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  #161  
Old December 22nd 18, 11:28 AM posted to rec.bicycles.tech
[email protected]
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Posts: 824
Default Power on hills.

On Saturday, December 22, 2018 at 12:52:51 AM UTC+1, Andre Jute wrote:
On Friday, December 21, 2018 at 8:25:37 PM UTC, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really isn't a large loss unless you're playing for real small power such as that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ρ A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
ρ = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200 lbs is huge. KE = ½mv²


Thanks for putting the numbers to my argument, Tom.

Andre Jute
DESIGNING AND BUILDING SPECIAL CARS; Batsford, London; Bentley, Boston


I just reading an aero special article in TOUR magazin. Position on the bike can save you 54 watts or gaining 3.9 km/hr going from riding on the tops to riding in the drops. Clothes can make a difference of up to 27 Watt or gaining 2,3 km/hr in speed. An aero bike saves you 16 Watt or gaining 1.4 km/hr. This is all at a speed of 35 km/r, a speed not unrealistic for a lot of us. So putting some estimated numbers in a formula doesn't do the trick IMO.

Lou
Ads
  #162  
Old December 22nd 18, 03:54 PM posted to rec.bicycles.tech
Ralph Barone[_4_]
external usenet poster
 
Posts: 853
Default Power on hills.

wrote:
On Saturday, December 22, 2018 at 12:52:51 AM UTC+1, Andre Jute wrote:
On Friday, December 21, 2018 at 8:25:37 PM UTC, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really
isn't a large loss unless you're playing for real small power such as
that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ρ A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
ρ = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200
lbs is huge. KE = ½mv²


Thanks for putting the numbers to my argument, Tom.

Andre Jute
DESIGNING AND BUILDING SPECIAL CARS; Batsford, London; Bentley, Boston


I just reading an aero special article in TOUR magazin. Position on the
bike can save you 54 watts or gaining 3.9 km/hr going from riding on the
tops to riding in the drops. Clothes can make a difference of up to 27
Watt or gaining 2,3 km/hr in speed. An aero bike saves you 16 Watt or
gaining 1.4 km/hr. This is all at a speed of 35 km/r, a speed not
unrealistic for a lot of us. So putting some estimated numbers in a
formula doesn't do the trick IMO.

Lou


It works better when you do the math properly and get the units right. :-)

  #163  
Old December 22nd 18, 05:21 PM posted to rec.bicycles.tech
[email protected]
external usenet poster
 
Posts: 824
Default Power on hills.

On Saturday, December 22, 2018 at 3:54:20 PM UTC+1, Ralph Barone wrote:
wrote:
On Saturday, December 22, 2018 at 12:52:51 AM UTC+1, Andre Jute wrote:
On Friday, December 21, 2018 at 8:25:37 PM UTC, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really
isn't a large loss unless you're playing for real small power such as
that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ρ A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
ρ = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200
lbs is huge. KE = ½mv²

Thanks for putting the numbers to my argument, Tom.

Andre Jute
DESIGNING AND BUILDING SPECIAL CARS; Batsford, London; Bentley, Boston


I just reading an aero special article in TOUR magazin. Position on the
bike can save you 54 watts or gaining 3.9 km/hr going from riding on the
tops to riding in the drops. Clothes can make a difference of up to 27
Watt or gaining 2,3 km/hr in speed. An aero bike saves you 16 Watt or
gaining 1.4 km/hr. This is all at a speed of 35 km/r, a speed not
unrealistic for a lot of us. So putting some estimated numbers in a
formula doesn't do the trick IMO.

Lou


It works better when you do the math properly and get the units right. :-)


Yeah, that is what you get using Mickey Mouse units. Using feet and ending up with Watt? WTF. Don't you have a Mickey Mouse unit for power?

Lou
  #164  
Old December 22nd 18, 06:57 PM posted to rec.bicycles.tech
Mark J.
external usenet poster
 
Posts: 840
Default Power on hills.

On 12/22/2018 8:21 AM, wrote:
On Saturday, December 22, 2018 at 3:54:20 PM UTC+1, Ralph Barone wrote:
wrote:
On Saturday, December 22, 2018 at 12:52:51 AM UTC+1, Andre Jute wrote:
On Friday, December 21, 2018 at 8:25:37 PM UTC, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really
isn't a large loss unless you're playing for real small power such as
that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ρ A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
ρ = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200
lbs is huge. KE = ½mv²

Thanks for putting the numbers to my argument, Tom.

