CdA guestimates sought

Hi All,

Today I did some time trialling and a kind soul sent me pics.

Anyone care to hazard a guess as to what a reasonable CdA would be for
me as shown:

http://arbitrary.org/tt1.jpg
http://arbitrary.org/tt2.jpg

Joseph

On May 7, 4:00 pm, "
wrote:
Hi All,

Today I did some time trialling and a kind soul sent me pics.

Anyone care to hazard a guess as to what a reasonable CdA would be for
me as shown:

http://arbitrary.org/tt1.jpghttp://a...ry.org/tt2.jpg

Joseph

I see the spaghetti has slithered away from the shrubs in fear when it
saw your menacing bulk hurtling down the road!!!

Looks pretty good to this squid. Pretty flat back. Nice aero stubble.
Now you just need a lycra bota bag of vino for the finish.

wrote:

Hi All,

Today I did some time trialling and a kind soul sent me pics.

Anyone care to hazard a guess as to what a reasonable CdA would be for
me as shown:

http://arbitrary.org/tt1.jpg
http://arbitrary.org/tt2.jpg

Joseph

Your frontal area A can be approximated as
A = (0.85) x H x W (43)
where A is the area in square feet, H is the height of your frontal
profile on the bike and W is the width or your frontal profile on the
bike. You might split the frontal profile into a) the wheel(s) and
mechanics below the pedal which can be precisely calculated, b) your
body from pedal to the top of your shoulders which can be factored at
0.85 as above, and c) your head which can be factored at 0.70.

The aerodynamic coefficient Cd (or Cx for Europeans) is a
dimensionless coefficient. Perhaps someone can guess it for a
bicyclist closely enough by experience. I can't, so let's approach it
from the angle of better-known facts: you were trialling, so you know
your speed. And, more generally, we already know how much power a
cyclist generates.

Let's just say for the moment that your bike weighs nothing, so that
all your exertion is used to overcome aerodynamic drag. Then
Cd = (BHP x146,600)/(Av^3) (139)
where BHP is brake horsepower and v is speed in mph. One Watt is
0.001341 BHP. If you want to work in Watts the formula becomes:
Cd = (W x 197)/(Av^3)
where W is Watts. Fill in some numbers, then find CdA as the product
of Cd and A. You might want to refine this Cd number by first
subtracting from the power you expend whatever is required to overcome
friction the bike and your mass creates between tyre and road.

There you go. CdA a lot more closely calculated from known data than
netstimates by Uncle Tom Cobley & All. Just for the record, you look
so pacey, I guess your CdA is 0.62, roughly the same as a Caterham
(nee Lotus) Seven (ask Tom if you don't know what it is -- he's bound
to lust after one) and only about fifty percent worse than a blunt VW
Microbus which has a CdA of 0.42.

The righthand numbers in brackets are the page numbers on which the
formulae are found in:
Andre Jute: Designing and Building Special Cars, 1985, B T Batsford,
London
This book also describes a coastdown method of determining Cd; if
someone were to offer me a generally accepted coefficient of friction
for bicycles (it's about 0.014 for a car with four radial ply tyres),
I'd be happy to adapt the formula for bicycle use.

HTH.

E&OE.

Andre Jute
Visit Andre's books at: http://members.lycos.co.uk/fiultra/THE%20WRITER'S%20HOUSE.html

On May 7, 2:00*pm, "
wrote:
Hi All,

Today I did some time trialling and a kind soul sent me pics.

Anyone care to hazard a guess as to what a reasonable CdA would be for
me as shown:

http://arbitrary.org/tt1.jpghttp://a...ry.org/tt2.jpg

Joseph

No.

There's no point in trying this unless you do it right,
and those pictures won't let you do that. Too small,
not head-on, no reference length next to you.

This is like riding up hills for power estimates; you
don't really need fancy equipment, but you need to
invest time and thought. In this case, that means
setting up your TT bike in a trainer, a plain backdrop,
a length reference, and a camera to photograph
you that is far enough away so perspective isn't
a problem.

Kraig Willett (onetime RBR/RBT poster) wrote some
articles about doing this for the Prime Alliance guys.
The Prime Alliance article was on the Bike.com website
and is probably gone unless the Internet Archive
has it. However, there is at least one article on
Kraig's website that should give some ideas:
http://www.biketechreview.com/aerodynamics/kw.htm
also
http://www.biketechreview.com/aerody...fat_helmet.htm
has another picture of the setup.
Google may turn up more ideas.

This only measures A, not C_D. Just assume it's
about 0.5. You can't get much more precise than that
without a wind tunnel.

Ben

On May 8, 1:30*am, Andre Jute wrote:
wrote:
Hi All,

Today I did some time trialling and a kind soul sent me pics.

