Any brains out there?

Bob said:

Some physics needed here...
If a bike and rider are travelling in a straght line on a flat surface with
a 90 degree cross-wind of x value, is the cross-wind slowing the rider?

Yes. Drag is proportional to the square of the airspeed. If the
rider is moving at 1 unit (speed) and the wind is also at 1 unit,
the rider is moving through the air at 1.414 units (approx, sq-root
of 2) at 45% to his motion over the ground. This means the drag (at
45%) is proportional to 2 units, in contrast to the still air case
where he is experiencing drag proportional to 1 unit, directly
against him. The 2 unit drag at 45% contributes a component against the
direction of motion of 2*cos(45) or 2*(0.707) or 1.414. So your
lateral wind at the same speed as the rider raises drag by 41%.

The wind *is* usually against you.

Brendan
--
Brendan Halpin, Department of Sociology, University of Limerick, Ireland
Tel: w +353-61-213147 f +353-61-202569 h +353-61-338562; Room F2-025 x 3147
[email hidden] http://www.ul.ie/sociology/brendan.halpin.html

Bob said:

Some physics needed here...
If a bike and rider are travelling in a straght line on a flat surface with
a 90 degree cross-wind of x value, is the cross-wind slowing the rider?

To first order, no. A wind at 90 degrees has no component in the
direction of travel.

--
Tony

"A facility for quotation covers the absence of original thought" Lord
Peter Wimsey (Dorothy L. Sayers)

On second thoughts, provided he doesn't mind being driven off course, I
think it is actually speeding him up as his total speed will be a
vector of his original speed plus the speed that the wind pushes him
sideways. If he wants to keep on his original course then he'll have to
steer into the wind but I haven't got my head round how that affects
things yet...

And if he's steering into the wind to keep on a straight course then
not all of his energy is going into propelling him forwards so I was
wrong, the wind will slow him down. (I could be totaly wrong though,
especially at this time of night).

Brendan Halpin said:

Bob said:

Some physics needed here...
If a bike and rider are travelling in a straght line on a flat surface with
a 90 degree cross-wind of x value, is the cross-wind slowing the rider?

Yes. Drag is proportional to the square of the airspeed. If the
rider is moving at 1 unit (speed) and the wind is also at 1 unit,
the rider is moving through the air at 1.414 units (approx, sq-root
of 2) at 45% to his motion over the ground. This means the drag (at
45%) is proportional to 2 units, in contrast to the still air case
where he is experiencing drag proportional to 1 unit, directly
against him. The 2 unit drag at 45% contributes a component against the
direction of motion of 2*cos(45) or 2*(0.707) or 1.414. So your
lateral wind at the same speed as the rider raises drag by 41%.

The wind *is* usually against you.

Brendan

Ermmm no. Lets take your logic and a stationary bicycle. 1 unit side
wind and 0 units of moving gives 0.707 units of wind at 45 degrees.
This squared produces 1 unit of drag at 45 degrees which resolved as cos
45 produces a component against the direction of motion of 0.707. i.e
you have just argued that a side wind on a stationary bicycle produces a
head wind. Which is clearly nonsense because so is the logic that
created it.

I do agree though that the wind in usually against you ;-)

--
Tony

"A facility for quotation covers the absence of original thought" Lord
Peter Wimsey (Dorothy L. Sayers)

Bob said:

If a bike and rider are travelling in a straght line on a flat surface
with a 90 degree cross-wind of x value, is the cross-wind slowing the
rider?

Yes.

It's your triangle of forces, innit. If you've got wind pushing you from
the side, but you're making good a course that goes straight forward, the
effort you're exerting must be pointing somewhere between the wind and the
direction of travel. Which is harder than if there's no wind, and you can
just pedal in the direction you want to go.

Maybe, anyway. If you start taking dot products of those forces with the
speed you're actually moving at, i'm not sure it comes out as more work.