Andre Jute
DESIGNING AND BUILDING SPECIAL CARS; Batsford, London; Bentley, Boston

I just reading an aero special article in TOUR magazin. Position on the
bike can save you 54 watts or gaining 3.9 km/hr going from riding on the
tops to riding in the drops. Clothes can make a difference of up to 27
Watt or gaining 2,3 km/hr in speed. An aero bike saves you 16 Watt or
gaining 1.4 km/hr. This is all at a speed of 35 km/r, a speed not
unrealistic for a lot of us. So putting some estimated numbers in a
formula doesn't do the trick IMO.

Lou


It works better when you do the math properly and get the units right. :-)


Yeah, that is what you get using Mickey Mouse units. Using feet and ending up with Watt? WTF. Don't you have a Mickey Mouse unit for power?


Sure, /Horsepower/. 1 hp is about 750 watts. More precisely, 1 hp is
the power to lift 550 lbs (vertically) at a speed of 1 foot per second.
Or you can mess with BTU if you're so inclined (I'd guess not).

Mark J.
  #165  
Old December 22nd 18, 08:59 PM posted to rec.bicycles.tech
Frank Krygowski[_4_]
external usenet poster
 
Posts: 10,538
Default Power on hills.

On 12/22/2018 11:21 AM, wrote:
On Saturday, December 22, 2018 at 3:54:20 PM UTC+1, Ralph Barone wrote:
wrote:
On Saturday, December 22, 2018 at 12:52:51 AM UTC+1, Andre Jute wrote:
On Friday, December 21, 2018 at 8:25:37 PM UTC, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really
isn't a large loss unless you're playing for real small power such as
that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ρ A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
ρ = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200
lbs is huge. KE = ½mv²

Thanks for putting the numbers to my argument, Tom.

Andre Jute
DESIGNING AND BUILDING SPECIAL CARS; Batsford, London; Bentley, Boston

I just reading an aero special article in TOUR magazin. Position on the
bike can save you 54 watts or gaining 3.9 km/hr going from riding on the
tops to riding in the drops. Clothes can make a difference of up to 27
Watt or gaining 2,3 km/hr in speed. An aero bike saves you 16 Watt or
gaining 1.4 km/hr. This is all at a speed of 35 km/r, a speed not
unrealistic for a lot of us. So putting some estimated numbers in a
formula doesn't do the trick IMO.

Lou


It works better when you do the math properly and get the units right. :-)


Yeah, that is what you get using Mickey Mouse units. Using feet and ending up with Watt? WTF. Don't you have a Mickey Mouse unit for power?


Maybe for small amounts, we could use mousepower? It seems aesthetically
related to horsepower.

The U.S. system is so picturesque! Distance in furlongs or chains or
feet; weight in a couple different types of pounds and/or ounces; volume
in gallons or barrels or hogsheads, etc...

And you Euro guys have boring conversion factors - nothing but tens all
up and down the scale. We get lots of interesting ones to remember, and
I'm not even talking about SI to U.S. units. I'm just talking about the
conversions _within_ our system!

I wonder how many people in the U.S. know which U.S. units convert to
other U.S. units by multiplying by 231, or 5280, or 33,000, or 128, or
16, (or alternately 14.58, which is related to 7000), or 778.2, or 36,
or 3.

(I admit, I had to look two of those up.)

--
- Frank Krygowski
  #166  
Old December 22nd 18, 09:12 PM posted to rec.bicycles.tech
AMuzi
external usenet poster
 
Posts: 13,447
Default Power on hills.

On 12/22/2018 1:59 PM, Frank Krygowski wrote:
On 12/22/2018 11:21 AM, wrote:
On Saturday, December 22, 2018 at 3:54:20 PM UTC+1, Ralph
Barone wrote:
wrote:
On Saturday, December 22, 2018 at 12:52:51 AM UTC+1,
Andre Jute wrote:
On Friday, December 21, 2018 at 8:25:37 PM UTC,
wrote:
At low speeds - those below 100 mph of so, aerodynamic
drag really
isn't a large loss unless you're playing for real
small power such as
that developed by a human over relatively long periods
of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ρ A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
ρ = Density of fluid (liquid or gas), kg/m3. (dry air
at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car
which is about 2200
lbs is huge. KE = ½mv²

Thanks for putting the numbers to my argument, Tom.

