How do I measure wind resistance on my daily bike commute?
#1
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How do I measure wind resistance on my daily bike commute?
I use my Iphone with Distance Meter which measures my path, my speed and time, but it obviously can't measure wind resistance. How do I factor this in for my exercise logs? For example, on the way home it took 5 extra minutes (the wind was brutal), and I'm sure I burned MORE calories, but I have no way of measuring. How is this done?
BTW, I use mapmyfitness to log my workouts.
BTW, I use mapmyfitness to log my workouts.
#2
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You could check weather.com to get an approximation of what the wind speed is supposed to be.
Or, you could buy an anemometer and mount it 33ft in the air, as recommended. https://www.weathershack.com/education/anemometer.html
That still won't tell you what the actual windspeed was/is on your ride.
You could buy a handheld anemometer and stop at intervals during your ride to check the windspeed.
https://www.inspeed.com/anemometers/default.asp
Or, you could buy an anemometer and mount it 33ft in the air, as recommended. https://www.weathershack.com/education/anemometer.html
That still won't tell you what the actual windspeed was/is on your ride.
You could buy a handheld anemometer and stop at intervals during your ride to check the windspeed.
https://www.inspeed.com/anemometers/default.asp
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consider the extra calories burned as a 'bonus' to your usual burn. Kinda like rounding your checking account down to the nearest dollar. It should add up nicely.
and if I see anyone commuting with ILTBs' anemometer I will point and laugh...
and if I see anyone commuting with ILTBs' anemometer I will point and laugh...
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You could tie a string to it and let the iPhone hang down like a plumb bob from your handle bars. Then write an app to measure the deflection from vertical that the tilt sensor picks up. Calibrate the deflection to the corresponding air speed and subtract the ground speed, and that will leave you with the wind's speed for any direction you travel. I'm sure the physics are correct on this. And it would only cost you the price of the string.
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If you want to get technical, you could use this equation:
F = 1/2pv^2(Cd)A
or
F = 1/2 p(v1+vw)^2(Cd)A
Where...
p is the density of air
v is your velocity relative to the air which would be your velocity plus the velocity of a head wind
v1 is your velocity
vw is head wind velocity
Cd is the drag coefficient which would be very hard to calculate. Its a constant that is between 0 and 1. 1 being the case where all the air stops when it hits you. If I had to guess I'd say for a rider in the drops it would probably be close to .2-.3, maybe lower.
A is the area of the orthographic projection of your body. that would be very difficult to calculate so to simplify (and introduce some more error) you could try just using your surface area.
This will give you the force applied by wind resistance (drag).
If you assume a constant velocity of 15mph (~6.7m/s), and take the density of air to be 1.204 kg/m^3 you can calculate the difference in force required based on different head wind speeds.
So, for 15mph bike speed and 0mph wind speed you get:
Fdrag1 ~ 27*A*(Cd)
and for 15mph bike speed and 15mph head wind you get:
Fdrag2 ~ 81*A*(Cd)
Then using P=F*v
change in P = (change in F)*v
or
change in P = (81-27)*A*Cd*v
since power is measured in Joules per second, you can find the number of extra calories required since 1 Joule is .239 calories. Keep in mind that the calories you are talking about are actually kilocalories or 1000 calories.
so the extra work required between 0 wind speed and 15mph head wind when traveling at 15mph:
W = 54*A*Cd*v*t
to account for the calories you are looking for:
W = 54*A*Cd*v*t*.239/1000
or
W = .0129*A*Cd*v*t
also, since velocity times time is distance, you can say:
W = .0129*A*Cd*x
So for traveling 10 miles (16093 meters) the extra work needed would be:
W = 207*A*Cd
If we assume Cd is very low (.2):
W = 41.4*A
If we assume Cd is higher (.3):
W = 62.1*A
and once more when Cd is .5:
W = 103.5*A
so, assuming I didn't make any calculation errors (its pretty late and I should be in bed) over a 10 mile ride, traveling at 15mph, you would burn 40-100 (times your area) more calories if there is a direct head wind of 15mph compared to no wind.
So, change out the total mileage, your speed, and head wind speed (be sure to convert to meters and meters per second) and you should have about as accurate a value (well a range, but close enough) as you can get without some ridiculous equipment.
