Aero? Downwind?
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2010-05-24 7:53 PM |
Expert 2852 Pfafftown, NC | Subject: Aero? Downwind? Seems kinda unusual to ask. On one of my training rides, recently, I have a 3mi. stretch that's a gradual upgrade the entire distance. I caught it, downwind. I avg'd faster in that uphill portion than I did anywhere else on the ride. If I'd been riding a tri bike (and I hope to be, soon)......I wondered if I'd have been better off riding in aero or more upright? Do you guys/gals take advantage of your tailwind by staying in....or by coming out of aero position? Does it make much of a difference? Discernable? Thanks. |
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2010-05-24 8:01 PM in reply to: #2879556 |
Pro 6011 Camp Hill, Pennsylvania | Subject: RE: Aero? Downwind? I think the only way it would make sense to come out of aero in that situation would be if the tail wind was faster than you were able to ride. In other words, as long as you're going faster than the wind, you will gain benefits from being more aerodynamic, because you're cutting through the air. |
2010-05-24 8:29 PM in reply to: #2879556 |
Champion 7136 Knoxville area | Subject: RE: Aero? Downwind? aero. A corvette is more aero than a mini van going 100mph, 10mph, 0mph, upwind, downwind, whatever. |
2010-05-24 11:25 PM in reply to: #2879556 |
Champion 9407 Montague Gold Mines, Nova Scotia | Subject: RE: Aero? Downwind? I've modelled this in the past and the number is really high where it becomes benefical to situp versus riding aero. I forget the exact number but, assuming you can keep power to the pedals, it was around 100km/h tailwind before the "sail" effect kicks in. Shane |
2010-05-25 10:09 AM in reply to: #2879556 |
Extreme Veteran 590 Sioux Falls, SD | Subject: RE: Aero? Downwind? I have the advantage of my butt being a sail, so when I ride aero with the wind I have the best of both worlds. |
2010-05-25 2:01 PM in reply to: #2879621 |
Veteran 812 | Subject: RE: Aero? Downwind? Leegoocrap - A corvette is more aero than a mini van going 100mph, 10mph, 0mph, upwind, downwind, whatever. Actually, if the two vehicles are at at dead stop in a 50 mile an hour breeze, the van will be moved along _much_ faster than the corvette. As to the OP a couple of things. If you have a tail wind equaling your speed, then sitting up should be (in theory) equally effecient as the aero position. If you've got a 35 mph tail wind, then you should be seeing some real gains. But probably more importantly to realize wind resistance increases exponentially. There is almost no resistance at 5mph. The difference in wind resistance between 20 mph and 25 mph is huge. So, sitting up when cycling 20 mph with a 15 mph tail wind (net head wind 5 mph) there is very little cost to it. Going the other direction, you have a big problem. Even a 5 mph head wind can slow you down a ton as your net-wind speed goes from 20 mph to 25 mph. |
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2010-05-25 2:06 PM in reply to: #2881202 |
Resident Curmudgeon 25290 The Road Back | Subject: RE: Aero? Downwind? mrcurtain - 2010-05-25 2:01 PM Leegoocrap - A corvette is more aero than a mini van going 100mph, 10mph, 0mph, upwind, downwind, whatever. Actually, if the two vehicles are at at dead stop in a 50 mile an hour breeze, the van will be moved along _much_ faster than the corvette. Huh? If the two vehicles are at a dead stop won't they both be at a dead stop? |
2010-05-25 2:48 PM in reply to: #2881212 |
Veteran 812 | Subject: RE: Aero? Downwind? the bear - 2010-05-25 12:06 PM mrcurtain - 2010-05-25 2:01 PM Leegoocrap - A corvette is more aero than a mini van going 100mph, 10mph, 0mph, upwind, downwind, whatever. Actually, if the two vehicles are at at dead stop in a 50 mile an hour breeze, the van will be moved along _much_ faster than the corvette. Huh? If the two vehicles are at a dead stop won't they both be at a dead stop? If the brakes are left off, then both of the vehicles will be pushed along by a 50mph tail wind (or 80mph or whatever it take to get them moving). The van should be pushed along much faster. Thus showing that aerodynamics is counter to performance with a strong enough tail wind. |
2010-05-25 2:52 PM in reply to: #2881343 |
Champion 9600 Fountain Hills, AZ | Subject: RE: Aero? Downwind? mrcurtain - 2010-05-25 1:48 PM Thus showing that aerodynamics is counter to performance with a strong enough tail wind. No it's not. |
2010-05-25 3:43 PM in reply to: #2881202 |
