headwinds

wle said:

1. assume a person can maintain 20mph on a flat road.

if there is a steady headwind of 20mph, does that totally stop him?

Not unless there's an ice cream store along way.

Quoted message said:

2. now assume there is a hill which a person can only manage to go 5
mph on.

would a 5 mph headwind stop that person?

Not if there's an ice cream store at top of hill.

Quoted message said:

i would guess the answer to 1 is yes and the answer to 2 is no, but

a. that doesn;t make sense and

b. is that right, and why or why not?

No, because cyclists PEDAL and not just coast.

Bill "30 seconds of life gone" S.

"S o r n i" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

wle said:

1. assume a person can maintain 20mph on a flat road.

if there is a steady headwind of 20mph, does that totally stop him?

Not unless there's an ice cream store along way.

Quoted message said:

2. now assume there is a hill which a person can only manage to go 5
mph on.

would a 5 mph headwind stop that person?

Not if there's an ice cream store at top of hill.

Quoted message said:

i would guess the answer to 1 is yes and the answer to 2 is no, but

a. that doesn;t make sense and

b. is that right, and why or why not?

No, because cyclists PEDAL and not just coast.

Bill "30 seconds of life gone" S.

Bill, you are very amusing.

Dave "Needs more sushi" Thompson

wle said:

1. assume a person can maintain 20mph on a flat road.

if there is a steady headwind of 20mph, does that totally stop him?

2. now assume there is a hill which a person can only manage to go 5 mph on.

would a 5 mph headwind stop that person?

i would guess the answer to 1 is yes and the answer to 2 is no, but

No, the answer is no in both cases.

Quoted message said:

a. that doesn;t make sense and

b. is that right, and why or why not?

Let's just look at case 1 and assume that the only
resistance is due to air drag, i.e. no rolling
resistance or mechnical losses.
The air drag force is proportional to the square
of the relative air velocity and the amount of power
needed to overcome the drag is equal to the force
times the velocity of the bicycle. So in the case
where he's cycling at 20 mph on a calm day he has to
generate power equal to some constant times 20^3;or
C x 8000. Now on the windy day the relative wind
velocity will be (20 + V) and his speed is V, so the
power needed will be C x (20 + V)^2 x V. For this to
equal the power used before (8000 C), his velocity into
the headwind will be 9.31 mph.

Peter said:

wle said:

1. assume a person can maintain 20mph on a flat road.

if there is a steady headwind of 20mph, does that totally stop
him?

2. now assume there is a hill which a person can only manage to go 5
mph on.

would a 5 mph headwind stop that person?

i would guess the answer to 1 is yes and the answer to 2 is no, but

No, the answer is no in both cases.

Quoted message said:

a. that doesn;t make sense and

b. is that right, and why or why not?

Let's just look at case 1 and assume that the only
resistance is due to air drag, i.e. no rolling
resistance or mechnical losses.
The air drag force is proportional to the square
of the relative air velocity and the amount of power
needed to overcome the drag is equal to the force
times the velocity of the bicycle. So in the case
where he's cycling at 20 mph on a calm day he has to
generate power equal to some constant times 20^3;or
C x 8000. Now on the windy day the relative wind
velocity will be (20 + V) and his speed is V, so the
power needed will be C x (20 + V)^2 x V. For this to
equal the power used before (8000 C), his velocity into
the headwind will be 9.31 mph.

I thought my ice cream remarks implied all this.

Bill "matter of terminology" S.

wle said:

1. assume a person can maintain 20mph on a flat road.

if there is a steady headwind of 20mph, does that totally stop him?

2. now assume there is a hill which a person can only manage to go 5 mph on.

would a 5 mph headwind stop that person?

i would guess the answer to 1 is yes and the answer to 2 is no, but

a. that doesn;t make sense and

b. is that right, and why or why not?

wle.

The first mile an hour of headwind on the flat takes away 2/3 miles an hour of speed. Subsequent
mph's of wind take away less. The constraint is rider power, force times bike speed,
which on the flat is airspeed times bikespeed. Airspeed is bike speed plus headwind speed.

Since bike speed is a factor of the power, the power goes to zero as the bike speed goes to
zero, meaning that you always have enough power to advance if you go slowly enough.

