The reference you cited said...
Surprisingly, the received frequency is the absolute one: f '' = f whatever the angle. However, the observer should be aware that this happens because he is moving along with the emitter. The sound waves are truly undergoing the Doppler effect, but he cannot detect the frequency change any more because the virtual one cancels the regular one.
This is no different than saying that there is no observed Doppler effect, which is what I've been saying all along.
The animation at the top of the referenced page illustrates this perfectly. Look at the rate of arrival of the wavefronts at A, B, C, and D. They're all the same.
Maybe using a plane flying up-wind or down-wind is too close to the phenomenon we're tring to understand (sound wave propagation through air (moving and still).
Maybe a clearer understanding of what's happening can be seen by imagining a boat on a river. If the boat is traveling a 5 knots, relative to the water as measured on the boat, upstream and the river is flowing at 3 knots, the boat is traveling at ~2 knots· upstream relative to a given point on the shore and if we measure the time it takes to travel between two points on the shore, we can calculate the boat's speed. If the boat then·comes about and heads downstream, again at 5 knots relative to the water, the observed speed of the boat (the time it takes to transit the two points on shore) would be ~8 knots.
Because we know that the boat is always traveling at 5 knots relative to the water its in (like we know the speed of sound in air), we can calculate the speed and direction of the medium (air or water) by calculating the difference between the time it takes our boat or packet of sound to travel a·fixed distance compared to what is should have·taken in still air/water.
If we are on shore and observe that the boat is traveling past us at 8 knots (we know that all boats on the river must maintain a speed of exactly 5 knots as measured on the boat), we know that it is going downstream and that the·river is flowing at 3 knots. Conversely, if we observe that the boat is traveling past us at 2 knots, then we know that the boat is traveling upstream and that the river is flowing at 3 knots.
can·we please explain it in even simplier terms, that is, just using sound and air.
Let's simplify this first - we know we need more than one pair, but lets say we have this simple set up:
UsTx ---->>>> UsRx
Where Us = Ultrasonic, and Tx/Rx are the usual definitions.
What does the moving air do to the *sound* in these three configurations:
1. Air moves in the same direction as Tx to Rx, that is ---->>>
2. Air moves the opposite direction to Tx and Rx, that is <<<
3. Air cuts across the Tx-Rx line an a right angle.
My 10 cents:
1. Air moves in the same direction as Tx to Rx, that is ---->>> The wavelength increases (and frequency decreases). Effect is cancelled at Rx.
2. Air moves the opposite direction to Tx and Rx, that is <<<
The wavelenth decreases (and frequency increases). Effect is cancelled at Rx.
3. Air cuts across the Tx-Rx line an a right angle. It's velocity (direction) changes
The sound creates a pressure wave as it travels, increasing and decreasing pressure in a rythmic behaviour.
The loudness (amplitude) of the sound decreases with time. Loudness decreases as the square of the distance between emitter and receiver.
The sound does not do anything to the air other than the local passing oscillations that are present when the wave travels (e.g. the thermal state of the air is preserved).
Sound is like any other wave, it transports energy from one place to another.
What distinguishes sound from other waves is the pressure nature of it. Sound is called longitudal propagation (the wave motion is back and forth), as opposed to electromagnetic or water waves which are transverse waves (they move up and down).
Here is an article from Everyday Practical Electronics magazine that contains a pretty good "Theory of Operation"
for ultrasonic wind speed measurement.
I'll take another stab at it. Think of a pressure disturbance, a wavefront and don't worry about the frequency. Its velocity with respect to the fixed tx and rx points is the sum of, S, the speed of sound, plus or minus the movement of the air mass, W.
1. (S + W) * Tf = D ' speed of sound S plus wind speed W covers distance D in time Tf 2. (S - W) * Tr = D ' speed of sound S minus wind speed W covers distance D in time Tr
The trick with the anemometer is that it sends a pulse first in one direction and then the other, and measures the time each way. When you solve each of the above equations for a 1/t value and then subtract, you find, 1/Tf - 1/Tr = (S + W) / D - (S -W) /D = 2 * W / D
So W = 0.5 * D * (1/Tf - 1/Tr)
The article Duffer pointed out is really neat, but I was surprised that they took the difference between the two time measurements, rather than the difference of the inverses.
For the third case, the line from tx to rx is perpendicular to the wind. Consider the following diagram:
The disturbance originates at the tx point shown in red, and the disturbance propagates outward in circles around the center point. Buthe center point for the circles moves with the air mass and does not stay at tx. The air mass shown in blue has moved and carried the center point a distance to the right in the diagram, and the circle shown in blue is the disturbance at the instant it reaches the rx point, which has remained fixed in relation to the original tx point. It is important to understand the principle of advection. Withn the frame of reference of the moving air mass, the disturbance propagates outwards in circles at the speed of sound around the moving center point. That is why we have to use terms like S+W and S-W when dealing with arms on the fixed anemometer.
In this case there is geometry of a triangle:
The wind vector in this case is at a right angle to the line connecting tx to rx. That baseline, D, and the distance W * Tp that the wind carries the origin point in the moving air mass, form a right triangle with the distance S * Tp the sound has to travel as the hypotenuse. So that can be solved for the transit time Tp. as shown. Note that this is longer than the time it would take for sound to travel from tx to rx in still air. And there will be no difference in time between the two directions.
