1. Radar just uses radio waves,electromagnetic waves,but how does it works?Or how does the speed trap radar that our friendly cops use work?

2. Originally Posted by Starlight
Radar just uses radio waves,electromagnetic waves,but how does it works?Or how does the speed trap radar that our friendly cops use work?
Uses Doppler effect to measure car speed. Here

3. Originally Posted by Starlight
Radar just uses radio waves,electromagnetic waves,but how does it works?Or how does the speed trap radar that our friendly cops use work?
It uses the Doppler effect to detect speed. It's all described here Radar gun - Wikipedia, the free encyclopedia

4. Radar is a RF transmitter and receiver. If they are at the same relative speed, it can be used to measure distances based on how long it takes for the waves to travel to the other object in space and return. In the case of speed detection, if the objects are traveling at different speeds, the received will I think have a different frequency than the transmitted waves. Then it is calculated I believe based on the wavelengh differential between the transmitted waves and the reflected wave that are received back. Basically this can measure how fast the two objects are approaching or receding ffrom one another. I just made this up on the spot, but I think it is basically correct. It is an interesting question and I should think some members here could chime in with some math examples, and formulas on this if they should wish to.

5. Originally Posted by mayflow
Radar is a RF transmitter and receiver. If they are at the same relative speed, it can be used to measure distances based on how long it takes for the waves to travel to the other object in space and return. In the case of speed detection, if the objects are traveling at different speeds, the received will I think have a different frequency than the transmitted waves. Then it is calculated I believe based on the wavelengh differential between the transmitted waves and the reflected wave that are received back. Basically this can measure how fast the two objects are approaching or receding ffrom one another. I just made this up on the spot, but I think it is basically correct.

The above is a first step at doing things correctly. See if you can take the next step:

1. Try writing the equations for measuring the distance between two objects at rest wrt each other.
2. Try writing the equations for measuring the relative speed between two objects in motion wrt each other.
3. For extra points, see how you would derive the equations for measuring the distance between two objects in motion wrt each other as a function of time and their relative speed.

It is an interesting question and I should think some members here could chime in with some math examples, and formulas on this if they should wish to.
You would benefit a lot more if you tried to do it yourself. I gave you a "like", to encourage you to do the right thing.

6. Originally Posted by xOx
The above is a first step at doing things correctly. See if you can take the next step:

1. Try writing the equations for measuring the distance between two objects at rest wrt each other.
2. Try writing the equations for measuring the relative speed between two objects in motion wrt each other.
3. For extra points, see how you would derive the equations for measuring the distance between two objects in motion wrt each other as a function of time and their relative speed.
I am not really all that interested right now in doing that. I just said how I thnk the principle of it would seem to work from my viewpoint. That is why I mentioned that some others here may wish to chime in with the math and formulas. I first thought you mistyped "with" in point 1 (wrt) but when repeated in pont 2, I figured it must be some sort of acronym, so I figure you are meaning "with relationsip to"?

In point 1 you probably have to look at the speed of light in some units (like maybe meters per second) and figure out how long it took to receive the sent signal back, and say "it would have traveled this distance in this time frame" then divide that distance by two as the light had to go in the same distance twice. The speed of approach or separation would I think be a lot more complex because now you would need to figure out the differences in wavelengths I suppose by deciphering that each wavelength would travel a different distance in the same unit of time. Am I close, professor?

7. Originally Posted by mayflow
I am not really all that interested right now in doing that. I just said how I thnk the principle of it would seem to work from my viewpoint. That is why I mentioned that some others here may wish to chime in with the math and formulas. I first thought you mistyped "with" in point 1 (wrt) but when repeated in pont 2, I figured it must be some sort of acronym, so I figure you are meaning "with relationsip to"?

In point 1 you probably have to look at the speed of light in some units (like maybe meters per second) and figure out how long it took to receive the sent signal back, and say "it would have traveled this distance in this time frame" then divide that distance by two as the light had to go in the same distance twice. The speed of approach or separation would I think be a lot more complex because now you would need to figure out the differences in wavlengths I suppose by deciphering that each wavelength would travel a different distance in the same unit of time. Am I close, professor?
"wrt" = with respect to.
In order to learn physics you will need to learn how to translate the prose into math since the language of physics is math.
Try point 1, it is the easiest of all. You have the prose right. How about trying to translate it into math?

8. Well, if I simplify it I would say that if the sent signals traveled at 300 mph, and it took an hour for the signal to return, it would have had to travel 150 miles twice so the object was 150 miles away. If this is correct, can you give me an equivalent formula, and explain to me in prose how you derived it?

9. Originally Posted by mayflow
Well, if I simplify it I would say that if the sent signals traveled at 300 mph, and it took an hour for the signal to return, it would have had to travel 150 miles twice so the object was 150 miles away. If this is correct, can you give me an equivalent formula, and explain to me in prose how you derived it?
Yes, it is correct. Now, the next step for you is to learn to do the above symbolically, such that it works with any speed, time and distance.
It takes the time to do the roundtrip.
Therefore the objects are separated by a distance

OK, try tackling point 2. I'll give you a headstart:

It is emitted at frequency
It comes back at frequency where is the relative speed that you want to detect. Find out

10. Is this like saying that miles = miles per hour times hours? Mph*h=M? (divided by 2 as it was sent and reflected and had to travel twice the distance). Point two, we can maybe tackle later, as I need to do some other things now and also get some sleep. It is an interesting discussion to me.