Subject: RE: Aero? Downwind? I'm getting a power tap shortly. I can test this on a stretch of road I regularly ride down and backs on with a steady 20-25 mph wind. |
2010-05-25 4:22 PM in reply to: #2881343 |
Pro 6011 Camp Hill, Pennsylvania | Subject: RE: Aero? Downwind? Thus showing that aerodynamics is counter to performance with a strong enough tail wind. Doesn't that just show that size does matter? |
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2010-05-25 4:29 PM in reply to: #2881343 |
Champion 9407 Montague Gold Mines, Nova Scotia | Subject: RE: Aero? Downwind? mrcurtain - 2010-05-25 4:48 PM Thus showing that aerodynamics is counter to performance with a strong enough tail wind. Here are the equations that will allow you to calculate the power required to move a bicycle on a flat course. I will leave the algebra as an exercise to the reader Power to overcome rolling resistance P1 = mass * acceleration due to gravity * Crr * velocity_road Power to overcome aerodynamic resistance P2 = 1/2 * air density * CdA * velocity_wind^2 * velocity_road Mass is total mass of the system in kg Acceleration due to gravity is 9.81m/s^2 Crr - .0100 would be a good guess velocity_road is what your speedometer would read in m/s air density - I would use 1.22 CdA - cross sectional area exposed to the wind - try .3 for aero position and .5 for road position velocity_wind is the relative wind speed ETA - the total power to move the bicycle would be P = P1 + P2 (measured in watts) I don't have the calculations at hand, but IIRC for a typical rider (75kg) and bike (10kg) riding at 200W, I *think* the breakeven point was with a tailwind of about 100km/h. Shane Edited by gsmacleod 2010-05-25 4:30 PM |
2010-05-25 5:39 PM in reply to: #2881720 |
Veteran 183 Bellingham, WA | Subject: RE: Aero? Downwind? gsmacleod - 2010-05-25 4:29 PMHere are the equations that will allow you to calculate the power required to move a bicycle on a flat course. I will leave the algebra as an exercise to the reader :) Power to overcome rolling resistance P1 = mass * acceleration due to gravity * Crr * velocity_road Power to overcome aerodynamic resistance P2 = 1/2 * air density * CdA * velocity_wind^2 * velocity_road Mass is total mass of the system in kg Acceleration due to gravity is 9.81m/s^2 Crr - .0100 would be a good guess velocity_road is what your speedometer would read in m/s air density - I would use 1.22 CdA - cross sectional area exposed to the wind - try .3 for aero position and .5 for road position velocity_wind is the relative wind speed ETA - the total power to move the bicycle would be P = P1 + P2 (measured in watts) I don't have the calculations at hand, but IIRC for a typical rider (75kg) and bike (10kg) riding at 200W, I *think* the breakeven point was with a tailwind of about 100km/h. Shane Because I need a break at work, I actually sat down to work through the algebra. Help me out with some of it, please. The first equation is linear. Any assumptions around Crr will linearly impact the equation. In real life, how variable is this? Is it likely to be off by +/- 10%? If so, then the power number is +/- 10% as well . . . Does the velocity_road factor include the contribution from the wind? I have a hard time thinking what a reasonable assumption would be for the velocity_road factor since its really hard to guess what speed I'd be riding at without air resistance. I have a hard time reconciling the second equation. If I put in my relative air velocity at zero or near zero, the equation becomes zero (or just about). But logically, this doesn't make sense. You always have measureable resistance when moving through a fluid, even if the fluid isn't moving. In my head, I'm thinking the difference between HR's hit in Rockies stadium vs. Wrigley Field. So, please help me out with what you mean by "relative numbers." Since these are vectors, I feel like I need a point of reference. (For the record, I strongly suspect that you are correct, just wanted a chance to practice with Equation Solver in Excel.) |
2010-05-25 6:04 PM in reply to: #2881720 |