I think the subtraction of times vs. velocities can be explained by an implicit approximation arising from the speed of sound (s) being much greater than the speed of the wind (w):
····tfwd = k / (s + w)
····trev = k / (s - w)
····trev - tfwd = k / (s - w) - k / (s + w)
····= (k (s + w) - k (s - w)) / (s2 - w2) = 2 k w / (s2 - w2)
···· ~ 2 k w / s2, since s >> w.
The unfortunate thing about this approximation, though, is that it retains the factor s, which can vary with temperature and humidity.
That problem is addressed very nicely in the article that I referenced. The average of·'pings' in opposite directions represents the speed of sound (s) under the current conditions, thus making the setup self-calibrating.
John Becker's PIC code for this project is available from quasarelectronics.com, project number 3168. The speed acquisition takes place in a routine called SONICTX and in the ISR, and the averaging and calculations in a routine called GETSPEED. The "north" transit time is subtracted from the "south" transit time, and the absolute value of the difference goes into the average, the purpose of which is to smooth out the noise. The code does not compensate for temperature effects:
; [b]from John Becker, 3168_351.asm[/b]
; A 20^C change in temp = 3.5% error from above which is taken as
; insignificant and not correct for.
; N.B. temp change is the main cause of changes in sound speed
; at 0^C = 332 m/s
; at 20^C = 344 m/s
; at 100^C = 386 m/s
; (effects of humidity and barometric pressure are small by comparison)
It is true that the error is not too bad, and even less when as Phil showed the approximate dependence is 1/s2. But the point is that professional instruments subtract the inverse of the times, in which case the speed of sound cancels out completely. No compensation necessary.
Once the wind speed is determined accurately, the transit times can be used to calculate the temperature of the air. It is sometimes difficult to measure air temperature accurately with a standard thermometer, due to effects of radiation, and "sonic temperature" is useful in its own right. Fancy sonic anemometers measure wind in three dimenions and are used in sophisticated studies of carbon dioxide flux. For example: www.atmos.anl.gov/ABLE/ecor.html
I was curious how the math would come out for arbitrary angles between the wind and the line connecting the ultrasonic tx and rx. In particular, does the speed of sound still drop out of the equations when the forward and backward 1/transit times are subtracted, just as it does for the simpler cases? And is it hard to compute both speed and direction from an arrangement of 3 or 4 sensors?
The formula for each individual transit time is knarly, as seen in the following diagram, with transit time Tp, separation D, wind speed W, sound speed S, and angle theta between W and D. The point labeled tx transmits an ultrasonic pulse and the pulse arrives at the receiver at rx, a time Tp later.
A similar solution applies to the reverse transmission path, transit time Tq with only a change in sign in one of the terms. Luckily, the subtraction of forward and reverse transit times results in a lot of cancellation, as follows:
Again, the speed of sound drops out when the inverses are subtracted, very convenient These formulas reduce to the earlier ones presented when theta=0 and when theta=90 degrees. Clearly results from two different angles can be combined to find easily both wind speed and wind direction.
Comments
The animation at the top of the referenced page illustrates this perfectly. Look at the rate of arrival of the wavefronts at A, B, C, and D. They're all the same.
-Phil
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Tia'Shar Manetheren
Maybe a clearer understanding of what's happening can be seen by imagining a boat on a river. If the boat is traveling a 5 knots, relative to the water as measured on the boat, upstream and the river is flowing at 3 knots, the boat is traveling at ~2 knots· upstream relative to a given point on the shore and if we measure the time it takes to travel between two points on the shore, we can calculate the boat's speed. If the boat then·comes about and heads downstream, again at 5 knots relative to the water, the observed speed of the boat (the time it takes to transit the two points on shore) would be ~8 knots.
Because we know that the boat is always traveling at 5 knots relative to the water its in (like we know the speed of sound in air), we can calculate the speed and direction of the medium (air or water) by calculating the difference between the time it takes our boat or packet of sound to travel a·fixed distance compared to what is should have·taken in still air/water.
If we are on shore and observe that the boat is traveling past us at 8 knots (we know that all boats on the river must maintain a speed of exactly 5 knots as measured on the boat), we know that it is going downstream and that the·river is flowing at 3 knots. Conversely, if we observe that the boat is traveling past us at 2 knots, then we know that the boat is traveling upstream and that the river is flowing at 3 knots.
Duffer
can·we please explain it in even simplier terms, that is, just using sound and air.
Let's simplify this first - we know we need more than one pair, but lets say we have this simple set up:
UsTx ---->>>> UsRx
Where Us = Ultrasonic, and Tx/Rx are the usual definitions.
What does the moving air do to the *sound* in these three configurations:
1. Air moves in the same direction as Tx to Rx, that is ---->>>
2. Air moves the opposite direction to Tx and Rx, that is <<<
3. Air cuts across the Tx-Rx line an a right angle.
(And does the sound do anything to the air? )
I think this would help a lot.
thanks
- Howard
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1. Air moves in the same direction as Tx to Rx, that is ---->>> The wavelength increases (and frequency decreases). Effect is cancelled at Rx.
