3 ms·
I think there's a limited market for them, because the value of aerodynamics depends heavily on how you ride. Wind resistance increases with the square of the
by mrob 8y ago
I think there's a limited market for them, because the value of aerodynamics depends heavily on how you ride.
Wind resistance increases with the square of the speed. Pros can ride at 30mph average on fast stages, where managing wind resistance is huge. A typical road bike commuter might average 15mph. All else being equal that's 1/4th the wind resistance (1/8th the power needed to overcome it), which is still big, but not the overwhelmingly important concern it is for the pros. A commuter on a utility bike who doesn't want to sweat might average only 10mph, so 1/9th the wind resistance (1/27th the power needed to overcome it). At that speed it doesn't feel so important.
For the pro, speed is the top priority, but the commuter also has to consider comfort, reliability, and visibility of other traffic (recumbents put your head lower down so it's harder for you to see over cars, and it's harder for them to see you). On heavily congested roads, e.g. central London, traffic might prevent you from going fast enough for wind resistance to be a major concern. If you commute over steep hills then reduced weight might benefit you more. But if you have a long and flat commute with light traffic, I think a super-efficient commuter bike would be valuable, and it's a pity that UCI rules discourage development of super-efficient bike technology.
EDIT: added power values as per HankB99's comment.
- nerdponx 8y agoThis is the kind of reasoning I was looking for. Thank you!
- HankB99 8y ago>Wind resistance increases with the square of the speed. To be pedantic, wind resistance varies with the cube of the speed. That's my recollection and seems to be used in the formula used to calculate power required at https://www.exploratorium.edu/cycling/aerodynamics1.html https://www.exploratorium.edu/cycling/aerodynamics1.html.
- mrob 8y agoDrag force does vary with the square of the speed, but power needed to overcome that drag force varies with the cube of the speed. From Wikipedia[0]: "Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome aerodynamic drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW).[16] With a doubling of speed the drag (force) quadruples per the formula. Exerting 4 times the force over a fixed distance produces 4 times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, 4 times the work done in half the time requires 8 times the power." Power output is directly felt by the rider, so it's the more relevant comparison here, and your comment is more practically useful. [0] https://en.wikipedia.org/wiki/Drag_(physics)#Power https://en.wikipedia.org/wiki/Drag_(physics)#Power
- HankB99 8y agoOh... right. Work done is force * distance so distance varies linearly with speed.