Climbing a hill complicates the rule with an additional linear term, bike speed remains a factor,
and you can thus climb any hill in addition, if you go slowly enough. In fact I recommend it
as a way to learn to climb hills - the trick isn't conditioning but pacing. When you get good,
you can pace to go up a hill at the fastest non-tiring speed, rather than dawdling with
unnecessary slowness or burning out with excessive speed. Conditioning is nice, but won't
ever replace that talent.
--
Ron Hardin
[email hidden]

On the internet, nobody knows you're a jerk.

Ron Hardin said:

The first mile an hour of headwind on the flat takes away 2/3 miles an hour of speed. Subsequent
mph's of wind take away less. The constraint is rider power, force times bike speed,
which on the flat is airspeed times bikespeed. Airspeed is bike speed plus headwind speed.

airspeed squared time bike speed.

Bad editing.

When you go twice as fast, you hit twice as much air twice as hard, = four times the force.

Bike speed times that gives power.
--
Ron Hardin
[email hidden]

On the internet, nobody knows you're a jerk.

On 20 Sep 2004 10:43:34 -0700, [email hidden] (wle)

Quoted message said:

1. assume a person can maintain 20mph on a flat road.

if there is a steady headwind of 20mph, does that totally stop him?

2. now assume there is a hill which a person can only manage to go 5 mph on.

would a 5 mph headwind stop that person?

i would guess the answer to 1 is yes and the answer to 2 is no, but

a. that doesn;t make sense and

b. is that right, and why or why not?

wle.

Dear WLE,

In both cases, the rider slows down, but does not stop:

http://www.kreuzotter.de/english/espeed.htm

Plugging 232 watts and hands on the tops into this bicycle
speed calculator predicts 20.0 mph on windless level ground.

Now add a 20 mph headwind. The speed drops to 9.9 mph with
an apparent headwind of 29.9 mph.

For the hill, plug in the same 232 watts and an 11.92% grade
and you should see 5 mph.

Now add the 20 mph headwind on the hill, and the speed drops
to 3.9 mph.

The calculator is hardly perfect (crank the wind up to 200
mph and you'll see the idealized rider stubbornly inching
forward instead of being blown into the air), but it's a
roughly accurate predictor for reasonable speeds and
ordinary winds. (To be fair to the calculator's programmer,
this is a beastly problem.)

The calculator's prediction also squares with our
experience--most of us have cursed a 20 mph headwind on
level ground, but we weren't stopped dead in our tracks. We
just slow down. (We get more tired because we have to pedal
at the normal level of output for twice as long to cover the
same distance.)

Why?

The main reason is apparent wind. Our rider putting out 232
watts has his effort matched by rolling resistance and wind
drag when he reaches 20 mph--which means that he experiences
an apparent wind of 20 mph.

If we add a 20 mph headwind, then the rider faces an
apparent 40 mph wind.

Since he puts out only 232 watts, he must slow down until
the drag from the apparent wind (plus the base rolling
resistance, which also declines as ground speed drops) again
matches his output of 232 watts at around 9.9 mph ground
speed and 29.9 mph apparent wind speed.

In very rough terms, forces matched when the ground speed
was cut 50% and the apparent wind speed was increased 50%.
Wind drag is actually not linear with speed, so this is a
bit misleading, but the general principle applies.

Carl Fogel

On Mon, 20 Sep 2004 13:08:36 -0600, [email hidden]

Quoted message said:

On 20 Sep 2004 10:43:34 -0700, [email hidden] (wle)

Quoted message said:

1. assume a person can maintain 20mph on a flat road.

if there is a steady headwind of 20mph, does that totally stop him?

2. now assume there is a hill which a person can only manage to go 5 mph on.

would a 5 mph headwind stop that person?

i would guess the answer to 1 is yes and the answer to 2 is no, but

a. that doesn;t make sense and

b. is that right, and why or why not?

wle.

Dear WLE,

In both cases, the rider slows down, but does not stop:

http://www.kreuzotter.de/english/espeed.htm

Plugging 232 watts and hands on the tops into this bicycle
speed calculator predicts 20.0 mph on windless level ground.

Now add a 20 mph headwind. The speed drops to 9.9 mph with
an apparent headwind of 29.9 mph.

