How much faster is that new aero bike?

[Mark McM]

Originally Posted by marciero

For a given pedal force, a lower gear results in greater force at the road. In the headwind case, it would be possible for the rider to gear down such that they apply the same pedal force at the same cadence as in the non headwind case, but with a greater force at the road. Are they not doing the same work in either case and hence same power? Work per time turning the pedals, or moving the bike, is the same.

Careful not to mix up force, energy/work, and power. Energy is force applied over a distance. Power is force applied at a given speed (or to put it another way, power is the rate of energy delivery). With the same apparent wind (road speed + headwind), the drag force is the same. but the power is not the same. Because the rider force applied to the road, the power is the drive force applied to the road times the road speed, the road speed being lower in the headwind case. The aero power formula is:

Power = Force x Velocity = [ Cda x Rho x (Vroad + Vheadwind)^2 ] x Vroad

With the headwind case, the drag force is the same, but the road speed is lower, hence it takes less power to ride 15 mph into a 10 mph wind than it does to ride 25 mph in still air, even though the drag force is the same.

Consider: It is true that if the rider gears down they deliver more force to the road for the same pedal force. But think about that for a moment - if the rider applies the same pedal force and cadence with the same drag force (same apparent wind), then if they gear down in the headwind case they generate more drive force than the drag force. This will cause the rider to accelerate, until the apparent wind increases enough for the drag force to equal the drive force. Which is why I indicated that at the rider power necessary to go 25 mph in still air, the rider will go faster than 15 mph with a 10 mph headwind.

If we use the value Vr for the road speed in the headwind case, then using the same power in both the still air and headwind cases we find:

P = Cda x Rho x [ (25 mph)^2 ] x (25 mph) = Cda x Rho x [ (Vr + 10 mph)^2 ] x Vr

or

15,625 = Vr^3 + 20 x Vr^2 + 100 x Vr

Solving for Vr gives a speed of 18.8 mph when riding into the 10 mph headwind.

Likewise, when you ride with a tailwind, your speed does not increase as much as the tailwind - when you apply the same power to go 25 mph in still air, you will not go 35 mph with a 10 mph tail wind. The power equation for the 10 mph tailwind becomes:

P = Cda x Rho x [ (25mph)^2 ] x (25 mph) = Cda x Rho x [ (Vr - 10 mph)^2 ] x Vr

or

15,625 = Vr^3 - 20 x Vr^2 + 100 x Vr

Vr in this case is 32.1 mph.

[Mark McM]

Originally Posted by fa63

The power required to overcome rolling resistance is proportional to "ground speed", whereas the power required to overcome aerodynamic drag is proportional to "air speed" which includes the effects of the headwind.

Yes, the difference in rolling resistance at different road speeds further confounds the issue, but even when looking at the just the aero drag force (see above), a headwind does not slow a rider by an amount equal to the headwind speed. And because the aero component of total power is greater in a headwind, aerodynamics becomes more important in a headwind.

[NHAero]

Originally Posted by fa63

I don't know how fast your typical rides are, but power is proportional to speed cubed, so if you are averaging 16 mph instead of 25 mph, then the difference is only (16/25)^3 = 0.26 (26%) of 25 Watts, which would be 6.5 Watts. Then it would indeed be hardly noticeable.

This is comforting to this older rider, for whom a 16 mph average ride is a good day. I can stick with my non-aero bikes!

[fa63]

Your forum username does NOT check out then

But yeah, aerodynamics really come more into play on faster rides (although you will save more time if you ride slower and therefore out riding for longer, but that is a whole other discussion).

Originally Posted by NHAero

This is comforting to this older rider, for whom a 16 mph average ride is a good day. I can stick with my non-aero bikes!

[Baron Blubba]

Originally Posted by fa63

I don't know how fast your typical rides are, but power is proportional to speed cubed, so if you are averaging 16 mph instead of 25 mph, then the difference is only (16/25)^3 = 0.26 (26%) of 25 Watts, which would be 6.5 Watts. Then it would indeed be hardly noticeable.

Frequently averaging 23-24+ on flattish group rides. Mountain rides less, but even those rides have really fast rolling valley sections where I’m moving (solo) at 23+ averages over a 30 mile stretch.
I have as many pr’s on my Ritchey as on my Melee/TCR/Propel, and keeping up on those fast group rides is not any more difficult on either bike.
Aero and weight benefits exist, obviously, but in real world riding on good bikes, it’s at a pennies-per-dollar rate, so to speak, compared to what the lab tests and marketing would have you believe.

