Headwinds: equivalent gradient

MountainMonster

5* wind up.

chris_bass

thegreatdivide wrote:

Maglia Rosa wrote:

Is that your best effort?

Very poor

I did say it before you even bumped your gums. And if someone cheats and have their victories erased from the record books such as Lance has then you cannot call them winners. If you are so fickle as to then believe that everyone who ever won is a doper also then why do you follow this sport? Go take up speed knitting or something.

FFS where did you come from? Just what this forum needs, another prize bell end.

nah, just all the drugs giving him mood swings

rower63

fatsmoker wrote:

So as I battled through a 20mph headwind over about 6 miles on the way home from work today I wondered what sort of gradient that would be similar to going up. The wind had me in the 34T front ring and somewhere around the middle of the rear. About 4%?

To return to the OP's question.
The generally-used formulae for estimating power loss and drag in cycling suggest that the "equivalent gradient" of a cyclist travelling at a speed of v m/s with a (frontal portion of the) headwind of u m/s is given by the formula

gradient = tan(arcsin(rho x Cd x A / (2 x M x g) x (v + u)^2))

which for small angles or gradients approximates to

gradient ~ (rho x Cd x A / (2 x M x g) x (v + u)^2)

where rho = air density; Cd = coefficient of drag (0.9ish for on the hoods, 0.65 for pro on TT bike); A is projected frontal area of bike+rider; M = mass of bike+rider; g = acceleration due to gravity;

In other words, equivalent gradient will vary with the square of total apparent wind.

edit: all in all not very useful until you try an example, I'll admit
Using your original numbers, assuming you are 80kg and on the drops and struggling along at 14mph into a 20mph wind, the equivalent gradient is not that great, only about 0.4%.
But thinking about it it does make sense: you can get along reasonably OK at say 14mph into that sort of headwind and even stiffer, whereas as soon as you hit theseemingly low single percentages, say 4%, you slow down dramatically.

fatsmoker

Thank you Rower63. On that basis I won't be attempting Mt Ventoux any time soon.

Matthewfalle

The wind is our friend - it makes us strong.

fatsmoker

It made me sleep well last night. Shattered when I got home

chris_bass

fatsmoker wrote:

It made me sleep well last night. Shattered when I got home

I'll second that! looks like being worse tonight too! oh joy!

FJS

rower63 wrote:

fatsmoker wrote:

So as I battled through a 20mph headwind over about 6 miles on the way home from work today I wondered what sort of gradient that would be similar to going up. The wind had me in the 34T front ring and somewhere around the middle of the rear. About 4%?

To return to the OP's question.
The generally-used formulae for estimating power loss and drag in cycling suggest that the "equivalent gradient" of a cyclist travelling at a speed of v m/s with a (frontal portion of the) headwind of u m/s is given by the formula

gradient = tan(arcsin(rho x Cd x A / (2 x M x g) x (v + u)^2))

which for small angles or gradients approximates to

gradient ~ (rho x Cd x A / (2 x M x g) x (v + u)^2)

where rho = air density; Cd = coefficient of drag (0.9ish for on the hoods, 0.65 for pro on TT bike); A is projected frontal area of bike+rider; M = mass of bike+rider; g = acceleration due to gravity;

In other words, equivalent gradient will vary with the square of total apparent wind.

edit: all in all not very useful until you try an example, I'll admit
Using your original numbers, assuming you are 80kg and on the drops and struggling along at 14mph into a 20mph wind, the equivalent gradient is not that great, only about 0.4%.
But thinking about it it does make sense: you can get along reasonably OK at say 14mph into that sort of headwind and even stiffer, whereas as soon as you hit theseemingly low single percentages, say 4%, you slow down dramatically.

There's a Flemish website that has attempted to calculate this : http://www.fietsica.be/de_wind.htm (google translate will help)

meanredspider

rower63 wrote:

Using your original numbers, assuming you are 80kg and on the drops and struggling along at 14mph into a 20mph wind, the equivalent gradient is not that great, only about 0.4%.
But thinking about it it does make sense: you can get along reasonably OK at say 14mph into that sort of headwind and even stiffer, whereas as soon as you hit theseemingly low single percentages, say 4%, you slow down dramatically.

I have to say, this doesn't feel correct. Your example gives an effective airspeed of 34mph. At speeds of 30mph and above, a significant proportion of your power is going into defeating air resistance. So, with your numbers, it's equivalent to travelling in still air at about 30mph. Now, most people could ride up a 0.4% slope at 14mph but not that many can sustain 30mph on the flat. I'd say a 20mph headwind (true headwind with no shelter) is nearer 4% gradient than 0.4%. (I've not yet looked at the Flemish site)

iPete

I think this would provide an excellent business case to any other halves for the purchase of a power meter, take all this guess work out.