Master 1681 Rural Ontario | Subject: RE: Aero? Downwind? gsmacleod - 2010-05-25 5:29 PM mrcurtain - 2010-05-25 4:48 PM Thus showing that aerodynamics is counter to performance with a strong enough tail wind. Here are the equations that will allow you to calculate the power required to move a bicycle on a flat course. I will leave the algebra as an exercise to the reader Power to overcome rolling resistance P1 = mass * acceleration due to gravity * Crr * velocity_road Power to overcome aerodynamic resistance P2 = 1/2 * air density * CdA * velocity_wind^2 * velocity_road Mass is total mass of the system in kg Acceleration due to gravity is 9.81m/s^2 Crr - .0100 would be a good guess velocity_road is what your speedometer would read in m/s air density - I would use 1.22 CdA - cross sectional area exposed to the wind - try .3 for aero position and .5 for road position velocity_wind is the relative wind speed ETA - the total power to move the bicycle would be P = P1 + P2 (measured in watts) I don't have the calculations at hand, but IIRC for a typical rider (75kg) and bike (10kg) riding at 200W, I *think* the breakeven point was with a tailwind of about 100km/h. Shane From basic vector analysis, a 100km/hr absolute tailwind will provide a PUSH unless your relative ground speed is >100km/hr. The OP's original issue was about aero position and tailwind. When the tailwind exceeds your groundspeed, you are better off sitting up UNLESS you generate more power bent-over. If the tailwind is lower than your ground speed, then its a function of your power-generation curve depending on position and relative wind-velocity. As a pilot and professional engineer I feel pretty confident with this reasoning. As a rule of thumb for me, if the tailwind is >30km/hr, I'm sitting up, 15-30km/hr I'm in a relaxed position, and below 15km/hr I'm aero.... Mark Galanter |
2010-05-25 6:10 PM in reply to: #2881202 |
Champion 7136 Knoxville area | Subject: RE: Aero? Downwind? mrcurtain - 2010-05-25 3:01 PM Leegoocrap - A corvette is more aero than a mini van going 100mph, 10mph, 0mph, upwind, downwind, whatever. Actually, if the two vehicles are at at dead stop in a 50 mile an hour breeze, the van will be moved along _much_ faster than the corvette. What does that have to do with how aerodynamic something is? |
2010-05-25 6:27 PM in reply to: #2881892 |
Champion 9600 Fountain Hills, AZ | Subject: RE: Aero? Downwind? mgalanter - 2010-05-25 5:04 PM The OP's original issue was about aero position and tailwind. When the tailwind exceeds your groundspeed, you are better off sitting up UNLESS you generate more power bent-over. Power wasn't the question and that remains debatable. The concept of the "sail effect" has been discussed before. Edited by bryancd 2010-05-25 6:28 PM |
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2010-05-25 7:03 PM in reply to: #2881858 |
Champion 9407 Montague Gold Mines, Nova Scotia | Subject: RE: Aero? Downwind? FeS - 2010-05-25 7:39 PM gsmacleod - 2010-05-25 4:29 PMI will leave the algebra as an exercise to the reader Because I need a break at work, I actually sat down to work through the algebra. Help me out with some of it, please. You realize that you aren't actually supposed to do the exercise left for the reader; that's code for good luck and I hope you have lots of paper and pencils Seriously though, the first equation as you mention is easy. It scales linearly and within the speeds that a cyclist could expect would be valid. For the second equation though, things get very messy. As you state, the v_wind requires velocities in order to complete. So, a cyclist going at 10m/s with a 20m/s headwind would give a v_wind of 30m/s and the same cyclist with a 20m/s tailwind would give a v_wind of 10m/s. So equation 2 becomes: P2 = 1/2 * air density * CdA * (velocity_road - velocity_wind)^2 * velocity_road This then resolves as: P2 = 1/2 * air density * CdA * (velocity_road^3 - 2velocity_road^2*velocity_wind + velocity_road*velocity_wind^2) So when you combine P1 and P2 you have a cubic function to solve. Assuming that the cyclist can maintain power at high speeds then you end up with a very high v_road before it becomes benefical to situp and let the sail effect take over. The homerun issue you mention is addressed in air density equation; the air density at Mile High Stadium is about 1.05 instead of 1.22. This decreased air density is also why some cycling records are attempted at altitude (and Calgary has a very fast speedskating oval) because the decreased air density is more beneficial than the performance loss due to the reduction of the partial pressure of oxygen; at least to a point. Shane Edited by gsmacleod 2010-05-25 7:03 PM |
2010-05-25 7:06 PM in reply to: #2881892 |
Champion 9407 Montague Gold Mines, Nova Scotia | Subject: RE: Aero? Downwind? mgalanter - 2010-05-25 8:04 PM From basic vector analysis, a 100km/hr absolute tailwind will provide a PUSH unless your relative ground speed is >100km/hr. As long as you can keep power to the pedals, up to 100km/h tailwind, keep pedalling. Basically, for all reasonable wind speeds you would cycle in, if you have a tailwind you need to pedal harder so that you have a relative headwind. Shane |
