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#172
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Power on hills.
On Saturday, December 22, 2018 at 7:33:29 PM UTC-5, Mark J. wrote:
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) 43,560 square feet per acre. (Why on earth do I have that memorized?) And I've come across volumes of water measured in acre-feet. - Frank Krygowski |
#173
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Power on hills.
On Saturday, December 22, 2018 at 8:14:13 PM UTC-5, Ralph Barone wrote:
Mark J. wrote: 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) Some of these are guesses (I avoided consulting Google): 231 cubic inches per gallon? 5280 feet per mile 33,000 ft-lb/min per HP 128 ounces per gallon (US not English) 16 ounces per pint (or ounces per pound) 778.2 grains per pound? 36 inches per yard 3 feet per yard Seven out of eight is pretty darn good! (There are 7000 grains in a pound, IIRC.) And there are also 3 teaspoons in a tablespoon. If I were in charge, I'd have made it a lot more scientific. I'd have said there are Pi teaspoons in a tablespoon. ;-) - Frank Krygowski |
#174
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Power on hills.
On Saturday, December 22, 2018 at 6:06:38 PM UTC-5, John B. Slocomb wrote:
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 = 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 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. Some years ago, the guys at _Bicycle Quarterly_ paid for wind tunnel testing of a rider on their favorite type of bike. That bike is unlike what most of us probably ride. IIRC, wide tires (probably wider than 32 mm), but drop bars. IIRC they tested changes in rider position, but also presence or absence of fenders, handlebar bag, etc. As I recall they found the strong controlling factor was frontal area. They said, based on their measurements of frontal area, that the drag coefficient didn't change significantly. However, I don't recall what they tested or concluded about clothing. I know loose, flappy clothing adds drag, and probably without a measurable increase in frontal area. I've said for a long time the most significant thing a rider of an upright bike can do for speed is to ride an aero bar when feasible. The increase in speed is much more than just changing to the drops. - Frank Krygowski |
#175
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Power on hills.
On Sat, 22 Dec 2018 20:39:03 -0800 (PST), Frank Krygowski
wrote: On Saturday, December 22, 2018 at 6:06:38 PM UTC-5, John B. Slocomb wrote: 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. Some years ago, the guys at _Bicycle Quarterly_ paid for wind tunnel testing of a rider on their favorite type of bike. That bike is unlike what most of us probably ride. IIRC, wide tires (probably wider than 32 mm), but drop bars. IIRC they tested changes in rider position, but also presence or absence of fenders, handlebar bag, etc. As I recall they found the strong controlling factor was frontal area. They said, based on their measurements of frontal area, that the drag coefficient didn't change significantly. However, I don't recall what they tested or concluded about clothing. I know loose, flappy clothing adds drag, and probably without a measurable increase in frontal area. I've said for a long time the most significant thing a rider of an upright bike can do for speed is to ride an aero bar when feasible. The increase in speed is much more than just changing to the drops. - Frank Krygowski I believe that the RAAM guys use aero bars and I read that one of the reasons is that they support the upper body on the elbows and, they say, it is less tiring to ride long distances that way. As well, of course, the lower wind resistance. cheers, John B. |
#176
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Power on hills.
On Sat, 22 Dec 2018 20:24:38 -0800 (PST), Frank Krygowski
wrote: On Saturday, December 22, 2018 at 7:33:29 PM UTC-5, Mark J. wrote: 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 ? 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... 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) 43,560 square feet per acre. (Why on earth do I have that memorized?) And I've come across volumes of water measured in acre-feet. - Frank Krygowski I think it is all what you are used to :-) One of our senior accountants was British. I asked him if accounting with the British system - penny, shilling, pound and guinea was a problem and he said "not if you grew up with it" :-) And, again in G.B., the old folks counted sheep "yan, tan, tethera, pethera, pimp..." cheers, John B. |
#177
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Power on hills.
On 12/22/2018 8:24 PM, Frank Krygowski wrote:
On Saturday, December 22, 2018 at 7:33:29 PM UTC-5, Mark J. wrote: 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) 43,560 square feet per acre. (Why on earth do I have that memorized?) And I've come across volumes of water measured in acre-feet. Here in the land of hydro-power and mountain reservoirs (Oregon), acre-feet are in pretty common usage, I think. Pretty sure I've seen them in the newspapers fairly regularly. Mark J. |
#178
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Power on hills.
On Sun, 23 Dec 2018 17:10:40 -0800, "Mark J."
wrote: On 12/22/2018 8:24 PM, Frank Krygowski wrote: On Saturday, December 22, 2018 at 7:33:29 PM UTC-5, Mark J. wrote: 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 ? 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... 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) 43,560 square feet per acre. (Why on earth do I have that memorized?) And I've come across volumes of water measured in acre-feet. Here in the land of hydro-power and mountain reservoirs (Oregon), acre-feet are in pretty common usage, I think. Pretty sure I've seen them in the newspapers fairly regularly. Mark J. Most trades or industries have their own esoteric language. Irrigation is often described in acre-feet, horses race over furlongs, bicyclists describe their power output in something other then the traditional "horse power" :-) cheers, John B. |
#179
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Power on hills.
On 12/23/2018 7:42 PM, John B. Slocomb wrote:
On Sun, 23 Dec 2018 17:10:40 -0800, "Mark J." wrote: On 12/22/2018 8:24 PM, Frank Krygowski wrote: On Saturday, December 22, 2018 at 7:33:29 PM UTC-5, Mark J. wrote: 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 ? 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... 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) 43,560 square feet per acre. (Why on earth do I have that memorized?) And I've come across volumes of water measured in acre-feet. Here in the land of hydro-power and mountain reservoirs (Oregon), acre-feet are in pretty common usage, I think. Pretty sure I've seen them in the newspapers fairly regularly. Mark J. Most trades or industries have their own esoteric language. Irrigation is often described in acre-feet, horses race over furlongs, bicyclists describe their power output in something other then the traditional "horse power" :-) cheers, John B. You left out gear inches. -- Andrew Muzi www.yellowjersey.org/ Open every day since 1 April, 1971 |
#180
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Power on hills.
John B. Slocomb wrote:
On Sun, 23 Dec 2018 17:10:40 -0800, "Mark J." wrote: On 12/22/2018 8:24 PM, Frank Krygowski wrote: On Saturday, December 22, 2018 at 7:33:29 PM UTC-5, Mark J. wrote: 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 ? 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... 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) 43,560 square feet per acre. (Why on earth do I have that memorized?) And I've come across volumes of water measured in acre-feet. Here in the land of hydro-power and mountain reservoirs (Oregon), acre-feet are in pretty common usage, I think. Pretty sure I've seen them in the newspapers fairly regularly. Mark J. Most trades or industries have their own esoteric language. Irrigation is often described in acre-feet, horses race over furlongs, bicyclists describe their power output in something other then the traditional "horse power" :-) cheers, John B. Horsepower is simply too embarrassing of a unit to be used by any cyclist with an ego. We suffer badly by comparison with horses. That's why Watts are so popular, although I wonder why ergs per second never caught on. |
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