses. I don't think most people understand any of this, more now than ever.
http://www.recumbents.com/wisil/Mart...%20cycling.pdf

Originally Posted by Kerry Irons
Here I am, decades after graduating engineering school, and absent looking it up I could not have told you the difference between a power law function and an exponential. I had to look it up.
...
The term "exponential" is in fact pretty vague. There is a big difference if that exponent is 4 compared to if it is 1/2........
"L'enfer, c'est les autres"

Originally Posted by No Time Toulouse
The term "exponential" is in fact pretty vague.
Not if used correctly. Another reason to use precise terminology to communicate clearly.

Most people when they use the term "exponential" assume a value of 2 or more. They don't really know what it means, much, and around here, there is not much precise terminology in use.

Originally Posted by asgelle
Not if used correctly. Another reason to use precise terminology to communicate clearly.
Yes. It's difficult to communicate with someone about mathematics if they don't speak the (precise) language of mathematics, and they continually use the colloquial definition of mathematical terms.

Originally Posted by duriel
Most people when they use the term "exponential" assume a value of 2 or more.
You really shouldn't be speaking for most people.
Neither should I but it's hard to fathom most people thinking any value at all when they hear the term exponential. It's used to express a certain rate of change, up or down, not a value and in my experience most people do understand that.

RoadBikeReview Member
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Clarifying maths
As a mathematician, I would like to emphasise the point that an exponential function of v is one where v is in the exponent, like a^v, and a cubic function of v (such as v^3) is an example of one where it is not. And shame on any engineer who uses the mathematical term incorrectly: it is not something I would expect of my engineer friends!
Also, as a simple observation, it is the energy taken to overcome air resistance over a fixed distance (in the air!) that is quadratic in speed (while power is cubic in speed, you also get there more quickly).

Originally Posted by Kerry Irons
Haven't been able to find an email address. Do you just respond to VeloNews editorial or is there something more specific?
Send it to [email protected] and just tell them to get it to Lennard. Yeah, it's kinda weird they don't have a staff contact link on their web site. I've written him a couple times and received prompt responses, but that was years ago. In fact, I actually spoke to him on the phone at length in response to an email I sent him.
Do it.

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Originally Posted by duriel
... so what is the equation for power, with aero drag & mech loses. I don't think most people understand any of this, more now than ever.
P = (Vg*W*(K1+G) + K2*(Va)^3)/375
Where P is in horsepower, Vg is ground speed (mph), W is bike/rider weight in pounds, G is the grade, and Va is the rider's speed through the air (mph). Grade is feet or altitude gain per foot of horizontal distance, and while often expressed in per cent, in this equation is used as a decimal (a 6% grade is 0.06). K1 is a lumped constant for all frictional losses (tires, bearings, chain) and units conversion, and is generally reported with a value of 0.0053. K2 is a lumped constant for aerodynamic drag and is generally reported with a value of 0.0083. Note that power to overcome friction and gravity is proportional only to rider weight and ground speed. Power to overcome wind drag is proportional to the cube of the air speed.
For reference, 1 hphr = 641 "calories" delivered to the pedals, 1 hp = 746 watts, 1 calorie = 4.186 kj. Here, all calories are kgcalories, "big" calories, or "food calories." The human body runs at about 24% efficiency for a relatively fit athlete, so to deliver 1 hp (746 watts) to the pedals requires the body to consume about 2700 calories, more than 4 times the 641 "calories" delivered to the pedals.

Originally Posted by duriel
Most people when they use the term "exponential" assume a value of 2 or more. They don't really know what it means, much, and around here, there is not much precise terminology in use.
I think when "most people" use the term, they are simply thinking of something increasing much faster than linear and in an everincreasing fashion. That said, I don't think "most people" use the term at all. The concept of exponents is not something that enters the consciousness of "most people".

Originally Posted by \"none\"
Yes, we are all used to recumbent riders telling us how much more aerodynamically efficient they are than us "wedgie" riders. Funny how I have passed by a recumbent perhaps 5 times in my decades of riding. Any chance that the rider position so compromises the ability to produce power that it can't compensate for the improved aerodynamics? Not that you would ever get a recumbent rider to admit, so it must be that recumbent riders are all just slow.

Originally Posted by Kerry Irons
Yes, we are all used to recumbent riders telling us how much more aerodynamically efficient they are than us "wedgie" riders. Funny how I have passed by a recumbent perhaps 5 times in my decades of riding. Any chance that the rider position so compromises the ability to produce power that it can't compensate for the improved aerodynamics? Not that you would ever get a recumbent rider to admit, so it must be that recumbent riders are all just slow.
Maybe those 5 were on their easy days and didn't know that they were in a race.
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