Bloody is, though.

tom

--
Space Travel is Another Word for Love!

Brendan Halpin said:

Bob said:

Some physics needed here...
If a bike and rider are travelling in a straght line on a flat
surface with a 90 degree cross-wind of x value, is the cross-wind
slowing the rider?

Yes. Drag is proportional to the square of the airspeed. If the
rider is moving at 1 unit (speed) and the wind is also at 1 unit,
the rider is moving through the air at 1.414 units (approx, sq-root
of 2) at 45% to his motion over the ground. This means the drag (at
45%) is proportional to 2 units, in contrast to the still air case
where he is experiencing drag proportional to 1 unit, directly
against him. The 2 unit drag at 45% contributes a component against
the direction of motion of 2*cos(45) or 2*(0.707) or 1.414. So your
lateral wind at the same speed as the rider raises drag by 41%.

That's drag. Now what about lift or thrust? Can the rider and bike act
like a sail and get forwards motion from sideways force?

~PB

Tony Raven said:

Brendan Halpin said:

Bob said:

Some physics needed here...
If a bike and rider are travelling in a straght line on a flat

surface with

Quoted message said:

Quoted message said:

Quoted message said:

a 90 degree cross-wind of x value, is the cross-wind slowing the

rider?

Quoted message said:

Quoted message said:

Yes. Drag is proportional to the square of the airspeed. If the
rider is moving at 1 unit (speed) and the wind is also at 1 unit,
the rider is moving through the air at 1.414 units (approx, sq-root
of 2) at 45% to his motion over the ground. This means the drag (at
45%) is proportional to 2 units, in contrast to the still air case
where he is experiencing drag proportional to 1 unit, directly
against him. The 2 unit drag at 45% contributes a component against

the

Quoted message said:

Quoted message said:

direction of motion of 2*cos(45) or 2*(0.707) or 1.414. So your
lateral wind at the same speed as the rider raises drag by 41%.

The wind *is* usually against you.

Brendan

Ermmm no.

Just goes to show that if you ask a straightforward question on usenet,
you can usually rely on getting the right answer...and several wrong
ones!

Quoted message said:

Lets take your logic and a stationary bicycle. 1 unit side
wind and 0 units of moving gives 0.707 units of wind at 45 degrees.

Um..how? Wind is a vector, and 1 from the side and 0 from ahead, sums
to....wait for it....1 unit from the side. So the air speed is 1 unit,
the drag is proportional to 1^2, but this neither slows down or speeds
up the (stationary) rider cos it is perpendicular to the direction of
(non-)travel.

Brendan's answer is correct.

James

My personal opinion is that the only extra force required for forward motion
would be to make up the losses caused by friction. The wind may be strong
enough to move the bike laterally and this would need to be made up by
riding back to the original line of direction.
Some component of this would be against the wind.

A similar effect would be riding on the straight of a velodrome where the
track is inclined 12 - 13 degrees.
Takes some thinking.

"James Annan" wrote

Quoted message said:

Quoted message said:

Quoted message said:

If a bike and rider are travelling in a straght line on a
flat surface with a 90 degree cross-wind of x value,
is the cross-wind slowing the rider?

< snip >

Quoted message said:

Wind is a vector, and 1 from the side and 0 from ahead,
sums to....wait for it....1 unit from the side. So the air speed
is 1 unit, the drag is proportional to 1^2, but this neither slows
down or speeds up the (stationary) rider cos it is perpendicular
to the direction of (non-)travel.

Brendan's answer is correct.

I don't disagree with anything you have said about the wind speed vector but
....

I can't help thinking that experience tells me it's harder to cycle in a 30
mile side wind than in perfectly calm conditions.

Perhaps there is something a bit more complex than simple vector wind speeds
to consider, such as the turbulence caused by the cross wind. I'm out of my
depth now, so I'll shut up ;-)

--
Dave Lowther
E-mail [email hidden]
Drop the rconham to reply by e-mail.
Web http://www.dlowther.demon.co.uk/local.htm

Tony Raven said:

Ermmm no. Lets take your logic and a stationary bicycle.