2010-05-25 7:27 PM in reply to: #2881892 |
Extreme Veteran 502 Washington | Subject: RE: Aero? Downwind? mgalanter - 2010-05-25 6:04 PM gsmacleod - 2010-05-25 5:29 PM mrcurtain - 2010-05-25 4:48 PM Thus showing that aerodynamics is counter to performance with a strong enough tail wind. Here are the equations that will allow you to calculate the power required to move a bicycle on a flat course. I will leave the algebra as an exercise to the reader Power to overcome rolling resistance P1 = mass * acceleration due to gravity * Crr * velocity_road Power to overcome aerodynamic resistance P2 = 1/2 * air density * CdA * velocity_wind^2 * velocity_road Mass is total mass of the system in kg Acceleration due to gravity is 9.81m/s^2 Crr - .0100 would be a good guess velocity_road is what your speedometer would read in m/s air density - I would use 1.22 CdA - cross sectional area exposed to the wind - try .3 for aero position and .5 for road position velocity_wind is the relative wind speed ETA - the total power to move the bicycle would be P = P1 + P2 (measured in watts) I don't have the calculations at hand, but IIRC for a typical rider (75kg) and bike (10kg) riding at 200W, I *think* the breakeven point was with a tailwind of about 100km/h. Shane From basic vector analysis, a 100km/hr absolute tailwind will provide a PUSH unless your relative ground speed is >100km/hr. The OP's original issue was about aero position and tailwind. When the tailwind exceeds your groundspeed, you are better off sitting up UNLESS you generate more power bent-over. If the tailwind is lower than your ground speed, then its a function of your power-generation curve depending on position and relative wind-velocity. As a pilot and professional engineer I feel pretty confident with this reasoning. As a rule of thumb for me, if the tailwind is >30km/hr, I'm sitting up, 15-30km/hr I'm in a relaxed position, and below 15km/hr I'm aero.... Mark Galanter I agree with this post. For a tailwind exceeding your ground speed, it seems as if you'll gain free power by increasing your cda. Shane gives the equation: P = P1 + P2 Where P1 is the power required to overcome rolling resistance, P2 is the total power to overcome air resistance, and P is the total power required to move at velocity_road If we look closer at P2: P2 = 1/2 * air density * CdA * velocity_wind^2 * velocity_road Where velocity_wind is the relative wind speed which is determinable by a vector equation. but in the case of an absolute tailwind it becomes velocity_road - tailwind ( a cross tailwind would be velocity_road - cos(angle)*tailwind, and a headwind being velocity_road + cos(angle)*tailwind ) Therefore if tailwind > velocity_road, P2 goes negative and your total power becomes P = P1 + (-P2) or P = P1 - P2 Which means that P2 will decrease the amount of power required to go at velocity_road. Increasing the absolute value of P2 means more free power. Unfortunately the only variable in the P2 equation we can do anything about is CdA, which means sitting up. So if, tailwind > velocity_road, it seems you'll get free power by sitting up. Edited by ionlylooklazy 2010-05-25 7:31 PM |
2010-05-25 7:38 PM in reply to: #2882014 |
Champion 9407 Montague Gold Mines, Nova Scotia | Subject: RE: Aero? Downwind? ionlylooklazy - 2010-05-25 9:27 PM So if, tailwind > velocity_road, it seems you'll get free power by sitting up. But you'll go faster by staying aero and continuing to push the pedals. Shane |
2010-05-25 7:48 PM in reply to: #2882045 |
Extreme Veteran 502 Washington | Subject: RE: Aero? Downwind? gsmacleod - 2010-05-25 7:38 PM ionlylooklazy - 2010-05-25 9:27 PM So if, tailwind > velocity_road, it seems you'll get free power by sitting up. But you'll go faster by staying aero and continuing to push the pedals. Shane why is that? it seems like by the equation, if you sit up and continue to pedal, it seems like as long as relative_velocity_wind is < 0 the wind is giving you free watts, so increasing CdA increases your power generated by the wind on your back. as soon as relative_veloctiy_wind > 0, then you want to decrease CdA to minimize the power to overcome wind resistance. but i do think that relative_velocity_wind < 0 is rarely the case for most of the field, except for maybe BOP'ers on windy days Edited by ionlylooklazy 2010-05-25 8:06 PM |
