Calculate average wind velocity
I've got an anemometer to measure wind speed.
Its ratio is $50Hz = 15.5m/s$, so my Arduino counts pulses in 3100ms, which should give a count of $100 = 10m/s$.
I can display this on a dashboard on the wall, but I'm more interested in an average.
(I'm just using this for fun, I'm not controlling the traffic on a bridge or in an airport)
But what makes more sense?
I can count pulses for $(19*3.1) = 59.8 text{seconds}$ and divide by 190 to get an average for 1 minute.
Or the same for 10 minutes.
Or I could have a memory of the last p samples so I add the sample to a register, and divide with p, then at Tp subtract it from the register again, so I get a running average over the last minutes.
I could also take a weighted average of the current value and the current average as the next current average, which would emulate the charging and discharging of a capacitor. Something like
$$overline{V} = frac{V + (n-1)overline{V}prev}{n}$$
Or should I take into the account that energy of the wind is proportional to the speed cubed?
So 1 minute of 10m/s and 1 minute of 20 m/s have the same energy of
$$ sqrt[3]frac{10^{3} + 20^{3}}{2} approx 16.5 $$
average
asked Jan 4 at 22:49
LenneLenne
1273
|
show 2 more comments
I've got an anemometer to measure wind speed.
Its ratio is $50Hz = 15.5m/s$, so my Arduino counts pulses in 3100ms, which should give a count of $100 = 10m/s$.
I can display this on a dashboard on the wall, but I'm more interested in an average.
(I'm just using this for fun, I'm not controlling the traffic on a bridge or in an airport)
But what makes more sense?
I can count pulses for $(19*3.1) = 59.8 text{seconds}$ and divide by 190 to get an average for 1 minute.
Or the same for 10 minutes.
Or I could have a memory of the last p samples so I add the sample to a register, and divide with p, then at Tp subtract it from the register again, so I get a running average over the last minutes.
I could also take a weighted average of the current value and the current average as the next current average, which would emulate the charging and discharging of a capacitor. Something like
$$overline{V} = frac{V + (n-1)overline{V}prev}{n}$$
Or should I take into the account that energy of the wind is proportional to the speed cubed?
So 1 minute of 10m/s and 1 minute of 20 m/s have the same energy of
$$ sqrt[3]frac{10^{3} + 20^{3}}{2} approx 16.5 $$
average
asked Jan 4 at 22:49
LenneLenne
1273
1
The energy of the wind is proportional to the speed squared. The power of the wind is proportional to the speed cubed.
– Acccumulation
Jan 4 at 22:56
So would it give more sense to calculate the RMS than the average?
– Lenne
Jan 4 at 23:57
2
It depends on why you want an average. If you want to know how far a free-floating balloon might drift in an hour, you want average velocity. If you want to know how much electricity a wind turbine might generate in an hour, you want average power (I think--anyway not just average speed).
– David K
Jan 5 at 0:03
What kind of average would be the right to determine if it was safe to cross a bridge in a storm?
– Lenne
Jan 5 at 0:08
You might experiment with "exponential smoothing". It's easy to compute, flexible, and is often used to smooth time series. en.wikipedia.org/wiki/Exponential_smoothing
– awkward
Jan 5 at 13:32
|
show 2 more comments
I've got an anemometer to measure wind speed.
Its ratio is $50Hz = 15.5m/s$, so my Arduino counts pulses in 3100ms, which should give a count of $100 = 10m/s$.
I can display this on a dashboard on the wall, but I'm more interested in an average.
(I'm just using this for fun, I'm not controlling the traffic on a bridge or in an airport)
But what makes more sense?
I can count pulses for $(19*3.1) = 59.8 text{seconds}$ and divide by 190 to get an average for 1 minute.
Or the same for 10 minutes.
Or I could have a memory of the last p samples so I add the sample to a register, and divide with p, then at Tp subtract it from the register again, so I get a running average over the last minutes.
I could also take a weighted average of the current value and the current average as the next current average, which would emulate the charging and discharging of a capacitor. Something like
$$overline{V} = frac{V + (n-1)overline{V}prev}{n}$$
Or should I take into the account that energy of the wind is proportional to the speed cubed?
So 1 minute of 10m/s and 1 minute of 20 m/s have the same energy of
$$ sqrt[3]frac{10^{3} + 20^{3}}{2} approx 16.5 $$
average
asked Jan 4 at 22:49
LenneLenne
1273
I've got an anemometer to measure wind speed.
Its ratio is $50Hz = 15.5m/s$, so my Arduino counts pulses in 3100ms, which should give a count of $100 = 10m/s$.
I can display this on a dashboard on the wall, but I'm more interested in an average.
(I'm just using this for fun, I'm not controlling the traffic on a bridge or in an airport)
But what makes more sense?
I can count pulses for $(19*3.1) = 59.8 text{seconds}$ and divide by 190 to get an average for 1 minute.
Or the same for 10 minutes.