Andre Jute
DESIGNING AND BUILDING SPECIAL CARS; Batsford, London;
Bentley, Boston

I just reading an aero special article in TOUR magazin.
Position on the
bike can save you 54 watts or gaining 3.9 km/hr going
from riding on the
tops to riding in the drops. Clothes can make a
difference of up to 27
Watt or gaining 2,3 km/hr in speed. An aero bike saves
you 16 Watt or
gaining 1.4 km/hr. This is all at a speed of 35 km/r, a
speed not
unrealistic for a lot of us. So putting some estimated
numbers in a
formula doesn't do the trick IMO.

Lou


It works better when you do the math properly and get the
units right. :-)


Yeah, that is what you get using Mickey Mouse units. Using
feet and ending up with Watt? WTF. Don't you have a Mickey
Mouse unit for power?


Maybe for small amounts, we could use mousepower? It seems
aesthetically related to horsepower.

The U.S. system is so picturesque! Distance in furlongs or
chains or feet; weight in a couple different types of pounds
and/or ounces; volume in gallons or barrels or hogsheads,
etc...

And you Euro guys have boring conversion factors - nothing
but tens all up and down the scale. We get lots of
interesting ones to remember, and I'm not even talking about
SI to U.S. units. I'm just talking about the conversions
_within_ our system!

I wonder how many people in the U.S. know which U.S. units
convert to other U.S. units by multiplying by 231, or 5280,
or 33,000, or 128, or 16, (or alternately 14.58, which is
related to 7000), or 778.2, or 36, or 3.

(I admit, I had to look two of those up.)


+1
Let's toast with half a gill of whisky.

p.s. standard commodity pork price is in bellies, i.e.,
headless gutted frozen carcasses, traded in 20-ton lots.

--
Andrew Muzi
www.yellowjersey.org/
Open every day since 1 April, 1971


  #167  
Old December 23rd 18, 12:06 AM posted to rec.bicycles.tech
John B. Slocomb
external usenet poster
 
Posts: 805
Default Power on hills.

On Sat, 22 Dec 2018 02:28:15 -0800 (PST), wrote:

On Saturday, December 22, 2018 at 12:52:51 AM UTC+1, Andre Jute wrote:
On Friday, December 21, 2018 at 8:25:37 PM UTC, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really isn't a large loss unless you're playing for real small power such as that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ? A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
? = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200 lbs is huge. KE = mv


Thanks for putting the numbers to my argument, Tom.

Andre Jute
DESIGNING AND BUILDING SPECIAL CARS; Batsford, London; Bentley, Boston


I just reading an aero special article in TOUR magazin. Position on the rbike can save you 54 watts or gaining 3.9 km/hr going from riding on the tops to riding in the drops. Clothes can make a difference of up to 27 Watt or gaining 2,3 km/hr in speed. An aero bike saves you 16 Watt or gaining 1.4 km/hr. This is all at a speed of 35 km/r, a speed not unrealistic for a lot of us. So putting some estimated numbers in a formula doesn't do the trick IMO.

Lou


One route I used to ride had a long slightly downhill run on the way
home - perhaps a half kilometer and not very steep. It was at the top
of a fairly steep climb so I used to top the hill and then just sit
there and coast - probably 30 kph. Changing from riding on the tops to
riding on the drops gained about 1 kph. I used to change to the drops,
note the increase in speed and then go back to the tops and note the
decrease.

cheers,

John B.


  #168  
Old December 23rd 18, 12:11 AM posted to rec.bicycles.tech
John B. Slocomb
external usenet poster
 
Posts: 805
Default Power on hills.

On Sat, 22 Dec 2018 14:59:33 -0500, Frank Krygowski
wrote:

On 12/22/2018 11:21 AM, wrote:
On Saturday, December 22, 2018 at 3:54:20 PM UTC+1, Ralph Barone wrote:
wrote:
On Saturday, December 22, 2018 at 12:52:51 AM UTC+1, Andre Jute wrote:
On Friday, December 21, 2018 at 8:25:37 PM UTC, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really
isn't a large loss unless you're playing for real small power such as
that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ? A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
? = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200
lbs is huge. KE = mv

Thanks for putting the numbers to my argument, Tom.

Andre Jute
DESIGNING AND BUILDING SPECIAL CARS; Batsford, London; Bentley, Boston

I just reading an aero special article in TOUR magazin. Position on the
bike can save you 54 watts or gaining 3.9 km/hr going from riding on the
tops to riding in the drops. Clothes can make a difference of up to 27
Watt or gaining 2,3 km/hr in speed. An aero bike saves you 16 Watt or
gaining 1.4 km/hr. This is all at a speed of 35 km/r, a speed not
unrealistic for a lot of us. So putting some estimated numbers in a
formula doesn't do the trick IMO.