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2010-05-25 8:06 PM in reply to: #2882062 |
Champion 9407 Montague Gold Mines, Nova Scotia | Subject: RE: Aero? Downwind? ionlylooklazy - 2010-05-25 9:48 PM why is that? it seems like by the equation, if you sit up and continue to pedal, it seems like as long as relative_velocity_wind is < 0 the wind is giving you free watts, so increasing CdA increases your power generated by the wind on your back. as soon as relative_veloctiy_wind > 0, then you want to decrease CdA to minimize the power to overcome wind resistance. I reran the numbers; here's the chart:
Assuming you can keep putting power to the pedals you can see even with the tailwind, it takes less power to maintain the same speed while aero. About 100km/h tailwind is where it starts to make sense to sit up and sail. This assumes 87kg rider/bike, 1.22 density, 0.008 Crr and .3 (ok aero position) and .5 (good road position) for CdA. ETA - speeds are all km/h Shane Edited by gsmacleod 2010-05-25 8:07 PM |
2010-05-25 9:03 PM in reply to: #2881720 |
Master 1681 Rural Ontario | Subject: RE: Aero? Downwind? Power to overcome rolling resistance P1 = mass * acceleration due to gravity * Crr * velocity_road Power to overcome aerodynamic resistance P2 = 1/2 * air density * CdA * velocity_wind^2 * velocity_road Mass is total mass of the system in kg Acceleration due to gravity is 9.81m/s^2 Crr - .0100 would be a good guess velocity_road is what your speedometer would read in m/s air density - I would use 1.22 CdA - cross sectional area exposed to the wind - try .3 for aero position and .5 for road position velocity_wind is the relative wind speed ETA - the total power to move the bicycle would be P = P1 + P2 (measured in watts) Shane The equation for P2 should read: P2 = Force of Air Drag x Velocity of air = 1/2 * air density * CdA * velocity_wind^3 (this equation will hold true for non-laminar airflow where Re > 1000) Road velocity is inconcequential for P2. From this we can take your data (m=75kg, Cda = 0.3 aero, 0.5 upright, etc.) and plot it. The plot show two scenarios. In both we will set and maintain Road Velocity at 30km/hr (8.33 m/s) - this means our rider needs 61 wats to overcome rolling resistance. The only variables are Position (Cda) and Apparent Wind Velocity. You can see that below +3m/s apparent wind (true tailwind of 5m/s or 20km/hr causing an apparent headwind of 3m/s or 11km/hr) it stops making much differance if you are in Aero position or Upright. Once the tailwind is at 14m/s (50km/hr) it creates an apparent wind of -6m/s and at this point you might as well sit up and stop pedalling. man, this thread really brought out the engineer nerd in me.... |
2010-05-25 9:08 PM in reply to: #2882045 |
Champion 6503 NOVA - Ironic for an Endurance Athlete | Subject: RE: Aero? Downwind? Let me simplify. Unless you plan on riding the entire bike portion in aero, downwind is a better time to sit up than when you ride into the wind. |
2010-05-26 7:21 AM in reply to: #2882199 |
Champion 9407 Montague Gold Mines, Nova Scotia | Subject: RE: Aero? Downwind? mgalanter - 2010-05-25 11:03 PM The equation for P2 should read: P2 = Force of Air Drag x Velocity of air = 1/2 * air density * CdA * velocity_wind^3 (this equation will hold true for non-laminar airflow where Re > 1000) Road velocity is inconcequential for P2. Not quite; I believe your equation works for an object in flight but for something rolling along the ground, the equation I posted is correct. This is due to the fact that power is work over time and work is the dot product of force and distance. The force is calculated as 1/2 * air density * CdA * v_wind^2 and the distance is a result of the v_road not the v_wind. Once the tailwind is at 14m/s (50km/hr) it creates an apparent wind of -6m/s and at this point you might as well sit up and stop pedalling. My assumption is that the cyclist never stops pedalling and is able to keep power to the pedals regardless of the speed. Instead of setting your speed to be constant (8.33m/s) instead calculate the speed required for a given wattage. So if a cyclist can maintain 200W (or 150W or 100W or whatever) the real question is how fast does the tailwind need to be before sitting up is an advantage over staying aero. While you are correct that with a large tailwind you can situp and maintain a constant 30km/h, if you keep pedalling with 200W (in my model) you will go faster than you would sitting up and pedalling at 200W. Shane |
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