Go Back  Bike Forums > Bike Forums > General Cycling Discussion
Reload this Page >

Headwind and Power

Notices
General Cycling Discussion Have a cycling related question or comment that doesn't fit in one of the other specialty forums? Drop on in and post in here! When possible, please select the forum above that most fits your post!

Headwind and Power

Old 01-02-22, 01:02 PM
  #51  
tomato coupe
Senior Member
 
Join Date: Jul 2009
Posts: 2,964

Bikes: Colnago, Van Dessel, Factor, Cervelo, Ritchey

Mentioned: 5 Post(s)
Tagged: 0 Thread(s)
Quoted: 1845 Post(s)
Liked 2,920 Times in 1,211 Posts
Originally Posted by Racing Dan View Post
This thread has devolved in to nonsense and Im out of here as I said in #27
If it has devolved into nonsense, it's only because people are trying to find some explanation that will resonant with you so that you can understand why you are wrong. (And, yes, you are wrong.)
tomato coupe is offline  
Old 01-02-22, 01:02 PM
  #52  
PeteHski
Senior Member
 
PeteHski's Avatar
 
Join Date: May 2021
Posts: 3,199
Mentioned: 3 Post(s)
Tagged: 0 Thread(s)
Quoted: 1506 Post(s)
Liked 1,622 Times in 1,034 Posts
Originally Posted by Racing Dan View Post
Omg! Just stop :-) This was addressed several times before. Including by me in #13 and Im sure in other posts too since you all seem to throw this at me, even if I made no claim to the contrary.

This thread has devolved in to nonsense and Im out of here as I said in #27
People are just trying to explain to you why your assertion about the bike calculator being "wrong" was incorrect. Whether or not you want to accept that and perhaps learn something new is your choice. Your alternative is to believe that the bike calculator (of which there are many) and those people trying to explain really are wrong. On balance which do you think is more likely?
PeteHski is offline  
Old 01-02-22, 06:07 PM
  #53  
KerryIrons
Senior Member
 
Join Date: Mar 2012
Posts: 225
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 90 Post(s)
Likes: 0
Liked 124 Times in 69 Posts
Originally Posted by cyclezen View Post
so, in the spirit of the season - 'off-season' for some - a thread we can all present our As and Os...
...so, I've been wonderin, for so very long...
given the same rider, on the same bike
if you're riding at 10 mph (16 kph) into a 15 mph headwind (24 kph) , are you using the equal/same 'power' as if you were riding at 25 mph (40 kph) ???
is it an equal relationship, and is it linear ?
what other factors might influence this greatly ? - are rolling resistence and impedance nominal or significant players in this drama ?
So Bill Nyes _ The Science Guys in our midst - what is the reality?
not a scientist ? have an opinion? lets hear it !
My wishes for more power to all for the coming year
This discussion is not the place for opinions, it is the place for well-documented physics. The general formula for power to ride a bike is:

horse power = [Vg*W(.0053 + %G/100) + .0083(Va^3)]/375

calories/hr = [Vg*W(.0053 + %G/100) + .0083(Va^3)]*7.2

calories/hr = [V*W(.0053 + %G/100) + .0083(V^3)]*7.2

watts = [V*W(.0053 + %G/100) + .0083(V^3)]*2

where Vg is ground speed, Va is speed through the air (includes head/tail winds), W is bike + rider weight in lbs., and %G is grade in per cent. The factors listed here (0.0053 for friction + rolling resistance and 0.0083 for aerodynamic drag) are obviously not absolute. They will vary with efficiency of the tires and drive train, and with the aerodynamics of the bike + rider combination. Both of these assume a racing position on a racing bike. A clunker bike or a more efficient riding position will change these numbers, which are averages anyway. Power to overcome friction and gravity is proportional only to rider weight and ground speed. Power to overcome wind drag is proportional to the cube of the air speed. For reference, 1 hp = 2700 calories (because of human metabolic efficiency of 24%); 1 calorie = 0.276 watts; 1 hp = 746 watts. Here, all calories are kg-calories, or "food calories."
KerryIrons is offline  
Old 01-02-22, 06:23 PM
  #54  
ofajen
Cheerfully low end
 
ofajen's Avatar
 
Join Date: Jun 2020
Posts: 1,395
Mentioned: 2 Post(s)
Tagged: 0 Thread(s)
Quoted: 427 Post(s)
Liked 677 Times in 444 Posts
Originally Posted by KerryIrons View Post
watts = [V*W(.0053 + %G/100) + .0083(V^3)]*2

where Vg is ground speed, Va is speed through the air (includes head/tail winds), W is bike + rider weight in lbs., and %G is grade in per cent.
Thanks. Just pointing out that the third term, the power requirement for wind drag, should clarified as being proportional to Va**2 x Vg, in your nomenclature.

Otto
ofajen is offline  
Old 01-02-22, 08:29 PM
  #55  
PeteHski
Senior Member
 
PeteHski's Avatar
 
Join Date: May 2021
Posts: 3,199
Mentioned: 3 Post(s)
Tagged: 0 Thread(s)
Quoted: 1506 Post(s)
Liked 1,622 Times in 1,034 Posts
Originally Posted by ofajen View Post
Thanks. Just pointing out that the third term, the power requirement for wind drag, should clarified as being proportional to Va**2 x Vg, in your nomenclature.

Otto
Yep, I spotted that too. Drag force is proportional to the square of the airspeed and Power is proportional to Drag Force x Vg. When there is no headwind Va = Vg and the Power equation then simplifies to V cubed. It's actually this very fact that causes confusion when you start introducing headwind.
PeteHski is offline  

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off


Thread Tools
Search this Thread

Contact Us - Archive - Advertising - Cookie Policy - Privacy Statement - Terms of Service - Do Not Sell My Personal Information -

Copyright 2021 MH Sub I, LLC dba Internet Brands. All rights reserved. Use of this site indicates your consent to the Terms of Use.