I say this as someone with a vested interest in the opposite being true.

[fa63]

Good to know, thanks for sharing.

Originally Posted by Baron Blubba

Frequently averaging 23-24+ on flattish group rides. Mountain rides less, but even those rides have really fast rolling valley sections where I’m moving (solo) at 23+ averages over a 30 mile stretch.
I have as many pr’s on my Ritchey as on my Melee/TCR/Propel, and keeping up on those fast group rides is not any more difficult on either bike.
Aero and weight benefits exist, obviously, but in real world riding on good bikes, it’s at a pennies-per-dollar rate, so to speak, compared to what the lab tests and marketing would have you believe.

I say this as someone with a vested interest in the opposite being true.

[El Chaba]

Originally Posted by Baron Blubba

Frequently averaging 23-24+ on flattish group rides. Mountain rides less, but even those rides have really fast rolling valley sections where I’m moving (solo) at 23+ averages over a 30 mile stretch.
I have as many pr’s on my Ritchey as on my Melee/TCR/Propel, and keeping up on those fast group rides is not any more difficult on either bike.
Aero and weight benefits exist, obviously, but in real world riding on good bikes, it’s at a pennies-per-dollar rate, so to speak, compared to what the lab tests and marketing would have you believe.

I say this as someone with a vested interest in the opposite being true.

As a data point, I have found the same to be true. I would love to be able to buy 30 watts at 40 kmph. A great frame with great parts is what I find to be “fast” . My guess is that riders who learn to be fast learn how to get in an aerodynamic position as a part of the process. As long as the equipment allows for that and is reasonably aerodynamic, the importance of the aerodynamics of the equipment is WAY a down the list. I don’t discount the advantages of aero wheels, but they are generally oversold as well ( at until aero frames/bikes came along). I do rate supple, fast tires as a very important part of the equation, but I think you are also a bit on your own to figure out which tires they are.

[mattsbeers]

These conversations are exhausting. Aero bikes are faster…just a fact. Add that to aero clothing and helmet and even more so. If you don’t dig them, fine. It’s like asking is a new 911st really faster that my 1988 911? Yes, obviously but sometimes that’s not the point. We need to move on people.

[username]

Originally Posted by mattsbeers

These conversations are exhausting. Aero bikes are faster…just a fact. Add that to aero clothing and helmet and even more so. If you don’t dig them, fine. It’s like asking is a new 911st really faster that my 1988 911? Yes, obviously but sometimes that’s not the point. We need to move on people.

At the risk of further exhausting you, I'd put it slightly differently. I believe that aero bikes are, on balance, faster. But they're not likely to be faster enough for me, given the shape of my body and the kind of riding I do, to make much difference. Add to that, there's at least some chance that, given the above—I'm a slow pear—I'd be more comfortable on one of my antiquated bikes and therefore as fast or faster as I would be on a superbike.

Anyway, in the end, it seems to me that the bottom line is this: horses for courses. If someone wants to absolutely maximize their performance, they should absolutely get an aero bike, a skinsuit, and they should carve their body fat down to 6%. But even then, they're going to get smoked by someone whose VO2 threshold is, by the grace of the cycling gods, much higher than theirs. Genetics are one hell of a cruel mistress for most of us.

[username]

Adding one more thing: I think people should ride whatever they want. Cycling, for most of us, is supposed to be fun. If you're having fun riding the latest and greatest aero bike, that's awesome. I'm having more fun on my Kirk and my Peg than I've ever had riding before now. What's great is that even though you're going to cross the line first, we can both win.

[marciero]

Originally Posted by Mark McM

Careful not to mix up force, energy/work, and power. Energy is force applied over a distance. Power is force applied at a given speed (or to put it another way, power is the rate of energy delivery). With the same apparent wind (road speed + headwind), the drag force is the same. but the power is not the same. Because the rider force applied to the road, the power is the drive force applied to the road times the road speed, the road speed being lower in the headwind case. The aero power formula is:

Power = Force x Velocity = [ Cda x Rho x (Vroad + Vheadwind)^2 ] x Vroad

With the headwind case, the drag force is the same, but the road speed is lower, hence it takes less power to ride 15 mph into a 10 mph wind than it does to ride 25 mph in still air, even though the drag force is the same.