Anyone care to hazard a guess as to what a reasonable CdA would be for
me as shown:

http://arbitrary.org/tt1.jpg
http://arbitrary.org/tt2.jpg

Joseph

Your frontal area A can be approximated as
A = (0.85) x H x W * * * * * * * * * * * * * * * * * * (43)
where A is the area in square feet, H is the height of your frontal
profile on the bike and W is the width or your frontal profile on the
bike. You might split the frontal profile into a) the wheel(s) and
mechanics below the pedal which can be precisely calculated, b) your
body from pedal to the top of your shoulders which can be factored at
0.85 as above, and c) your head which can be factored at 0.70.

The aerodynamic coefficient Cd (or Cx for Europeans) is a
dimensionless coefficient. Perhaps someone can guess it for a
bicyclist closely enough by experience. I can't, so let's approach it
from the angle of better-known facts: you were trialling, so you know
your speed. And, more generally, we already know how much power a
cyclist generates.

Let's just say for the moment that your bike weighs nothing, so that
all your exertion is used to overcome aerodynamic drag. Then
Cd = (BHP x146,600)/(Av^3) * * * * * * * * * * * * * * * * * * (139)
where BHP is brake horsepower and v is speed in mph. One Watt is
0.001341 BHP. If you want to work in Watts the formula becomes:
Cd = (W x 197)/(Av^3)
where W is Watts. Fill in some numbers, then find CdA as the product
of Cd and A. You might want to refine this Cd number by first
subtracting from the power you expend whatever is required to overcome
friction the bike and your mass creates between tyre and road.

There you go. CdA a lot more closely calculated from known data than
netstimates by Uncle Tom Cobley & All. Just for the record, you look
so pacey, I guess your CdA is 0.62, roughly the same as a Caterham
(nee Lotus) Seven (ask Tom if you don't know what it is -- he's bound
to lust after one) and only about fifty percent worse than a blunt VW
Microbus which has a CdA of 0.42.

The righthand numbers in brackets are the page numbers on which the
formulae are found in:
Andre Jute: Designing and Building Special Cars, 1985, B T Batsford,
London
This book also describes a coastdown method of determining Cd; if
someone were to offer me a generally accepted coefficient of friction
for bicycles (it's about 0.014 for a car with four radial ply tyres),
I'd be happy to adapt the formula for bicycle use.

HTH.

E&OE.

Andre Jute
Visit Andre's books at: * *http://members.lycos.co.uk/fiultra/THE%20WRITER'S%20HOUSE.html

Excellent. One problem is that power is an unknown variable in this
case. As for your WAG of 0.62, I think that is way off. Closer to and
maybe even under 0.3.

An adapted coast-down formula would be great. Friction for racing bike
tires is probably reasonably estimated as 0.006.

Joseph

On May 8, 8:04*am, "
wrote:
On May 8, 1:30*am, Andre Jute wrote:



wrote:
Hi All,

Today I did some time trialling and a kind soul sent me pics.

Anyone care to hazard a guess as to what a reasonable CdA would be for
me as shown:

http://arbitrary.org/tt1.jpg
http://arbitrary.org/tt2.jpg

Joseph

Your frontal area A can be approximated as
A = (0.85) x H x W * * * * * * * * * * * * * * * * * * (43)
where A is the area in square feet, H is the height of your frontal
profile on the bike and W is the width or your frontal profile on the
bike. You might split the frontal profile into a) the wheel(s) and
mechanics below the pedal which can be precisely calculated, b) your
body from pedal to the top of your shoulders which can be factored at
0.85 as above, and c) your head which can be factored at 0.70.

The aerodynamic coefficient Cd (or Cx for Europeans) is a
dimensionless coefficient. Perhaps someone can guess it for a
bicyclist closely enough by experience. I can't, so let's approach it
from the angle of better-known facts: you were trialling, so you know
your speed. And, more generally, we already know how much power a
cyclist generates.

Let's just say for the moment that your bike weighs nothing, so that
all your exertion is used to overcome aerodynamic drag. Then
Cd = (BHP x146,600)/(Av^3) * * * * * * * * * * * * * * * * * * (139)
where BHP is brake horsepower and v is speed in mph. One Watt is
0.001341 BHP. If you want to work in Watts the formula becomes:
Cd = (W x 197)/(Av^3)
where W is Watts. Fill in some numbers, then find CdA as the product
of Cd and A. You might want to refine this Cd number by first
subtracting from the power you expend whatever is required to overcome
friction the bike and your mass creates between tyre and road.