A stationary bicycle is fixed to the floor.
Any forces will be transmitted through the floor mounts.

A side wind from 90 degrees on a stationary bicycle produces
a force at 90 degrees.

To get the same vectors as the initial problem, you need to add a fan
in front of the rider.

in message <[email hidden]>, Bob

(') said:

Some physics needed here...
If a bike and rider are travelling in a straght line on a flat surface
with a 90 degree cross-wind of x value, is the cross-wind slowing the
rider?

Not directly (in a physics exercise sense). But in the real world wind
is rarely laminar, and a strong turbulent crosswind gives rise to a lot
of buffeting which makes riding (and, particularly, maintaining a
straight line) much more difficult.

--
[email hidden] (Simon Brooke) http://www.jasmine.org.uk/~simon/

((DoctorWho)ChristopherEccleston).act();
uk.co.bbc.TypecastException: actor does not want to be typecast.
[adapted from autofile on /., 31/03/05]

in message <[email hidden]>, Brendan Halpin

(') said:

Bob said:

Some physics needed here...
If a bike and rider are travelling in a straght line on a flat
surface with a 90 degree cross-wind of x value, is the cross-wind
slowing the rider?

Yes. Drag is proportional to the square of the airspeed. If the
rider is moving at 1 unit (speed) and the wind is also at 1 unit,
the rider is moving through the air at 1.414 units (approx, sq-root
of 2) at 45% to his motion over the ground. This means the drag (at
45%) is proportional to 2 units, in contrast to the still air case
where he is experiencing drag proportional to 1 unit, directly
against him. The 2 unit drag at 45% contributes a component against
the direction of motion of 2*cos(45) or 2*(0.707) or 1.414. So your
lateral wind at the same speed as the rider raises drag by 41%.

Maybe, I'm not convinced. The tyres of a bike produce much more
effective lateral resistance than the keel of a yacht. While I agree
that drag exists the lateral drag is countered by heeling to windward
and I don't believe it directly causes the rider to do more work.

Quoted message said:

The wind *is* usually against you.

Oh, granted.

--
[email hidden] (Simon Brooke) http://www.jasmine.org.uk/~simon/

;; L'etat c'est moi -- Louis XVI
;; I... we... the Government -- Tony Blair

"David Lowther" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

I don't disagree with anything you have said about the wind speed vector
but ...

I can't help thinking that experience tells me it's harder to cycle in a
30 mile side wind than in perfectly calm conditions.

Perhaps there is something a bit more complex than simple vector wind
speeds to consider, such as the turbulence caused by the cross wind. I'm
out of my depth now, so I'll shut up ;-)

--
Dave Lowther
E-mail [email hidden]
Drop the rconham to reply by e-mail.
Web http://www.dlowther.demon.co.uk/local.htm

Harder to balance and maintain a straight line perhaps. So I'd guess
there'd be some secondary effects:

* You will cycle slightly further as you're less likely to maintain as
straight a line. (Of course this assumes a real world wind with gusts,
rather than a constant smooth airflow)
* There will be a tiny amount of extra load on the tyres due to the wind
and leaning into it.
* Due to the lean and sideways component, I'd guess there'd be some tire
scrub as well.

I'd think that the sum all the effects will be much less than cycling
directly into said sidewind, instead of at right angles to it.

Brendan Halpin said:

Bob said:

Some physics needed here...
If a bike and rider are travelling in a straght line on a flat

surface with

Quoted message said:

Quoted message said:

a 90 degree cross-wind of x value, is the cross-wind slowing the

rider?

Quoted message said:

Yes. Drag is proportional to the square of the airspeed. If the
rider is moving at 1 unit (speed) and the wind is also at 1 unit,
the rider is moving through the air at 1.414 units (approx, sq-root
of 2) at 45% to his motion over the ground.