F = 1/2pv^2(Cd)A
or
F = 1/2 p(v1+vw)^2(Cd)A
Where...
p is the density of air
v is your velocity relative to the air which would be your velocity plus the velocity of a head wind
v1 is your velocity
vw is head wind velocity
Cd is the drag coefficient which would be very hard to calculate. Its a constant that is between 0 and 1. 1 being the case where all the air stops when it hits you. If I had to guess I'd say for a rider in the drops it would probably be close to .2-.3, maybe lower.
A is the area of the orthographic projection of your body. that would be very difficult to calculate so to simplify (and introduce some more error) you could try just using your surface area.
This will give you the force applied by wind resistance (drag).
If you assume a constant velocity of 15mph (~6.7m/s), and take the density of air to be 1.204 kg/m^3 you can calculate the difference in force required based on different head wind speeds.
So, for 15mph bike speed and 0mph wind speed you get:
Fdrag1 ~ 27*A*(Cd)
and for 15mph bike speed and 15mph head wind you get:
Fdrag2 ~ 81*A*(Cd)
Then using P=F*v
change in P = (change in F)*v
or
change in P = (81-27)*A*Cd*v
since power is measured in Joules per second, you can find the number of extra calories required since 1 Joule is .239 calories. Keep in mind that the calories you are talking about are actually kilocalories or 1000 calories.
so the extra work required between 0 wind speed and 15mph head wind when traveling at 15mph:
W = 54*A*Cd*v*t
to account for the calories you are looking for:
W = 54*A*Cd*v*t*.239/1000
or
W = .0129*A*Cd*v*t
also, since velocity times time is distance, you can say:
W = .0129*A*Cd*x
So for traveling 10 miles (16093 meters) the extra work needed would be:
W = 207*A*Cd
If we assume Cd is very low (.2):
W = 41.4*A
If we assume Cd is higher (.3):
W = 62.1*A
and once more when Cd is .5:
W = 103.5*A
so, assuming I didn't make any calculation errors (its pretty late and I should be in bed) over a 10 mile ride, traveling at 15mph, you would burn 40-100 (times your area) more calories if there is a direct head wind of 15mph compared to no wind.
So, change out the total mileage, your speed, and head wind speed (be sure to convert to meters and meters per second) and you should have about as accurate a value (well a range, but close enough) as you can get without some ridiculous equipment.
Last edited by toThinkistoBe; 10-02-08 at 01:16 AM.
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When I tried keeping a log, I just entered "Very windy today".
A power meter would be most helpful if you are keeping track of your workouts. And it's +1000 on the OCP scale.
A power meter would be most helpful if you are keeping track of your workouts. And it's +1000 on the OCP scale.
#10
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Just go as hard as you can for the entire commute. If you get home 5 minutes early keep going!
#11
Senior Member
If you want to get technical, you could use this equation:
F = 1/2pv^2(Cd)A
or
F = 1/2 p(v1+vw)^2(Cd)A
Where...
p is the density of air
v is your velocity relative to the air which would be your velocity plus the velocity of a head wind
v1 is your velocity
vw is head wind velocity
Cd is the drag coefficient which would be very hard to calculate. Its a constant that is between 0 and 1. 1 being the case where all the air stops when it hits you. If I had to guess I'd say for a rider in the drops it would probably be close to .2-.3, maybe lower.
A is the area of the orthographic projection of your body. that would be very difficult to calculate so to simplify (and introduce some more error) you could try just using your surface area.
This will give you the force applied by wind resistance (drag).
If you assume a constant velocity of 15mph (~6.7m/s), and take the density of air to be 1.204 kg/m^3 you can calculate the difference in force required based on different head wind speeds.
So, for 15mph bike speed and 0mph wind speed you get:
Fdrag1 ~ 27*A*(Cd)
and for 15mph bike speed and 15mph head wind you get:
Fdrag2 ~ 81*A*(Cd)
Then using P=F*v
change in P = (change in F)*v
or
change in P = (81-27)*A*Cd*v
since power is measured in Joules per second, you can find the number of extra calories required since 1 Joule is .239 calories. Keep in mind that the calories you are talking about are actually kilocalories or 1000 calories.
so the extra work required between 0 wind speed and 15mph head wind when traveling at 15mph:
W = 54*A*Cd*v*t
to account for the calories you are looking for:
W = 54*A*Cd*v*t*.239/1000
or
W = .0129*A*Cd*v*t
also, since velocity times time is distance, you can say:
W = .0129*A*Cd*x
So for traveling 10 miles (16093 meters) the extra work needed would be:
W = 207*A*Cd
If we assume Cd is very low (.2):
W = 41.4*A
If we assume Cd is higher (.3):
W = 62.1*A
and once more when Cd is .5:
W = 103.5*A
so, assuming I didn't make any calculation errors (its pretty late and I should be in bed) over a 10 mile ride, traveling at 15mph, you would burn 40-100 (times your area) more calories if there is a direct head wind of 15mph compared to no wind.
