To all the aero gurus (cough, jackmott/tanhalt, cough)
How do I convert quoted savings from higher to lower speed?
Most companies test and report their aero savings as “saves x watts at 30mph/50km” or “our doodad is 100g lower drag compare to our competitor’s doodad”. Which sounds great, but power goes up exponentially as speed increases. So how to calculate the savings at 40km or 35km? I can’t average 50km/h for my rides. Thanks for the help.
If savings are quoted as time, you can just assume that the time saved at 30mph will be the same time saved at any speed. This isn’t exactly true, you will save a bit more time the slower you are, but pretty close to the same.
If savings are quoted in terms of watts, then you can just linearly interpolate. This means if you Save 5 watts at 30 mph, then the watt savings at 15mph is about half. Again, this isn’t entirely true, because as you note, power requirements are not linear with speed, but within normal cycling speeds it is ‘pretty close’
If you want exact answers, fire up analyticcycling.com
The funny thing with aero saving is, the longer time is ridden with aero advantag for a distance the bigger is the time gain. That is very interesting for non professionals …
A linear approximation will be fine for very close speeds, but if you’re comparing something like 20mph to a quoted “watts saved at 30mph” then you’ll want to take the cube of the ratio, which in this case would be (2/3)^3, which is approximately 0.3. So 10 watts saved at 30mph would only be around 3 watts at 20mph. It’s a little harder to do in your head than a simple linear approximation but it’s easy enough if you have a calculator/phone/computer handy.
The funny thing with aero saving is, the longer time is ridden with aero advantag for a distance the bigger is the time gain. That is very interesting for non professionals …
Time is linear, drag is not. How does that work? Going from 15mph to 30mph halves the time spent on course, it approximately increases drag by a factor of 4. The inverse is just as true, halving the speed drastically reduces drag at a rate far exceeding time lost.
This comes straight from wikipedia. So how are bikes exempt and why are we talking about interpreting anything linearly? That’s not how parabolic curves work.
A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula.
I should do a blog entry on this topic since it comes up monthly.
To fully convince yourself you need to do the math, computing time to cover a fixed distance given a change in CdA.
If you don’t actually do that math, and instead think about power requirements or intuition you just aren’t going to get it enough to be convinced.
Now you can do that by hand but if you don’t have the skills for that you can use various calculators like analyticcycling.com or cyclingpowerlab.com, or bestbikesplit.com
They all use the same underlying, simple physics equations.
So let us take a given scenario, a rider who does 300 watts with a CdA of .25 on a flat course at .004 crr and normal air density will go about:
42.62kph
Reduce CdA by .005 (equivlent to 50g drag@30mph) and speed goes to
42.89kph
An increase of 0.27kph
Now, same experiment for a lower power rider, just 200 watts, his speeds will be
36.63kph and 36.86kph
an increase of only .23 kph (less!)
But what about time saved over a fixed distance?
300 watt rider: 56:18 → 55:57 = 21 seconds saved
200 watt rider: 1:05:31 → 1:05:06 → 25 seconds saved (more!)
The funny thing with aero saving is, the longer time is ridden with aero advantag for a distance the bigger is the time gain. That is very interesting for non professionals …
Time is linear, drag is not. How does that work? Going from 15mph to 30mph halves the time spent on course, it approximately increases drag by a factor of 4. The inverse is just as true, halving the speed drastically reduces drag at a rate far exceeding time lost.
This comes straight from wikipedia. So how are bikes exempt and why are we talking about interpreting anything linearly? That’s not how parabolic curves work.
A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula.
The funny thing with aero saving is, the longer time is ridden with aero advantag for a distance the bigger is the time gain. That is very interesting for non professionals …
Time is linear, drag is not. How does that work? Going from 15mph to 30mph halves the time spent on course, it approximately increases drag by a factor of 4. The inverse is just as true, halving the speed drastically reduces drag at a rate far exceeding time lost.
This comes straight from wikipedia. So how are bikes exempt and why are we talking about interpreting anything linearly? That’s not how parabolic curves work.
A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula.
This explains it fairly well:
http://www.cervelo.com/en/engineering/thinking-and-processes/slow-vs-fast-riders.html
And then, if you’re still confused, watch this: https://www.youtube.com/watch?v=O-7g1kqYJAY&list=PLcmaLnqmqDnmn_bCR0RJ-soSCDKCKR97t&index=3
Thanx, that article I was remembering, too …
I should do a blog entry on this topic since it comes up monthly.
To fully convince yourself you need to do the math, computing time to cover a fixed distance given a change in CdA.
If you don’t actually do that math, and instead think about power requirements or intuition you just aren’t going to get it enough to be convinced.
Now you can do that by hand but if you don’t have the skills for that you can use various calculators like analyticcycling.com or cyclingpowerlab.com, or bestbikesplit.com
They all use the same underlying, simple physics equations.