The cross wind does not affect the speed of the rider through the air
at all. I think you are getting motion through the air mixed up with
relative motion. They are not always the same and in this example they
are not the same. Because the motion of the rider through the air does
not change with the cross wind the force it takes to overcome drag in
order to move thru the air does not change either.

The cross wind increases the relative airflow influencing the rider and
changes its direction. This increases drag and changes its direction.
In order to move thru the air you must overcome drag but what is just
as true is that in order to remain still in moving air (wind) you must
also overcome drag. The increased drag does not slow the motion or the
rider thru the air it tries to cause the sideways motion of the rider.
Drag is in the direction of the relative airflow that caused it and the
increased drag is caused by the component of the relative airflow that
is generated by the cross wind.

Overcoming this sideways drag force may cause a little more rolling
drag. Any lift say from solid dish wheels will have a forward
component. The chances of the wind being against you are the same as
the chances of it being with you. I am very optimistic about that.

This means the drag (at

Quoted message said:

45%) is proportional to 2 units, in contrast to the still air case
where he is experiencing drag proportional to 1 unit, directly
against him. The 2 unit drag at 45% contributes a component against

the

Quoted message said:

direction of motion of 2*cos(45) or 2*(0.707) or 1.414. So your
lateral wind at the same speed as the rider raises drag by 41%.

The wind *is* usually against you.

Brendan
--
Brendan Halpin, Department of Sociology, University of Limerick,

Ireland

Quoted message said:

Tel: w +353-61-213147 f +353-61-202569 h +353-61-338562; Room F2-025

x 3147

Quoted message said:

[email hidden]

http://www.ul.ie/sociology/brendan.halpin.html

Bob said:

Some physics needed here...
If a bike and rider are travelling in a straght line on a flat surface with
a 90 degree cross-wind of x value, is the cross-wind slowing the rider?

I'm not qualified to do the physics.

But every experience tells me cross winds slow you down.

BugBear

Pete Biggs said:

That's drag. Now what about lift or thrust? Can the rider and bike
act like a sail and get forwards motion from sideways force?

Paul Davies maintains that in the usual Tuesday Night Gale at Castle Combe
it's possible to lap around 2/3 of the circuit at 25 mph without having to
pedal.

The caveat being that you need to be using one of these:

URL:http://www.legslarry.beerdrinkers.co.uk/images/pix/BobSmith/Big/bobsmith_035.jpg

--
Dave Larrington - <http://www.legslarry.beerdrinkers.co.uk/>
While you were out at the Rollright Stones, I came and set fire to your
Shed.

bugbear bugbear@trim_papermule.co.uk_trim said:

But every experience tells me cross winds slow you down.

Happy winds speed you up though..

ob physics:

What you need to do is to work out what the resultant wind speed is
for the rider. That will indicate whether there is any effect or not.

...d

--
----------------------------------
David Martin PhD
Bioinformatics Scientific Officer
Wellcome Trust Biocentre, Dundee
+44 1382 348704
----------------------------------

David Lowther said:

"James Annan" wrote

Quoted message said:

Quoted message said:

>If a bike and rider are travelling in a straght line on a
>flat surface with a 90 degree cross-wind of x value,
>is the cross-wind slowing the rider?

< snip >

Quoted message said:

Wind is a vector, and 1 from the side and 0 from ahead,
sums to....wait for it....1 unit from the side. So the air speed
is 1 unit, the drag is proportional to 1^2, but this neither slows
down or speeds up the (stationary) rider cos it is perpendicular
to the direction of (non-)travel.

Brendan's answer is correct.

I don't disagree with anything you have said about the wind speed vector but
...

I can't help thinking that experience tells me it's harder to cycle in a 30
mile side wind than in perfectly calm conditions.

I don't know why you used "but" in there. Of course a 30mph side wind is
harder to cycle in. That's exactly what Brendan and I said.

James