11. Originally Posted by mayflow
Is this like saying that miles = miles per hour times hours? Mph*h=M? (divided by 2 as it was sent and reflected and had to travel twice the distance).
yes, it is.

12. I can't quote on this forum, and I could not copy you formula to paste it either - well, I would say that if Fr=Fo(1+2v/c), then v=(cFr-cFo)/2Fo If this incorrect feel free to explain, but where do the 1 and the 2v/c come from and what do they represent?

13. Originally Posted by mayflow
I can't quote on this forum, and I could not copy you formula to paste it either - well, I would say that if Fr=Fo(1+2v/c), then v=(cFr-cFo)/2Fo If this incorrect feel free to explain, but where do the 1 and the 2v/c come from and what do they represent?
Correct. You get another "like".

The origin of the formula is complicated (and the wiki derivation is incorrect, though they get the correct final result).
Here it goes:

The standard (non-relativistic) Doppler effect teaches us that a frequency sent by the radar gun is reflected as

(wiki says but this is wrong)

where is the speed of the receiver and is the speed of the source, both wrt. the road.

For :

where is the relative speed between the radar gun and the car. Putting it all together:

14. For distance: Radar, ladar, sonar and gps use time it takes for the signals to propagate. You know the speed of the signal, you know the time ergo you can calculate the distance.
For velocity: The received signal processing can either calculate the difference in time required for successive pulses, pings or flashes to arrive or measure the change in wavelength. Something moving farther away; the ping interval and the wavelength will increase compared to the original signal. Something moving closer; the ping interval and wavelength decreases from the original signal. The faster the relative velocities the greater the decrease or increase.
In the case of gps no signal is bounced off an object. The object (gps receiver) receives timed signals from three or more satellites and processes the signal into user friendly displays of maps, bearings, compasses, velocity etc.

15. Originally Posted by pikpobedy
For distance: Radar, ladar, sonar and gps use time it takes for the signals to propagate. You know the speed of the signal, you know the time ergo you can calculate the distance.
For velocity: The received signal processing can either calculate the difference in time required for successive pulses, pings or flashes to arrive or measure the change in wavelength. Something moving farther away; the ping interval and the wavelength will increase compared to the original signal. Something moving closer; the ping interval and wavelength decreases from the original signal. The faster the relative velocities the greater the decrease or increase.
In the case of gps no signal is bounced off an object. The object (gps receiver) receives timed signals from three or more satellites and processes the signal into user friendly displays of maps, bearings, compasses, velocity etc.
I think this all makes sense to me, but I am not knowing what the ping interval thing is all about?

At xOx, what is the point 3 task?

16. Originally Posted by mayflow
I think this all makes sense to me, but I am not knowing what the ping interval thing is all about?

At xOx, what is the point 3 task?
3. For extra points, see how you would derive the equations for measuring the distance between two objects in motion wrt each other as a function of time and their relative speed.

To get you started:

At point 1, you have measured the roundtrip T
At point 2, you have derived the relative speed v, from the frequencies.
You have now everything necessary to solve question 3.

17. Not sure what that means but is this it?

I substituted x for Fr, and Y for Fo - just simpler to me.

So, now we have Y*(1+2*v/c)=x
=y*1+Y*(2V/C)=X
then Y+2V*Y/C=X
So subtracting Y form both sides= we have to get v to one side, right?

we get 2vy/c=x-y
Divide by 2y and v=c*(x-y)/2y

I think we could have then gone v = 1/2*c(x-y)/y

but also v=(cx-cy)/2y where X = Fr and Y =Fo

18. Sonar ping it is the signal emitted by active sonar.
Ping interval is the time between each successive ping. If the object that reflects the ping is moving away from the emitter-receiver, the pings will bounce back with each later one taking a longer time than the previos, as such the ping interval is said to be increased. For objects moving toward the emitter-receiver the ping interval will be shorter because each successive ping has a shorter distance to travel. Moving closer or farther are relative to each other. If the target reflecting is stationary with respect to the earth and the signal source-receiver is moving, the consequences are the same.

19. Originally Posted by mayflow
Not sure what that means but is this it?

I substituted x for Fr, and Y for Fo - just simpler to me.

So, now we have Y*(1+2*v/c)=x
=y*1+Y*(2V/C)=X
then Y+2V*Y/C=X
So subtracting Y form both sides= we have to get v to one side, right?

we get 2vy/c=x-y
Divide by 2y and v=c*(x-y)/2y

I think we could have then gone v = 1/2*c(x-y)/y

but also v=(cx-cy)/2y where X = Fr and Y =Fo
No, you answered point 2 again. Point 3 is asking you how to figure out the DISTANCE between the incoming car and the radar. You have all the info, I am sure that you can do it.

