How much more work in the rain?

So as I'm riding in the rain with my son today, trying to avoid the
spray off his rear wheel (no fender), and then noticing how much water
is coming off my front fender, I start thinking about the obvious.

All that water, being transported *up* from the ground. That takes work.
The question is, how much water is being moved up, and how high? I can't
envision any sort of manner in which that effort is negated by an
opposing force.

My guess is that somebody has already done the math. Anyone here who's
already done the work?

Thanks-

--Mike-- Chain Reaction Bicycles
www.ChainReactionBicycles.com

On Sun, 17 Jan 2010 22:35:25 -0800, "Mike Jacoubowsky"
wrote:

So as I'm riding in the rain with my son today, trying to avoid the
spray off his rear wheel (no fender), and then noticing how much water
is coming off my front fender, I start thinking about the obvious.

All that water, being transported *up* from the ground. That takes work.
The question is, how much water is being moved up, and how high? I can't
envision any sort of manner in which that effort is negated by an
opposing force.

My guess is that somebody has already done the math. Anyone here who's
already done the work?

Thanks-

--Mike-- Chain Reaction Bicycles
www.ChainReactionBicycles.com

Dear Mike,

The spray is no more than the thin film of water that clings to a
smooth rubber tire and is then flung off as the tire accelerates.

When you're going 20 mph forward, the contact patch is doing roughly 0
mph at all times, down in the water.

After the wetted contact patch rotates a quarter turn (the back of the
tire, level with the axle), it has accelerated to 20 mph straight up.

When the water-covered contact patch reaches the top of the tire
(right over the axle), it's doing 40 mph forward, right into the
wind).

At 20 mph, a ~2100 mm circumference 700c tire spins at about 255 RPM.

Accelerating from 0 mph at the contact patch to 40 mph at the top of
the tire in about 1/500th of a second is why the thin film of water is
flung off the tire and blown back into your face as spray.

But a little spray goes a long way when it's in your face, so things
seem much worse than they really are. Very little water is actually
raised, no more than wets a tire.

Think about the effort required by a spray bottle, which is worked
with one finger, as opposed to two legs.

That's why you won't notice much of a speed loss on a wet road.

Another way to put it in perspective is to ask how much the rain
itself slows you down--all those drops of water hitting your whole
body and bicycle literally don't have much of an impact.

Of course, when the tire has to push significant amounts of water out
of the way to the sides (standing water, as opposed to a wet road),
then things do indeed become noticeable.

Cheers,

Carl Fogel

On Mon, 18 Jan 2010 00:18:40 -0700, wrote:

On Sun, 17 Jan 2010 22:35:25 -0800, "Mike Jacoubowsky"
wrote:

So as I'm riding in the rain with my son today, trying to avoid the
spray off his rear wheel (no fender), and then noticing how much water
is coming off my front fender, I start thinking about the obvious.

All that water, being transported *up* from the ground. That takes work.
The question is, how much water is being moved up, and how high? I can't
envision any sort of manner in which that effort is negated by an
opposing force.

My guess is that somebody has already done the math. Anyone here who's
already done the work?

Thanks-

--Mike-- Chain Reaction Bicycles
www.ChainReactionBicycles.com

Dear Mike,

The spray is no more than the thin film of water that clings to a
smooth rubber tire and is then flung off as the tire accelerates.

When you're going 20 mph forward, the contact patch is doing roughly 0
mph at all times, down in the water.

After the wetted contact patch rotates a quarter turn (the back of the
tire, level with the axle), it has accelerated to 20 mph straight up.

When the water-covered contact patch reaches the top of the tire
(right over the axle), it's doing 40 mph forward, right into the
wind).

At 20 mph, a ~2100 mm circumference 700c tire spins at about 255 RPM.

Accelerating from 0 mph at the contact patch to 40 mph at the top of
the tire in about 1/500th of a second is why the thin film of water is
flung off the tire and blown back into your face as spray.

But a little spray goes a long way when it's in your face, so things
seem much worse than they really are. Very little water is actually
raised, no more than wets a tire.

Think about the effort required by a spray bottle, which is worked
with one finger, as opposed to two legs.

That's why you won't notice much of a speed loss on a wet road.

Another way to put it in perspective is to ask how much the rain
itself slows you down--all those drops of water hitting your whole
body and bicycle literally don't have much of an impact.

Of course, when the tire has to push significant amounts of water out
of the way to the sides (standing water, as opposed to a wet road),
then things do indeed become noticeable.

Cheers,

Carl Fogel

Aargh! When will I learn to show enough of my work to avoid absurd
errors?

That's 255 tire RPM at 20 mph--

Per minute, not per second.