There you go. CdA a lot more closely calculated from known data than
netstimates by Uncle Tom Cobley & All. Just for the record, you look
so pacey, I guess your CdA is 0.62, roughly the same as a Caterham
(nee Lotus) Seven (ask Tom if you don't know what it is -- he's bound
to lust after one) and only about fifty percent worse than a blunt VW
Microbus which has a CdA of 0.42.

Of course I'm speaking in the above par of Cd, not of CdA; everyone
has so far understood what I intended rather than what taking what I
wrote at face value! That speaks well of their intelligence and
general knowledge of quite abstruse matters.

The righthand numbers in brackets are the page numbers on which the
formulae are found in:
Andre Jute: Designing and Building Special Cars, 1985, B T Batsford,
London
This book also describes a coastdown method of determining Cd; if
someone were to offer me a generally accepted coefficient of friction
for bicycles (it's about 0.014 for a car with four radial ply tyres),
I'd be happy to adapt the formula for bicycle use.

HTH.

E&OE.

Andre Jute
Visit Andre's books at: * *http://members.lycos.co.uk/fiultra/THE%20WRITER'S%20HOUSE.html

Excellent. One problem is that power is an unknown variable in this
case.

That's not an insuperable difficulty. We can approximate power pretty
closely with simple procedures and a bit of applied physics.

As for your WAG of 0.62, I think that is way off. Closer to and
maybe even under 0.3.

It wasn't even a wild ass guess, Joseph, more like a joke. Those
photographs aren't much chop for calculating the frontal area and you
didn't give us any further information. My assumptions, in the absence
of evidence, are that you and the bike would present a frontal aspect
of 8 sq ft, that you were riding at 20mph (clearly a very conservative
estimate!), and that you were expending the 200W often bandied around
on RBT. We need much more information, which means you will have to do
much more work, some of it tedious. (Jobst didn't get to be a great
engineer and cyclist for sitting on his skinny ass! It's your turn to
put in the hard graft.) The advantage of doing it this way is that the
uncertainties and estimates and guestimates are progressively reduced
until the final result is in every particular defensible as physics.

An adapted coast-down formula would be great.

I'm working on the formulation to determine power and a methodology
with just a bike and a road available to divide its dissipation
between friction and aerodynamic drag. (More precisely, I worked it
all out long since on cars I built and put it in my book; what I'm
working on now is adapting the method from cars to bikes.) Any chance
of you borrowing an accelerometer with divisions smaller than 0.1g up
to about 0.3g i.e. not the common aircraft type which is too coarse?
It would speed up your work immensely. Alternatively, a helmet camera
would already be boon to record the clock on the run -- most of the
wasted time is in stopping to take notes and starting up again; even a
pocket tape recorder would help.

Friction for racing bike
tires is probably reasonably estimated as 0.006.

Thanks for the constant; anyone wants to kibbitz it, speak up now or
it will go down in street lore as the gospel.

Andre Jute
http://members.lycos.co.uk/fiultra/B...20CYCLING.html

On May 8, 1:57*pm, Andre Jute wrote:
On May 8, 8:04*am, "



wrote:
On May 8, 1:30*am, Andre Jute wrote:

wrote:
Hi All,

Today I did some time trialling and a kind soul sent me pics.

Anyone care to hazard a guess as to what a reasonable CdA would be for
me as shown:

http://arbitrary.org/tt1.jpg
http://arbitrary.org/tt2.jpg

Joseph

Your frontal area A can be approximated as
A = (0.85) x H x W * * * * * * * * * * * * * * * * * * (43)
where A is the area in square feet, H is the height of your frontal
profile on the bike and W is the width or your frontal profile on the
bike. You might split the frontal profile into a) the wheel(s) and
mechanics below the pedal which can be precisely calculated, b) your
body from pedal to the top of your shoulders which can be factored at
0.85 as above, and c) your head which can be factored at 0.70.

The aerodynamic coefficient Cd (or Cx for Europeans) is a
dimensionless coefficient. Perhaps someone can guess it for a
bicyclist closely enough by experience. I can't, so let's approach it
from the angle of better-known facts: you were trialling, so you know
your speed. And, more generally, we already know how much power a
cyclist generates.

Let's just say for the moment that your bike weighs nothing, so that
all your exertion is used to overcome aerodynamic drag. Then
Cd = (BHP x146,600)/(Av^3) * * * * * * * * * * * * * * * * * * (139)
where BHP is brake horsepower and v is speed in mph. One Watt is
0.001341 BHP. If you want to work in Watts the formula becomes:
Cd = (W x 197)/(Av^3)
where W is Watts. Fill in some numbers, then find CdA as the product
of Cd and A. You might want to refine this Cd number by first
subtracting from the power you expend whatever is required to overcome
friction the bike and your mass creates between tyre and road.