Or I could have a memory of the last p samples so I add the sample to a register, and divide with p, then at Tp subtract it from the register again, so I get a running average over the last minutes.
I could also take a weighted average of the current value and the current average as the next current average, which would emulate the charging and discharging of a capacitor. Something like
$$overline{V} = frac{V + (n-1)overline{V}prev}{n}$$
Or should I take into the account that energy of the wind is proportional to the speed cubed?
So 1 minute of 10m/s and 1 minute of 20 m/s have the same energy of
$$ sqrt[3]frac{10^{3} + 20^{3}}{2} approx 16.5 $$
average
average
asked Jan 4 at 22:49
LenneLenne
1273
asked Jan 4 at 22:49
LenneLenne
1273
asked Jan 4 at 22:49
LenneLenne
1273
asked Jan 4 at 22:49
LenneLenne
1273
asked Jan 4 at 22:49
LenneLenne
1273
1273
1
The energy of the wind is proportional to the speed squared. The power of the wind is proportional to the speed cubed.
– Acccumulation
Jan 4 at 22:56
So would it give more sense to calculate the RMS than the average?
– Lenne
Jan 4 at 23:57
2
It depends on why you want an average. If you want to know how far a free-floating balloon might drift in an hour, you want average velocity. If you want to know how much electricity a wind turbine might generate in an hour, you want average power (I think--anyway not just average speed).
– David K
Jan 5 at 0:03
What kind of average would be the right to determine if it was safe to cross a bridge in a storm?
– Lenne
Jan 5 at 0:08
You might experiment with "exponential smoothing". It's easy to compute, flexible, and is often used to smooth time series. en.wikipedia.org/wiki/Exponential_smoothing
– awkward
Jan 5 at 13:32
|
show 2 more comments
1
The energy of the wind is proportional to the speed squared. The power of the wind is proportional to the speed cubed.
– Acccumulation
Jan 4 at 22:56
So would it give more sense to calculate the RMS than the average?
– Lenne
Jan 4 at 23:57
2
It depends on why you want an average. If you want to know how far a free-floating balloon might drift in an hour, you want average velocity. If you want to know how much electricity a wind turbine might generate in an hour, you want average power (I think--anyway not just average speed).
– David K
Jan 5 at 0:03
What kind of average would be the right to determine if it was safe to cross a bridge in a storm?
– Lenne
Jan 5 at 0:08
You might experiment with "exponential smoothing". It's easy to compute, flexible, and is often used to smooth time series. en.wikipedia.org/wiki/Exponential_smoothing
– awkward
Jan 5 at 13:32
1
1
The energy of the wind is proportional to the speed squared. The power of the wind is proportional to the speed cubed.
– Acccumulation
Jan 4 at 22:56
The energy of the wind is proportional to the speed squared. The power of the wind is proportional to the speed cubed.
– Acccumulation
Jan 4 at 22:56
So would it give more sense to calculate the RMS than the average?
– Lenne
Jan 4 at 23:57
So would it give more sense to calculate the RMS than the average?
– Lenne
Jan 4 at 23:57
2
2
It depends on why you want an average. If you want to know how far a free-floating balloon might drift in an hour, you want average velocity. If you want to know how much electricity a wind turbine might generate in an hour, you want average power (I think--anyway not just average speed).
– David K
Jan 5 at 0:03
It depends on why you want an average. If you want to know how far a free-floating balloon might drift in an hour, you want average velocity. If you want to know how much electricity a wind turbine might generate in an hour, you want average power (I think--anyway not just average speed).
– David K
Jan 5 at 0:03
What kind of average would be the right to determine if it was safe to cross a bridge in a storm?
– Lenne
Jan 5 at 0:08
What kind of average would be the right to determine if it was safe to cross a bridge in a storm?
– Lenne
Jan 5 at 0:08
You might experiment with "exponential smoothing". It's easy to compute, flexible, and is often used to smooth time series. en.wikipedia.org/wiki/Exponential_smoothing
– awkward
Jan 5 at 13:32
You might experiment with "exponential smoothing". It's easy to compute, flexible, and is often used to smooth time series. en.wikipedia.org/wiki/Exponential_smoothing
– awkward
Jan 5 at 13:32
|
show 2 more comments
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1
The energy of the wind is proportional to the speed squared. The power of the wind is proportional to the speed cubed.
– Acccumulation
Jan 4 at 22:56
So would it give more sense to calculate the RMS than the average?
– Lenne
Jan 4 at 23:57
2
It depends on why you want an average. If you want to know how far a free-floating balloon might drift in an hour, you want average velocity. If you want to know how much electricity a wind turbine might generate in an hour, you want average power (I think--anyway not just average speed).
– David K
Jan 5 at 0:03
What kind of average would be the right to determine if it was safe to cross a bridge in a storm?
– Lenne
Jan 5 at 0:08
You might experiment with "exponential smoothing". It's easy to compute, flexible, and is often used to smooth time series. en.wikipedia.org/wiki/Exponential_smoothing
– awkward
Jan 5 at 13:32