So about 4.25 tire spins per second (255/60), or 8.5 half-spins per
second.

About 1/8th of a second to accelerate from 0 mph to 40 mph, not
1/500th.

Cheers,

Carl Fogel

On Sun, 17 Jan 2010 22:35:25 -0800, Mike Jacoubowsky wrote:

So as I'm riding in the rain with my son today, trying to avoid the
spray off his rear wheel (no fender), and then noticing how much water
is coming off my front fender, I start thinking about the obvious.

All that water, being transported *up* from the ground. That takes work.
The question is, how much water is being moved up, and how high? I can't
envision any sort of manner in which that effort is negated by an
opposing force.


Humid air is less dense, so you may gain on the swing what you loose on the
roundabout.

On Jan 18, 6:35 am, "Mike Jacoubowsky"
wrote:
So as I'm riding in the rain with my son today, trying to avoid the
spray off his rear wheel (no fender), and then noticing how much water
is coming off my front fender, I start thinking about the obvious.

All that water, being transported *up* from the ground. That takes work.
The question is, how much water is being moved up, and how high? I can't
envision any sort of manner in which that effort is negated by an
opposing force.

My guess is that somebody has already done the math. Anyone here who's
already done the work?

Thanks-

--Mike-- Chain Reaction Bicycleswww.ChainReactionBicycles.com

Not much work, actually. Assuming the absence of confuddling factors
like knobbly tires with large-scooplike undercuts: That's a *very*
thin film of water turned into spray. Compare riding on a wet road,
with an apparent large amount of spray, to riding through standing
water at least rim-deep, where the displacement slows you and a bunch
of much heavier drops are not propelled nearly as far or high. In the
latter case, appreciable work is being done; in the wet road case i'd
hate to have to quantify a number to a critical audience.

In general, water is merely a thicker gas, and aerodynamic principles
can be applied. I'm not so sure that there would be any advantage even
in competitive circumstances to redesigning any bicycle component to
meet a wet road better. If there is anything to be gained, it would,
aero and naval architechture principles be along the lines of keeping
the thin layer of water attached longer rather than casting it off; if
this sounds counter-intutive, consider that clinging water prevents
more water being picked up and cast off, which is where the work is
performed, so that on average a very thin film of water clinging to
the tire for as long as possible is the less power-consuming option.
As I say, this is genuinely hairsplitting stuff.

Andre Jute
"The brain of an engineer is a delicate instrument which must be
protected against the unevenness of the ground." -- Wifredo-Pelayo
Ricart Medina

On Jan 18, 7:18 am, wrote:
On Sun, 17 Jan 2010 22:35:25 -0800, "Mike Jacoubowsky"





wrote:
So as I'm riding in the rain with my son today, trying to avoid the
spray off his rear wheel (no fender), and then noticing how much water
is coming off my front fender, I start thinking about the obvious.

All that water, being transported *up* from the ground. That takes work.
The question is, how much water is being moved up, and how high? I can't
envision any sort of manner in which that effort is negated by an
opposing force.

My guess is that somebody has already done the math. Anyone here who's
already done the work?

Thanks-

--Mike-- Chain Reaction Bicycles
www.ChainReactionBicycles.com

Dear Mike,

The spray is no more than the thin film of water that clings to a
smooth rubber tire and is then flung off as the tire accelerates.

When you're going 20 mph forward, the contact patch is doing roughly 0
mph at all times, down in the water.

After the wetted contact patch rotates a quarter turn (the back of the
tire, level with the axle), it has accelerated to 20 mph straight up.

When the water-covered contact patch reaches the top of the tire
(right over the axle), it's doing 40 mph forward, right into the
wind).

At 20 mph, a ~2100 mm circumference 700c tire spins at about 255 RPM.

Accelerating from 0 mph at the contact patch to 40 mph at the top of
the tire in about 1/500th of a second is why the thin film of water is
flung off the tire and blown back into your face as spray.

But a little spray goes a long way when it's in your face, so things
seem much worse than they really are. Very little water is actually
raised, no more than wets a tire.

Think about the effort required by a spray bottle, which is worked
with one finger, as opposed to two legs.

That's why you won't notice much of a speed loss on a wet road.

Another way to put it in perspective is to ask how much the rain
itself slows you down--all those drops of water hitting your whole
body and bicycle literally don't have much of an impact.

Of course, when the tire has to push significant amounts of water out
of the way to the sides (standing water, as opposed to a wet road),
then things do indeed become noticeable.

Cheers,

Carl Fogel

Dear Carl:

You must be on speed. Your mental arithmetic is out by a multiple of
60. Write it down on paper, dear boy.