So let us take a given scenario, a rider who does 300 watts with a CdA of .25 on a flat course at .004 crr and normal air density will go about:
42.62kph
Reduce CdA by .005 (equivlent to 50g drag@30mph) and speed goes to
42.89kph
An increase of 0.27kph
Now, same experiment for a lower power rider, just 200 watts, his speeds will be
36.63kph and 36.86kph
an increase of only .23 kph (less!)
So you go from 300 watts to 200 watts and assume the same drag? That makes no sense to me, unless you’re selling bikes.
I did math too, ignoring a lot of other resistance obviously.
F = CdA * Air Density x velocity ^2/2
Assume your .25 CdA * 1.23 * 10m/s ^2/2 (an optimistic age grouper speed at an optimistic CdA)
I get 15 newtons which must be multiplied again by the velocity to get 153 watts.
Do the same math at wind tunnel speeds (13.4 m/s) and you get over 368 watts.
My age grouper has slown down to 75% the original velocity and needs 41% of the original power.
If a helmet measures 5 watts saved at wind tunnel speed, there is no way that is linear. 5 watt savings in 368 is great but 5 watts out of 153 (or even 75% of 5) is inflating that number significantly. I would guess using this that instead of 5 watts advertised you get 2 while reducing velocity by only 25% from advertised. That isn’t linear.
If I’m wrong, please show me where. But I would ask that we keep the links to cervelo.com to a minimum, really the company that sells bikes is trying to prove that slow people should buy superbikes too? Gasp!
So you go from 300 watts to 200 watts and assume the same drag? That makes no sense to me, unless you’re selling bikes.
No, CdA is not an expression of an amount of drag. It is coefficient of drag times frontal area. At speeds between 20 and 30mph, the Cd of objects does not change, and of course, frontal area doesn’t change with speed, so CdA remains constant.
I will repeat again, if you want to convince yourself of this, compute the time saved over a fixed distance, for a change in CdA.
You didn’t do that. You computed the power requirement given a change in speed.
You seem to be stuck on my comment about linear interpolation for power, which I said in the beginning, is not really true. It is just a reasonable approximation across small changes in speed. However the time saved over a fixed distance given a change in CdA, really is MORE time saved for slower riders.
links to cervelo: 0
Yeah also important to keep in mind that while the slower rider saves more seconds, their time is improved by a lower percentage.
BUT, regardless, to close approximation, something that saves 10 seconds over 40k @ 30mph
saves about 10 seconds @20mph too
and that really is enough to make most decisions, if you need more rigor than that, fire up bestbikesplit and do course simulations
Calculate with this in mind … if you drive longer (time) with saving of same sources (less drag) you save more time … but not a podium 
Calculate with this in mind … if you drive longer (time) with savings of the same source (less drag) you save more time … but not a podium
I should do a blog entry on this topic since it comes up monthly.
To fully convince yourself you need to do the math, computing time to cover a fixed distance given a change in CdA.
If you don’t actually do that math, and instead think about power requirements or intuition you just aren’t going to get it enough to be convinced.
Now you can do that by hand but if you don’t have the skills for that you can use various calculators like analyticcycling.com or cyclingpowerlab.com, or bestbikesplit.com
They all use the same underlying, simple physics equations.
So let us take a given scenario, a rider who does 300 watts with a CdA of .25 on a flat course at .004 crr and normal air density will go about:
42.62kph
Reduce CdA by .005 (equivlent to 50g drag@30mph) and speed goes to
42.89kph
An increase of 0.27kph
Now, same experiment for a lower power rider, just 200 watts, his speeds will be
36.63kph and 36.86kph
an increase of only .23 kph (less!)
So you go from 300 watts to 200 watts and assume the same drag? That makes no sense to me, unless you’re selling bikes.
I did math too, ignoring a lot of other resistance obviously.
F = CdA * Air Density x velocity ^2/2
Assume your .25 CdA * 1.23 * 10m/s ^2/2 (an optimistic age grouper speed at an optimistic CdA)
I get 15 newtons which must be multiplied again by the velocity to get 153 watts.
Do the same math at wind tunnel speeds (13.4 m/s) and you get over 368 watts.
My age grouper has slown down to 75% the original velocity and needs 41% of the original power.
If a helmet measures 5 watts saved at wind tunnel speed, there is no way that is linear. 5 watt savings in 368 is great but 5 watts out of 153 (or even 75% of 5) is inflating that number significantly. I would guess using this that instead of 5 watts advertised you get 2 while reducing velocity by only 25% from advertised. That isn’t linear.
If I’m wrong, please show me where. But I would ask that we keep the links to cervelo.com to a minimum, really the company that sells bikes is trying to prove that slow people should buy superbikes too? Gasp!
And you’ve just proven once again why aero test results should be reported in terms of CdA…not watts…not “grams of drag”…etc. All that does is cause confusion. CdA allows for the user to calculate WHATEVER metrics they desire without additional information about the test.
Anyway, your confusion in this case is because you haven’t done ALL the math…keep going, you’re almost there…but first you may have to set aside your preconceived notions for a bit
The reason that I posted links to Cervelo is because they HAVE done all the math 
It’s all basic physics and math…not rocket surgery.