There you go. CdA a lot more closely calculated from known data than
netstimates by Uncle Tom Cobley & All. Just for the record, you look
so pacey, I guess your CdA is 0.62, roughly the same as a Caterham
(nee Lotus) Seven (ask Tom if you don't know what it is -- he's bound
to lust after one) and only about fifty percent worse than a blunt VW
Microbus which has a CdA of 0.42.

Of course I'm speaking in the above par of Cd, not of CdA; everyone
has so far understood what I intended rather than what taking what I
wrote at face value! That speaks well of their intelligence and
general knowledge of quite abstruse matters.



The righthand numbers in brackets are the page numbers on which the
formulae are found in:
Andre Jute: Designing and Building Special Cars, 1985, B T Batsford,
London
This book also describes a coastdown method of determining Cd; if
someone were to offer me a generally accepted coefficient of friction
for bicycles (it's about 0.014 for a car with four radial ply tyres),
I'd be happy to adapt the formula for bicycle use.

HTH.

E&OE.

Andre Jute
Visit Andre's books at: * *http://members.lycos.co.uk/fiultra/THE%20WRITER'S%20HOUSE.html

Excellent. One problem is that power is an unknown variable in this
case.

That's not an insuperable difficulty. We can approximate power pretty
closely with simple procedures and a bit of applied physics.

Power calcs on the way up the hill, CdA on the way down. Sounds like
fun.


As for your WAG of 0.62, I think that is way off. Closer to and
maybe even under 0.3.

It wasn't even a wild ass guess, Joseph, more like a joke. Those
photographs aren't much chop for calculating the frontal area and you
didn't give us any further information. My assumptions, in the absence
of evidence, are that you and the bike would present a frontal aspect
of 8 sq ft, that you were riding at 20mph (clearly a very conservative
estimate!), and that you were expending the 200W often bandied around
on RBT. We need much more information, which means you will have to do
much more work, some of it tedious. (Jobst didn't get to be a great
engineer and cyclist for sitting on his skinny ass! It's your turn to
put in the hard graft.) The advantage of doing it this way is that the
uncertainties and estimates and guestimates are progressively reduced
until the final result is in every particular defensible as physics.

Ah! Jokes. I've heard of those.

As for using an off the shelf power estimate, that is the issue. There
is a wide power range found in cyclists, and what often determines who
is faster. Which is of course important in a race!

An adapted coast-down formula would be great.

I'm working on the formulation to determine power and a methodology
with just a bike and a road available to divide its dissipation
between friction and aerodynamic drag. (More precisely, I worked it
all out long since on cars I built and put it in my book; what I'm
working on now is adapting the method from cars to bikes.) Any chance
of you borrowing an accelerometer with divisions smaller than 0.1g up
to about 0.3g i.e. not the common aircraft type which is too coarse?
It would speed up your work immensely. Alternatively, a helmet camera
would already be boon to record the clock on the run -- most of the
wasted time is in stopping to take notes and starting up again; even a
pocket tape recorder would help.

I've got a helmet cam, but I would stow it someplace out of the way.
Perhaps on the seatpost looking down at a watch and a finish line
painted on the road?

Friction for racing bike
tires is probably reasonably estimated as 0.006.

Thanks for the constant; anyone wants to kibbitz it, speak up now or
it will go down in street lore as the gospel.

Here is a compilation or 2 tests for Crr for various current racing
tires:

http://tour-de-france.velonews.com/article/12493

Those tests are on a smooth drum, which sorts the tires relative to
each other, but underestimates the resistance of a real road.

Joseph

On May 7, 3:00*pm, "
wrote:
Hi All,

Today I did some time trialling and a kind soul sent me pics.

Anyone care to hazard a guess as to what a reasonable CdA would be for
me as shown:

http://arbitrary.org/tt1.jpghttp://a...ry.org/tt2.jpg

Joseph

If ya ever want to get a really good CdA, ya better shave.

On May 8, 2:34*pm, "Qui si parla Campagnolo-www.vecchios.com"
wrote:
On May 7, 3:00*pm, "

wrote:
Hi All,

Today I did some time trialling and a kind soul sent me pics.

Anyone care to hazard a guess as to what a reasonable CdA would be for
me as shown:

http://arbitrary.org/tt1.jpghttp://a...ry.org/tt2.jpg

Joseph

If ya ever want to get a really good CdA, ya better shave.

Shaving is saved for when I can finally hang on and not get dropped in
a race. I feel sort of foolish otherwise.

Joseph

On May 7, 2:00 pm, "
wrote:
Anyone care to hazard a guess as to what a reasonable CdA would be for
me as shown:

Nice photos. Useless for estimating drag area.