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#161
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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 |
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#162
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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
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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
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Power on hills.
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#165
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Power on hills.
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#167
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Power on hills.
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#168
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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
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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
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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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