And you’ve just proven once again why aero test results should be reported in terms of CdA…not watts…not “grams of drag”…etc. All that does is cause confusion. CdA allows for the user to calculate WHATEVER metrics they desire without additional information about the test.
…
The reason that I posted links to Cervelo is because they HAVE done all the math
Which if you read what the OP wrote, is exactly what he’s asking “watts” “drag”.
As for Cervelo, maybe they did the math but they aren’t kind enough to share any of it, so I would call that marketing BS.
And you’ve just proven once again why aero test results should be reported in terms of CdA…not watts…not “grams of drag”…etc. All that does is cause confusion. CdA allows for the user to calculate WHATEVER metrics they desire without additional information about the test.
…
The reason that I posted links to Cervelo is because they HAVE done all the math
Which if you read what the OP wrote, is exactly what he’s asking “watts” “drag”.
As for Cervelo, maybe they did the math but they aren’t kind enough to share any of it, so I would call that marketing BS.
Do the math yourself (I have) and you’ll see that they’re provably correct.
It’s not marketing BS, it’s logic and fact.
No, CdA is not an expression of an amount of drag. It is coefficient of drag times frontal area. At speeds between 20 and 30mph, the Cd of objects does not change, and of course, frontal area doesn’t change with speed, so CdA remains constant.
Where do you find anything that backs this up? As I understand it, under 20mph and something like a biker approaches low reynolds numbers that do actually take into account velocity in CdA.
Thanks for not linking to Cervelo, wasn’t implying you did
Edit: this has to be, CdA has to approach zero as speed does. At some point, it has to stop being a constant.
Cd - coefficient of drag
A - area
We can agree that area does not change with velocity I think, even when velocity goes to zero.
The coefficient doesn’t change with speed at a given reynolds number. At 0, the coefficient can be 0.5, 100, whatever, once it is multiplied by velocity squared, drag force will be zero.
Shapes have a given drag coefficient, see:
http://en.wikipedia.org/wiki/Drag_coefficient#mediaviewer/File:14ilf1l.svg
Now that can change at high enough, or low enough ,velocities. But within the range of speeds cyclists ride around at (10 to 40mph) it is fairly constant, as is the coefficient of drag.
So the CdA - coefficient of drag times area - allows you to compute whatever you want from there. A change in CdA of .005 will save me how much power at 30mph? Will save me how much time at 400 watts? etc.
At 30mph you will save more watts, and more drag force, and will enjoy a greater increase in speed for a given CdA than you will at 20mph.
But you will save more time over a fixed distance at 20mph than 30mph.
Now, do we assume these things are true? Or do we know it from experiment?
Well we know it from experiment, and here are just a couple of examples:
A couple of the authors of that paper are frequent posters here
http://www.trainingandracingwithapowermeter.com/2010/10/challenge-to-cycling-aerodynamicists.html
And, perhaps most interesting recently, these concepts were applied to the new hour record, with Trek using the AlphaMantis track aero system:
The inventor of the track aero system, Andy Frocioni, is a frequent poster here.
Many of us have been using this kind of knowledge to good effect for years. It is pretty wonderful stuff.
No, CdA is not an expression of an amount of drag. It is coefficient of drag times frontal area. At speeds between 20 and 30mph, the Cd of objects does not change, and of course, frontal area doesn’t change with speed, so CdA remains constant.
Where do you find anything that backs this up? As I understand it, under 20mph and something like a biker approaches low reynolds numbers that do actually take into account velocity in CdA.
Thanks for not linking to Cervelo, wasn’t implying you did
Edit: this has to be, CdA has to approach zero as speed does. At some point, it has to stop being a constant.
As I understand it, under 20mph and something like a biker approaches low reynolds numbers that do actually take into account velocity in CdA.
Do the math. At 20mph, for a bike in air, Re=O(10^6).
As I understand it, under 20mph and something like a biker approaches low reynolds numbers that do actually take into account velocity in CdA.
Do the math. At 20mph, for a bike in air, Re=O(10^6).
If you look at the bike/rider as a whole, then a constant Cd over a range of velocities is a reasonable assumption, however there are quite possibly some interesting effects when you look at individual components of the system, as the size of the “part” is smaller than the whole. e.g. upper arms acting as cylinders perpendicular to the flow may just fall into that zone where Cd can vary quite a bit with Re at such air speeds.
Cd - coefficient of drag
A - area
…
In any case, for someone comparing tests from tunnels, then you are right that one can use a rule of thumb to make at least a first order reasonable approximation of what the impact on speed / power / aero is.
100g of drag
~= 1 second per km
~= 0.01m^2 CdA
As you say, there are those that will fall a bit either side of that ROT, but as a starting point for quickly getting your head around what the impact means, it’s pretty good.
Tom A. is on the money too as providing a CdA value is more universally useful but I can understand why tunnels provide the drag in grams as that’s what they actually measure.