2. Air moves the opposite direction to Tx and Rx, that is <<<
The wavelenth decreases (and frequency increases). Effect is cancelled at Rx.
3. Air cuts across the Tx-Rx line an a right angle. It's velocity (direction) changes
The sound creates a pressure wave as it travels, increasing and decreasing pressure in a rythmic behaviour.
The loudness (amplitude) of the sound decreases with time. Loudness decreases as the square of the distance between emitter and receiver.
The sound does not do anything to the air other than the local passing oscillations that are present when the wave travels (e.g. the thermal state of the air is preserved).
Sound is like any other wave, it transports energy from one place to another.
What distinguishes sound from other waves is the pressure nature of it. Sound is called longitudal propagation (the wave motion is back and forth), as opposed to electromagnetic or water waves which are transverse waves (they move up and down).
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Don't worry. Be happy
for ultrasonic wind speed measurement.
http://www.kitsrus.com/pdf/K168_article.pdf
Duffer
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1. (S + W) * Tf = D ' speed of sound S plus wind speed W covers distance D in time Tf
2. (S - W) * Tr = D ' speed of sound S minus wind speed W covers distance D in time Tr
The trick with the anemometer is that it sends a pulse first in one direction and then the other, and measures the time each way. When you solve each of the above equations for a 1/t value and then subtract, you find,
1/Tf - 1/Tr = (S + W) / D - (S -W) /D = 2 * W / D
So
W = 0.5 * D * (1/Tf - 1/Tr)
The article Duffer pointed out is really neat, but I was surprised that they took the difference between the two time measurements, rather than the difference of the inverses.
For the third case, the line from tx to rx is perpendicular to the wind. Consider the following diagram:
The disturbance originates at the tx point shown in red, and the disturbance propagates outward in circles around the center point. Buthe center point for the circles moves with the air mass and does not stay at tx. The air mass shown in blue has moved and carried the center point a distance to the right in the diagram, and the circle shown in blue is the disturbance at the instant it reaches the rx point, which has remained fixed in relation to the original tx point. It is important to understand the principle of advection. Withn the frame of reference of the moving air mass, the disturbance propagates outwards in circles at the speed of sound around the moving center point. That is why we have to use terms like S+W and S-W when dealing with arms on the fixed anemometer.
In this case there is geometry of a triangle:
The wind vector in this case is at a right angle to the line connecting tx to rx. That baseline, D, and the distance W * Tp that the wind carries the origin point in the moving air mass, form a right triangle with the distance S * Tp the sound has to travel as the hypotenuse. So that can be solved for the transit time Tp. as shown. Note that this is longer than the time it would take for sound to travel from tx to rx in still air. And there will be no difference in time between the two directions.
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Tracy Allen
www.emesystems.com
Post Edited (Tracy Allen) : 9/24/2009 3:18:37 AM GMT
I think the subtraction of times vs. velocities can be explained by an implicit approximation arising from the speed of sound (s) being much greater than the speed of the wind (w):
····tfwd = k / (s + w)
····trev = k / (s - w)
····trev - tfwd = k / (s - w) - k / (s + w)
····= (k (s + w) - k (s - w)) / (s2 - w2) = 2 k w / (s2 - w2)
···· ~ 2 k w / s2, since s >> w.
The unfortunate thing about this approximation, though, is that it retains the factor s, which can vary with temperature and humidity.
-Phil
That problem is addressed very nicely in the article that I referenced. The average of·'pings' in opposite directions represents the speed of sound (s) under the current conditions, thus making the setup self-calibrating.
Duffer
-Phil
It is true that the error is not too bad, and even less when as Phil showed the approximate dependence is 1/s2. But the point is that professional instruments subtract the inverse of the times, in which case the speed of sound cancels out completely. No compensation necessary.
Once the wind speed is determined accurately, the transit times can be used to calculate the temperature of the air. It is sometimes difficult to measure air temperature accurately with a standard thermometer, due to effects of radiation, and "sonic temperature" is useful in its own right. Fancy sonic anemometers measure wind in three dimenions and are used in sophisticated studies of carbon dioxide flux. For example:
www.atmos.anl.gov/ABLE/ecor.html
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Tracy Allen
www.emesystems.com
The formula for each individual transit time is knarly, as seen in the following diagram, with transit time Tp, separation D, wind speed W, sound speed S, and angle theta between W and D. The point labeled tx transmits an ultrasonic pulse and the pulse arrives at the receiver at rx, a time Tp later.
A similar solution applies to the reverse transmission path, transit time Tq with only a change in sign in one of the terms. Luckily, the subtraction of forward and reverse transit times results in a lot of cancellation, as follows:
Again, the speed of sound drops out when the inverses are subtracted, very convenient These formulas reduce to the earlier ones presented when theta=0 and when theta=90 degrees. Clearly results from two different angles can be combined to find easily both wind speed and wind direction.
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Tracy Allen
www.emesystems.com
Post Edited (Tracy Allen) : 9/26/2009 6:15:28 AM GMT