So, change out the total mileage, your speed, and head wind speed (be sure to convert to meters and meters per second) and you should have about as accurate a value (well a range, but close enough) as you can get without some ridiculous equipment.
F = 1/2pv^2(Cd)A
or
F = 1/2 p(v1+vw)^2(Cd)A
Where...
p is the density of air
v is your velocity relative to the air which would be your velocity plus the velocity of a head wind
v1 is your velocity
vw is head wind velocity
Cd is the drag coefficient which would be very hard to calculate. Its a constant that is between 0 and 1. 1 being the case where all the air stops when it hits you. If I had to guess I'd say for a rider in the drops it would probably be close to .2-.3, maybe lower.
A is the area of the orthographic projection of your body. that would be very difficult to calculate so to simplify (and introduce some more error) you could try just using your surface area.
This will give you the force applied by wind resistance (drag).
If you assume a constant velocity of 15mph (~6.7m/s), and take the density of air to be 1.204 kg/m^3 you can calculate the difference in force required based on different head wind speeds.
So, for 15mph bike speed and 0mph wind speed you get:
Fdrag1 ~ 27*A*(Cd)
and for 15mph bike speed and 15mph head wind you get:
Fdrag2 ~ 81*A*(Cd)
Then using P=F*v
change in P = (change in F)*v
or
change in P = (81-27)*A*Cd*v
since power is measured in Joules per second, you can find the number of extra calories required since 1 Joule is .239 calories. Keep in mind that the calories you are talking about are actually kilocalories or 1000 calories.
so the extra work required between 0 wind speed and 15mph head wind when traveling at 15mph:
W = 54*A*Cd*v*t
to account for the calories you are looking for:
W = 54*A*Cd*v*t*.239/1000
or
W = .0129*A*Cd*v*t
also, since velocity times time is distance, you can say:
W = .0129*A*Cd*x
So for traveling 10 miles (16093 meters) the extra work needed would be:
W = 207*A*Cd
If we assume Cd is very low (.2):
W = 41.4*A
If we assume Cd is higher (.3):
W = 62.1*A
and once more when Cd is .5:
W = 103.5*A
so, assuming I didn't make any calculation errors (its pretty late and I should be in bed) over a 10 mile ride, traveling at 15mph, you would burn 40-100 (times your area) more calories if there is a direct head wind of 15mph compared to no wind.
So, change out the total mileage, your speed, and head wind speed (be sure to convert to meters and meters per second) and you should have about as accurate a value (well a range, but close enough) as you can get without some ridiculous equipment.
Well, there you go. Don't you feel silly now for asking.
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I use my Iphone with Distance Meter which measures my path, my speed and time, but it obviously can't measure wind resistance. How do I factor this in for my exercise logs? For example, on the way home it took 5 extra minutes (the wind was brutal), and I'm sure I burned MORE calories, but I have no way of measuring. How is this done?
BTW, I use mapmyfitness to log my workouts.
BTW, I use mapmyfitness to log my workouts.
A personal relative system I use is on a 1 to 6 scale (I like for some reason, having seen it used in white water rafting). I assess how an unobstructed flag is flying on a pole of about 20 feet or so height. For example if it is gently wafted and easily falls back, that's a 2. If it flies and wavers, but doesn't fall back, that's about 3, and if it is flying pretty straight, that's a 4. Here in Boston I rarely get to 4. This system helps me record why I might have gone fast or slow on a given day. Sometimes I'm not sure if it's my energy level or the wind.
The strongest winds I've ridden in were out west. On a highway near the Grand Canyon where we had to hold tight to stay on a straight course, sometimes the wind would abate and we'd veer into the center of the road. Crossing a pass in San Berardino, we barely had to pedal with a strong tailwind.