Andre Jute
It takes only a fraction of a second to put your mind in gear

On 18 Jan, 11:52, _
wrote:


Humid air is less dense, so you may gain on the swing what you loose on the
roundabout.

That'll save the plastic surgeon a bit of work. How do I get to the
swing after decking on the roundabout?

On 1/18/2010 12:35 AM, Mike Jacoubowsky wrote:
So as I'm riding in the rain with my son today, trying to avoid the
spray off his rear wheel (no fender), and then noticing how much water
is coming off my front fender, I start thinking about the obvious.

All that water, being transported *up* from the ground. That takes work.
The question is, how much water is being moved up, and how high? I can't
envision any sort of manner in which that effort is negated by an
opposing force.

My guess is that somebody has already done the math. Anyone here who's
already done the work?

Thanks-

--Mike-- Chain Reaction Bicycles
www.ChainReactionBicycles.com




Slightly off-topic....

I spoke of my tire-building efforts on a web forum, and one fellow
responded that a "better" tire would have a circumferential groove in
the center rather than any kind of center rib or tread blocks, because
(apparently he has observed that) this type of center-section on a tire
prevented water from being thrown up.

As an example, he mentioned the "Tioga 1.75/1.95 slicks". Tioga doesn't
make a "slick" that I can tell, I guessed he meant the City Slicker
tire, which does have a center groove:
http://www.amazon.com/Tioga-City-Sli.../dp/B000C15FTI

??????

I mentioned that new-fangled invention--the fender, and he wouldn't have
any part of it.

I've had all kinds of tires on bikes over the years, don't remember much
difference in what water they tossed up except that a slick throws it
all up in one central stream, where something like knobbies gives you a
gentle shower from below.
~

On 18 Jan, 14:15, DougC wrote:
On 1/18/2010 12:35 AM, Mike Jacoubowsky wrote:



So as I'm riding in the rain with my son today, trying to avoid the
spray off his rear wheel (no fender), and then noticing how much water
is coming off my front fender, I start thinking about the obvious.

All that water, being transported *up* from the ground. That takes work..
The question is, how much water is being moved up, and how high? I can't
envision any sort of manner in which that effort is negated by an
opposing force.

My guess is that somebody has already done the math. Anyone here who's
already done the work?

Thanks-

--Mike-- * * Chain Reaction Bicycles
www.ChainReactionBicycles.com

Slightly off-topic....

I spoke of my tire-building efforts on a web forum, and one fellow
responded that a "better" tire would have a circumferential groove in
the center rather than any kind of center rib or tread blocks, because
(apparently he has observed that) this type of center-section on a tire
prevented water from being thrown up.

As an example, he mentioned the "Tioga 1.75/1.95 slicks". Tioga doesn't
make a "slick" that I can tell, I guessed he meant the City Slicker
tire, which does have a center groove:http://www.amazon.com/Tioga-City-Sli.../dp/B000C15FTI

??????

I mentioned that new-fangled invention--the fender, and he wouldn't have
any part of it.

I've had all kinds of tires on bikes over the years, don't remember much
difference in what water they tossed up except that a slick throws it
all up in one central stream, where something like knobbies gives you a
gentle shower from below.
~

I did use circumferential grooved 27" tyres as a teenager with shortie
guards. Cannot remember specifically being upset by the rain, I got
wet and put up with it I suppose. By creating the edge where the
water collects as the tyre leaves the road surface there is a
reduction in the area of contact for the volume of water and the water
will leave quicker as long as the groove is wide enough.

On Jan 17, 10:35*pm, "Mike Jacoubowsky"
wrote:
So as I'm riding in the rain with my son today, trying to avoid the
spray off his rear wheel (no fender), and then noticing how much water
is coming off my front fender, I start thinking about the obvious.

All that water, being transported *up* from the ground. That takes work.
The question is, how much water is being moved up, and how high? I can't
envision any sort of manner in which that effort is negated by an
opposing force.

My guess is that somebody has already done the math. Anyone here who's
already done the work?

I would assume that there are also traction losses that have to be
thrown in. Also throw in the drag created by fenders, if you use
fenders -- both aerodynamic and due to the pumping of water through
the fender. Depending on the temperature, your body may also be
working harder to heat itself, at least on the downhills. There is
also the added losses caused by slowing greatly on corners or areas of
questionable traction and then getting back up to speed. Loss of
lubrication on your chain (yes, water is a lubricant, but on a long
ride, my chain starts to feel like it is made of walnut shells).

I raced for many years in the spring not-very-classics in Oregon, most
run in the rain, and it was a lot of just coping -- coping with not
being able to see anything because of wheel spray, hypothermia down
hills, slow corners with jumps, crashes and crash avoidance, gritty
water bottles, etc. -- Jay Beattie.