For the hill, plug in the same 232 watts and an 11.92% grade
and you should see 5 mph.

Now add the 20 mph headwind on the hill, and the speed drops
to 3.9 mph.

The calculator is hardly perfect (crank the wind up to 200
mph and you'll see the idealized rider stubbornly inching
forward instead of being blown into the air), but it's a
roughly accurate predictor for reasonable speeds and
ordinary winds. (To be fair to the calculator's programmer,
this is a beastly problem.)

The calculator's prediction also squares with our
experience--most of us have cursed a 20 mph headwind on
level ground, but we weren't stopped dead in our tracks. We
just slow down. (We get more tired because we have to pedal
at the normal level of output for twice as long to cover the
same distance.)

Why?

The main reason is apparent wind. Our rider putting out 232
watts has his effort matched by rolling resistance and wind
drag when he reaches 20 mph--which means that he experiences
an apparent wind of 20 mph.

If we add a 20 mph headwind, then the rider faces an
apparent 40 mph wind.

Since he puts out only 232 watts, he must slow down until
the drag from the apparent wind (plus the base rolling
resistance, which also declines as ground speed drops) again
matches his output of 232 watts at around 9.9 mph ground
speed and 29.9 mph apparent wind speed.

In very rough terms, forces matched when the ground speed
was cut 50% and the apparent wind speed was increased 50%.
Wind drag is actually not linear with speed, so this is a
bit misleading, but the general principle applies.

Carl Fogel

And Peter's explanation is much shorter, clearer, and better
than mine.

Carl Fogel

(wle) said:

1. assume a person can maintain 20mph on a flat road.

if there is a steady headwind of 20mph, does that totally stop him?

No, but it will make getting somewhere mightly unpleasant and slow.
The energy consumed in going 20mph is not entirely due to air drag,
remember; at lower speeds, tire rolling resistance and driveline
losses may dominate; at a certain speed, drag becomes the principal
force. So, take the energy required to overcome the drag due to a
20mph headwind, subtract that from the energy consumed in going 20mph,
and you have a difference which would still be able to produce forward
motion.

Quoted message said:

2. now assume there is a hill which a person can only manage to go 5 mph on.

would a 5 mph headwind stop that person?

No, and unless the rider's frontal area is large, the speed reduction
would be minimal; the reason should be obvious. The speed is being
limited not by air drag but by gravity and the limitations of the
powerplant. The powerplant is assumed to be operating at full output
to maintain 5mph. At 5mph, the wind drag force is very small. Under
the stated conditions, unless the rider has a low power output (and
this isn't much of a hill), the 5mph forward velocity replects the
suplus capacity of the powerplant, and it is almost dead certain that
this power output will be far greater than the drag from a 5mph
headwind.

--
Typoes are a feature, not a bug.
Some gardening required to reply via email.
Words processed in a facility that contains nuts.

S o r n i said:

I thought my ice cream remarks implied all this.

Variables which are not explicitly declared may have random values at
instantiation.
--
Typoes are a feature, not a bug.
Some gardening required to reply via email.
Words processed in a facility that contains nuts.

Werehatrack said:

S o r n i said:

I thought my ice cream remarks implied all this.

Variables which are not explicitly declared may have random values at
instantiation.

So, like, chocolate, right?

Bill "just trying to clarify" S.

Ron Hardin said:

Ron Hardin said:

The first mile an hour of headwind on the flat takes away 2/3 miles an hour of speed. Subsequent
mph's of wind take away less. The constraint is rider power, force times bike speed,
which on the flat is airspeed times bikespeed. Airspeed is bike speed plus headwind speed.

airspeed squared time bike speed.

Bad editing.

When you go twice as fast, you hit twice as much air twice as hard, = four times the force.

Bike speed times that gives power.

The 2/3 mph loss for 1 mph of headwind rule-of-thumb just isn't
correct. A fairly accurate rule-of thumb is that you lose or gain
half the windspeed for light pure headwinds or tailwinds. For much
better accuracy, use something like the Analytic Cycling website.

Dave Lehnen

S o r n i said...

Quoted message said:

Bill "30 seconds of life gone" S.

Now with your post, that makes 42...oops, there goes 11 more for posting
this...

