Radar Guns are used to measure the speed of a moving object. The wave it emits experiences two Doppler shifts. How does the radar gun then use the difference in frequency of the initial wave it emits and the frequency of the final wave it receives to calculate the speed of the object? Could you derive this equation for me?
I thought it only experienced one doppler shift; when the wave reflects off of the target object. What is the second one?
Modern police radars do two independent measurements, one of a signal reflected from the target and a second in which they measure the speed of the ground relative to the radar gun. Thus, they can compute the ground speed of the target (which is what the traffic courts care about) whether the police car is moving or not. I think that's what OP is thinking about... If not, I don't understand the "two doppler shift" part either.
yes that's correct for vehicle mounted radar that can be used from a moving or stationary vehicle for a handheld unit there is only the one measurement and for both systems, the signal to and from the target undergoes only 1 doppler shift Dave
neither of these answer my question. there are two doppler shifts, the first occurs when the radar first emits a wave, the source is stationary and the observer is moving. the second occurs when the wave is reflected back, the source is moving and the observer is stationary. I am not talking about radar guns using the vehicles ground speed either. either way I may have to find the answer to these questions elsewhere
there are two doppler shifts, the first occurs when the radar first emits a wave, the source is stationary and the observer is moving. the second occurs when the wave is reflected back, the source is moving and the observer is stationary.
There's no Doppler shift when the radar emits the wave - the radar is not moving relative to the radar.
You'll find the equation in the wikipedia article on the Doppler effect: http://en.wikipedia.org/wiki/Doppler_effect You just have to remember that that if we're using a radar gun we're dealing with electromagnetic radiation, so the "velocity of the source relative to the medium" is zero and the "velocity of the target relative to the medium" is the velocity of the target relative to the source - which of course is exactly what you're looking for.
The radar gun emitting the wave is not subject to a doppler shift because it is stationary in its own frame and remains that way. The "observer" in this case is either stationary with the radar gun or stationary with the car, but not both.
If the above from nugatory does not satisfy your requirement(s) for the formula you are seeking, you may see the section “Unmodulated continuous-wave type” at http://en.wikipedia.org/wiki/Continuous-wave_radar
Here there seems to be an English language (semantic) misstatement causing the faulty conclusion that there are two Doppler shifts to be reached. Let me assign some different terms: 1. The “source” is the Police radar set that emits the microwaves. 2. The target is the speeding car. 3. The “observer” that receives the reflected energy is the same as the “source”. Operation: The radar transmits its continuous wave (CW) signal. That signal reflects off the speeding car (target), and since there is some relative velocity between that car and the "source", the reflected signal is Doppler-shifted. That signal is detected by the radar set and the difference frequency between the transmitted and received signal is proportional to the relative velocity between the source and the target. Conclusion: There is one Doppler shift.
generally does it? alright the wave emitted from a stationary radar gun to a moving car does experience two Doppler shifts. 1. when the source (radar gun) is stationary and the observer (car) is moving. 'A doppler effect also occurs when an observer moves towards a source, but here the wavelength does not change, instead it is the effective velocity that changes and leads to an apparent change in the frequency of the sound. When the observer moves towards the source of the sound' 2. as the wave is reflected to the radar gun, the observer now becomes the source and it moving, the source now becomes the observer and is stationary. 'When a source of sound moves a stationary observer hears an apparent shift in the frequency.' http://www.ic.sunysb.edu/Class/phy1...effect_when_sound_bounces_off_a_moving_object Now the difference between the initial emitted frequency and that which is finally recieved by the radar gun is used by the radar gun to calculate the speed of the car. which returns me to my initial question of HOW the radar gun does this.
alright. disregarding the doppler shift issue. can we return to my original question. what equation does the radar gun use calculate the speed of the vehicle? it uses the change in frequency right?
There is only one doppler shift, not two. For clarity, note that in our example the radar gun is both the source and the observer. The car is not measuring the wavelength of the radar. The signal is emitted and its wavelength is shifted one time as it is reflected off of the moving car. There is no need to ever change observer frames in this example, and it will only confuse you if you do so. The equation to find the new frequency can be found here: http://en.wikipedia.org/wiki/Doppler_radar#Frequency_variation Since the gun knows the emitted wavelength, and it measures the returned wavelength, it's a simple matter of finding V in the equation.
How can we "disregard" Doppler shift? That is precisely the effect that is used and it describes the change in frequency.