Consider: It is true that if the rider gears down they deliver more force to the road for the same pedal force. But think about that for a moment - if the rider applies the same pedal force and cadence with the same drag force (same apparent wind), then if they gear down in the headwind case they generate more drive force than the drag force. This will cause the rider to accelerate, until the apparent wind increases enough for the drag force to equal the drive force. Which is why I indicated that at the rider power necessary to go 25 mph in still air, the rider will go faster than 15 mph with a 10 mph headwind.

I was thinking of power as work per time, which is equivalent, but this makes sense. In each case the drag force is the same, so the force at at wheel to balance that would be the same. Same pedal force in same gear would do it, but now lower cadence, less work per time, less power. Same lesser power if gear down with lesser pedal force and same cadence.

[adub]

Originally Posted by Mark McM

Careful not to mix up force, energy/work, and power. Energy is force applied over a distance. Power is force applied at a given speed (or to put it another way, power is the rate of energy delivery). With the same apparent wind (road speed + headwind), the drag force is the same. but the power is not the same. Because the rider force applied to the road, the power is the drive force applied to the road times the road speed, the road speed being lower in the headwind case. The aero power formula is:

Power = Force x Velocity = [ Cda x Rho x (Vroad + Vheadwind)^2 ] x Vroad

With the headwind case, the drag force is the same, but the road speed is lower, hence it takes less power to ride 15 mph into a 10 mph wind than it does to ride 25 mph in still air, even though the drag force is the same.

Consider: It is true that if the rider gears down they deliver more force to the road for the same pedal force. But think about that for a moment - if the rider applies the same pedal force and cadence with the same drag force (same apparent wind), then if they gear down in the headwind case they generate more drive force than the drag force. This will cause the rider to accelerate, until the apparent wind increases enough for the drag force to equal the drive force. Which is why I indicated that at the rider power necessary to go 25 mph in still air, the rider will go faster than 15 mph with a 10 mph headwind.

If we use the value Vr for the road speed in the headwind case, then using the same power in both the still air and headwind cases we find:

P = Cda x Rho x [ (25 mph)^2 ] x (25 mph) = Cda x Rho x [ (Vr + 10 mph)^2 ] x Vr

or

15,625 = Vr^3 + 20 x Vr^2 + 100 x Vr

Solving for Vr gives a speed of 18.8 mph when riding into the 10 mph headwind.

Likewise, when you ride with a tailwind, your speed does not increase as much as the tailwind - when you apply the same power to go 25 mph in still air, you will not go 35 mph with a 10 mph tail wind. The power equation for the 10 mph tailwind becomes:

P = Cda x Rho x [ (25mph)^2 ] x (25 mph) = Cda x Rho x [ (Vr - 10 mph)^2 ] x Vr

or

15,625 = Vr^3 - 20 x Vr^2 + 100 x Vr

Vr in this case is 32.1 mph.

That's what I was going to say..

[hernium]

Originally Posted by GregL

^^^Excellent assessment ^^^ I’ll opine that if you’re road (or fast gravel) racing, aero frames are fast becoming a necessity. I’ve gone all in on aero and efficiency for racing (pretty much everything in your list above). The marginal gains do add up.

Greg

I would say that this all only really matters in the race. On a group ride, or training alone, power and position is all I look at these days. Maybe I should train like I race, but at the level I am at right now, what little I lose out on over an aero optimized equipment setup I can gain back over a few weeks of following a good workout plan.

[GregL]

Originally Posted by hernium

I would say that this all only really matters in the race. On a group ride, or training alone, power and position is all I look at these days. Maybe I should train like I race, but at the level I am at right now, what little I lose out on over an aero optimized equipment setup I can gain back over a few weeks of following a good workout plan.

I train in a similar manner. I leave the deep aero wheels, aero clothing, and aero helmet home on training days. Come race day, all the legal aero toys come out for play time.

Greg

[9tubes]

Paywall, so I can't confirm their assumptions. The OP wrote the comparison was to a "baseline Trek Emonda from 2015 - rim brakes, box rims, basically standard bike."

If that means using 32 round spokes, or even 24 round spokes, the 25 watt difference could be in the wheels alone.

Likewise, did they use the same tire setup for all? It's pretty easy to have a 25 watt difference in two tires.



.