View Single Post
  #107  
Old September 13th 17, 06:12 PM posted to rec.bicycles.tech
Radey Shouman
external usenet poster
 
Posts: 1,747
Default program to compute gears, with table

John B. writes:

On Tue, 12 Sep 2017 16:19:21 -0400, Radey Shouman
wrote:

John B. writes:

On Mon, 11 Sep 2017 08:23:48 -0700 (PDT), wrote:


[ ... ]

Yesterday I rode on a 35 mile ride. On the way out into a headwind I
averaged a little less than 14 mph. I had a cup of coffee while in
the city square the worst band I ever heard was making awful
noises. When I was in a band if we had played that badly on our
first try in a rehearsal we would have quit.

On the way back the wind had reversed and I had a hard time
maintaining 12 mph for most of the way. By the time I got home I was
exhausted. Do you think that I could improve my performance with an
11 or 12 speed?

I know my limits and it isn't playing as if I was Chris Froome.

Something I've always wondered about is how in the world can I ride an
out and back course and have a head wind both ways :-(


With some reasonable assumptions I think you can show that this is
actually true, in a sense. Suppose for example the wind is blowing at
right angles to your (perfectly straight) direction, and that it happens
to be blowing at exactly your ground speed, v.

The apparent wind will be at 45 degrees your heading, at a velocity of
sqrt(v^2 + v^2) = sqrt(2)*v.

For turbulent flow, the drag force is approximately proportional to the
square of the wind speed, so the drag force will be twice the drag force
you would see in still air, F. (At this point we have assumed a
cylindrical bike & rider, meaning that the coefficient of drag is the
same from the front as the side, since drag from the side is normally
greater, this is conservative).

Fortunately the drag force acts at 45 degrees to your course, so the
drag component that holds you back is
cos(45 deg)*F = (2/sqrt(2))*F
= sqrt(2)*F
~= 1.414 F

This is as true on the way out as it is on the way back, hence you
really do have an effective head wind both ways.


I was thinking of days when I ride what I call my short route. It is a
square loop in the city on which the two longer legs are essentially
due north and due south. I set out and on the south leg the wind was
directly in my face. then the "cross wind" leg, about 1 km and
protected by tall buildings and then the north bound leg. Again wind
directly in my face. There are traffic lights on both the north and
south legs and the wind doesn't stop blowing when I stopped at a red
light :-)


I suspect that a constant wind always slows one down over a closed loop.
Almost certainly true if you ride straight into the wind and then
straight back. Supposing you're a real hard ass, and ride the speed of
the wind both ways: The trip out has four times the drag force, the
trip back zero. Net over the course is twice the drag force.

--
Ads
 

Home - Home - Home - Home - Home