In article <[email hidden]>, [email hidden]
says...

Quoted message said:

Quoted message said:

In very rough terms, forces matched when the ground speed

was cut 50% and the apparent wind speed was increased 50%.
Wind drag is actually not linear with speed, so this is a
bit misleading, but the general principle applies.

Carl Fogel

Theory give a change in speed of about one half the wind speed for light to
moderate winds and a bicycle speed of 20mph. Here the report by someone who has
actually measured it: http://mikebentley.com/bike/harry/19-21.htm

His formula is speed = 13 + tailwind/3 where a headwind is a negative tailwind.
The whole travelog is worth reading.

Dave Lehnen said:

Quoted message said:

Bike speed times that gives power.

The 2/3 mph loss for 1 mph of headwind rule-of-thumb just isn't
correct. A fairly accurate rule-of thumb is that you lose or gain
half the windspeed for light pure headwinds or tailwinds. For much
better accuracy, use something like the Analytic Cycling website.

w=headwind speed v=bicycle speed P=power

v . (w+v)^2 = P

differentials, power constant

dv . (w+v)^2 + 2 . v . (v + w) . (dw + dv) = 0

set w = 0 to evaluate differential at small wind

dv . v^2 + 2 . v . (v) . (dw + dv) = 0

divide by v^2

dv + 2.dv + 2.dw = 0

collect

3 . dv + 2 . dw = 0

ratio

dv/dw = -2 / 3

--
Ron Hardin
[email hidden]

On the internet, nobody knows you're a jerk.

Werehatrack said:

(wle) said:

1. assume a person can maintain 20mph on a flat road.
if there is a steady headwind of 20mph, does that totally stop him?

No, but it will make getting somewhere mightly unpleasant and slow.
The energy consumed in going 20mph is not entirely due to air drag,
remember; at lower speeds, tire rolling resistance and driveline
losses may dominate;

No, this is a distraction from the real reason.

The real reason is that, while we generally think of power to overcome air
resistance as proportional to the cube of "speed", in fact it is
proportional to the square of airspeed times groundspeed, because the work
[1] is done by applying a force against the ground.

Hence clearly if groundspeed were zero, no power would be needed - the
OP would have done well to ask what would happen in a 30mph wind, why the
bicyclist is not helplessly blown backwards at 10mph. If the power output
in the flat calm is 8000k (8,000 is 20^3 and k is some constant) then into
the 20mph wind at speed v the power output is;
(20+v)(20+v)vk
and there exists some nonzero v for which this is an equal power output.

[This assumes that non-aero drag is trivial - something much closer to the
truth at 20mph or with like effort into a 20mph headwind.]

Quoted message said:

Quoted message said:

2. now assume there is a hill which a person can only manage to go 5 mph
on. would a 5 mph headwind stop that person?

Here of course the non-aero drag is highly significant, and the small
increase in aero drag only necessitates a small reduction in speed.

[1] I am using the term "work" in the formal applied mathematics sense of
force times distance.
--
David Damerell <[email hidden]> Distortion Field!

Ron Hardin said:

Dave Lehnen said:

Quoted message said:

Bike speed times that gives power.

The 2/3 mph loss for 1 mph of headwind rule-of-thumb just isn't
correct. A fairly accurate rule-of thumb is that you lose or gain
half the windspeed for light pure headwinds or tailwinds. For much
better accuracy, use something like the Analytic Cycling website.

w=headwind speed v=bicycle speed P=power

v . (w+v)^2 = P

differentials, power constant

dv . (w+v)^2 + 2 . v . (v + w) . (dw + dv) = 0

set w = 0 to evaluate differential at small wind

dv . v^2 + 2 . v . (v) . (dw + dv) = 0

divide by v^2

dv + 2.dv + 2.dw = 0

collect

3 . dv + 2 . dw = 0

ratio

dv/dw = -2 / 3

My mistake. That's what I get for using my faulty memory and
not doing the math myself to check. FWIW, you lose more than
half the windspeed until the wind gets up to 1.236 times the
calm air bike speed, ignoring rolling friction and other smaller
power losses. The 1.236 is sqrt(5) - 1, or twice the golden
section ratio.

Dave Lehnen