FJS

meanredspider wrote:

rower63 wrote:

Using your original numbers, assuming you are 80kg and on the drops and struggling along at 14mph into a 20mph wind, the equivalent gradient is not that great, only about 0.4%.
But thinking about it it does make sense: you can get along reasonably OK at say 14mph into that sort of headwind and even stiffer, whereas as soon as you hit theseemingly low single percentages, say 4%, you slow down dramatically.

I have to say, this doesn't feel correct. Your example gives an effective airspeed of 34mph. At speeds of 30mph and above, a significant proportion of your power is going into defeating air resistance. So, with your numbers, it's equivalent to travelling in still air at about 30mph. Now, most people could ride up a 0.4% slope at 14mph but not that many can sustain 30mph on the flat. I'd say a 20mph headwind (true headwind with no shelter) is nearer 4% gradient than 0.4%. (I've not yet looked at the Flemish site)

The Flemish site suggest 5% for 48 km/h (29 mph)

fatsmoker

Cool. I'll ask my missus if she minds me buggering off for a week in the Alps this summer.

meanredspider

iPete wrote:

I think this would provide an excellent business case to any other halves for the purchase of a power meter, take all this guess work out.

Yes - I meant to add that. In a typical windy N Holland day, I'm in the small ring putting out about 250W. I need to find out what that looks like up one of the hills here in Scotland. Frustratingly, airport security was marginal about whether I'd be allowed to transport the Stages in my hand baggage: it's an expensive risk to have it confiscated.

rower63

What's a factor of 10 amongst friends?

My apologies, redoing my calcs I come up with 5.5%, I did it originally literally on the back of an envelope and was a little surprised at the result, but then neglected to check it :oops:

itboffin

I'd second that on the 20mph at least 4% I can very easily ride to work at 20 mph avg fully loaded over a rolling 15 miles add a brisk headwind and strong gusts and its a strugglento hit 15, thats on a par with a long climb in summer.

ForumNewbie

itboffin wrote:

I'd second that on the 20mph at least 4% I can very easily ride to work at 20 mph avg fully loaded over a rolling 15 miles add a brisk headwind and strong gusts and its a strugglento hit 15, thats on a par with a long climb in summer.

itboffin, you must be very fast if you can commute at 20mph average very easily, while fully loaded, riding solo, for 15 miles on a rolling road. What would your average be going flat out on the same sort of route on a light bike with no luggage?

Sutton_Rider

To approximate the headwind component of a crosswind take your heading, lets say 360 deg or North, wind coming from 45 deg at 20mph. Now think of a clock 45 is 3/4 of an hour. Therefore the headwind component would be 3/4 of 20mph So 15mph. When you get over 60 degs it's mostly crosswind anyway. As I said its only an approximate. It works for landing aircraft.

pilot_pete

Sutton Rider wrote:

To approximate the headwind component of a crosswind take your heading, lets say 360 deg or North, wind coming from 45 deg at 20mph. Now think of a clock 45 is 3/4 of an hour. Therefore the headwind component would be 3/4 of 20mph So 15mph. When you get over 60 degs it's mostly crosswind anyway. As I said its only an approximate. It works for landing aircraft.

That rule of thumbs not works for a wind that is 45 deg off! And then only approximately. What if it was 30 deg off? Using your rule, 30 on a clock face is 1/2, therefore the headwind component would be 1/2 of 20, which is 10...which it is not!

There are plenty of free aviation apps available which work out crosswind and headwind components for pilots. I would use one of these myself, but there again I suppose I would as a pilot!

I did a loop (cycling) last week and on the way home I was up at my TT threshold and only managing 14mph on the flat. I know I am out of shape but still! That was one strong headwind.

PP

Sutton_Rider

Sutton Rider wrote:

To approximate the headwind component of a crosswind take your heading, lets say 360 deg or North, wind coming from 45 deg at 20mph. Now think of a clock 45 is 3/4 of an hour. Therefore the headwind component would be 3/4 of 20mph So 15mph. When you get over 60 degs it's mostly crosswind anyway. As I said its only an approximate. It works for landing aircraft.

Should read crosswind

mabbo

I assume rower 63 watches the big bang theory as well?
Living down here on the SE coast, it is always windy, always from the south west, and always in your face. All year. Get on with it.