Lou


It works better when you do the math properly and get the units right. :-)


Yeah, that is what you get using Mickey Mouse units. Using feet and ending up with Watt? WTF. Don't you have a Mickey Mouse unit for power?


Maybe for small amounts, we could use mousepower? It seems aesthetically
related to horsepower.

The U.S. system is so picturesque! Distance in furlongs or chains or
feet; weight in a couple different types of pounds and/or ounces; volume
in gallons or barrels or hogsheads, etc...


:-) My grandfather used to refer to distances in "rods".

And you Euro guys have boring conversion factors - nothing but tens all
up and down the scale. We get lots of interesting ones to remember, and
I'm not even talking about SI to U.S. units. I'm just talking about the
conversions _within_ our system!

I wonder how many people in the U.S. know which U.S. units convert to
other U.S. units by multiplying by 231, or 5280, or 33,000, or 128, or
16, (or alternately 14.58, which is related to 7000), or 778.2, or 36,
or 3.

(I admit, I had to look two of those up.)


cheers,

John B.


  #169  
Old December 23rd 18, 12:49 AM posted to rec.bicycles.tech
Andre Jute[_2_]
external usenet poster
 
Posts: 10,422
Default Power on hills.

On Saturday, December 22, 2018 at 8:12:47 PM UTC, AMuzi wrote:

p.s. standard commodity pork price is in bellies, i.e.,
headless gutted frozen carcasses, traded in 20-ton lots.


That's a racist measure, or at the very least an anti-German-dish diss and anti-English-dish diss. How can you make Trotters in Jelly if the feet are cut off? Or Pork Knuckles and Ginger Stew?

The pig doesn't start at the hock!

Andre Jute
http://coolmainpress.com/andrejutefoodindex.html
  #170  
Old December 23rd 18, 01:33 AM posted to rec.bicycles.tech
Mark J.
external usenet poster
 
Posts: 840
Default Power on hills.

On 12/22/2018 11:59 AM, Frank Krygowski wrote:
On 12/22/2018 11:21 AM, wrote:
On Saturday, December 22, 2018 at 3:54:20 PM UTC+1, Ralph Barone wrote:
wrote:
On Saturday, December 22, 2018 at 12:52:51 AM UTC+1, Andre Jute wrote:
On Friday, December 21, 2018 at 8:25:37 PM UTC,
wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really
isn't a large loss unless you're playing for real small power such as
that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ρ A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
ρ = Density of fluid (liquid or gas), kg/m3. (dry air at 70
degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200
lbs is huge.* KE = ½mv²

Thanks for putting the numbers to my argument, Tom.

Andre Jute
DESIGNING AND BUILDING SPECIAL CARS; Batsford, London; Bentley, Boston

I just reading an aero special article in TOUR magazin. Position on the
bike can save you 54 watts or gaining 3.9 km/hr going from riding on
the
tops to riding in the drops. Clothes can make a difference of up to 27
Watt or gaining 2,3 km/hr in speed. An aero bike saves you 16 Watt or
gaining 1.4 km/hr. This is all at a speed of 35 km/r, a speed not
unrealistic for a lot of us. So putting some estimated numbers in a
formula doesn't do the trick IMO.

Lou


It works better when you do the math properly and get the units
right. :-)


Yeah, that is what you get using Mickey Mouse units. Using feet and
ending up with Watt? WTF. Don't you have a Mickey Mouse unit for power?


Maybe for small amounts, we could use mousepower? It seems aesthetically
related to horsepower.

The U.S. system is so picturesque! Distance in furlongs or chains or
feet; weight in a couple different types of pounds and/or ounces; volume
in gallons or barrels or hogsheads, etc...

And you Euro guys have boring conversion factors - nothing but tens all
up and down the scale. We get lots of interesting ones to remember, and
I'm not even talking about SI to U.S. units. I'm just talking about the
conversions _within_ our system!

I wonder how many people in the U.S. know which U.S. units convert to
other U.S. units by multiplying by 231, or 5280, or 33,000, or 128, or
16, (or alternately 14.58, which is related to 7000), or 778.2, or 36,
or 3.


I used to excuse exchange students from having to learn many of these
(e.g. 5280), but I told US residents they were (currently) stuck with
the system, so yes, it might be on the exam.

Mark J.

PS - 640 (Acres in a square mile)

 




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