20. I think I may see what you are getting at here. We figured out how to measure distance using the measurement of the time it takes electromagnetic waves traveling at a constant speed. I will use thought construction on all of this because I only have high school math with simple algebra. So distance between two objects traveling at the same speed wrt each other (off topic slightly but I think if people on the forum spoke wrt each other, it would be a better learning and teaching environment, not to speak of more enjoyable as well) it is equal to the speed of the waves if we receive the waves from them times the time it took, and equal to 1/2 of that if we transmit them and receive their reflections, as they had to travel both ways.

Then we came up with a way to measure their velocities relative to each other because of the wavelength (frequency) changes so like we are gaining or losing distances between us commensurate with the time frame we measure the distance in reference to.

It seems to me the distance between the two objects must be in a rate of change if they are traveling at different speeds so now I would think distance is changing with time.

21. Originally Posted by mayflow
I think I may see what you are getting at here. We figured out how to measure distance using the measurement of the time it takes electromagnetic waves traveling at a constant speed. I will use thought construction on all of this because I only have high school math with simple algebra. So distance between two objects traveling at the same speed wrt each other (off topic slightly but I think if people on the forum spoke wrt each other, it would be a better learning and teaching environment, not to speak of more enjoyable as well) it is equal to the speed of the waves if we receive the waves from them times the time it took, and equal to 1/2 of that if we transmit them and receive their reflections, as they had to travel both ways.

Then we came up with a way to measure their velocities relative to each other because of the wavelength (frequency) changes so like we are gaining or losing distances between us commensurate with the time frame we measure the distance in reference to.

It seems to me the distance between the two objects must be in a rate of change if they are traveling at different speeds so now I would think distance is changing with time.
Ok,

Now try putting the above into equations. It is only one .....

22. Originally Posted by x0x
Ok,

Now try putting the above into equations. It is only one .....
You are the equation master, not me. I just simply solved a fairly simple equation you supplied me for a different variable. I did not create the equation. All I think is that the distance between two things moving at different speeds is always changing, and it is very likely in most cases that they are not constantly changing at the same rate, and they are usually not continually moving in the same trajectories.

23. Originally Posted by mayflow
You are the equation master, not me. I just simply solved a fairly simple equation you supplied me for a different variable. I did not create the equation.
Now , I am helping you to step up in your knowledge, by writing your own equations. I know you can do it.

All I think is that the distance between two things moving at different speeds is always changing,
Correct. But I made the problem simpler: the car is moving in a straight line towards the radar gun with speed v. Try writing the equation of the distance between gun and car as a function of time.

and it is very likely in most cases that they are not constantly changing at the same rate, and they are usually not continually moving in the same trajectories.
Straight line. Constant speed. Give it a try, you can do it.

24. I think that is calculus. I have not sudied calculus. If the car is approaching the radar gun at speed v, it means if it is approaching it traveling a certain distance in a certain time frame. So if the car is going 100mph, and the radar gun is relatively stationary, the distance between them must be dissapated at the same rate, namely 100mph.

25. No need for calculus for straight line constant velocity.
Calculus, however, would be extremely useful for continuous functions of varying speed, varying directions and both.

26. Originally Posted by mayflow
I think that is calculus. I have not sudied calculus.
It is basic algebra. Let me give you a hint:

-initially , the car and the radar gun are separated by a distance
-the car is approaching the radar gun with speed
What is the equation of the distance gun-car as a function of time, ?

If the car is approaching the radar gun at speed v, it means if it is approaching it traveling a certain distance in a certain time frame. So if the car is going 100mph, and the radar gun is relatively stationary, the distance between them must be dissapated at the same rate, namely 100mph.
Use symbols to produce the general formula. Physics is about generating symbolic formulas. You can do it.

27. So Xo is original distance. v is the speed closing the distance. v seems to be distance*time. At a point in time x must be equal to xo-distance traveled in that time frame. Xo-v=x?

28. Originally Posted by mayflow
So Xo is original distance. v is the speed closing the distance. v seems to be distance/time. At a point in time x must be equal to xo-distance traveled in that time frame. Xo-x/t=x? (I used x/t in place of v)
Not good. See if you can find your mistakes (there are several). Part of the learning exercise it finding your own mistakes.

30. for constant V ; V=(X2-X1)/t2-t1)

31. Originally Posted by mayflow
This is better , yet not perfect. is FIXED, it is the roundtrip time for the radar em ray. So, the correct formula is:

where is the elapsed time AFTER the ray has returned to the radar gun.

Now, knowing that, can you figure out as a function of ? This is the last step. Where was the car when the ray came back to the radar gun?

32. I am not sure if you want to add the time for the radar beam to get to the car, or get back to the car or return from the car or for the full round trip.

the time to get to the moving car I think would be Do/(v+c) and the time to return I think would be (Do-D1)/c

The whole radar ray round trip would be Do/(v+c)+(Do-D1)/c if that is correct.

In this case the t is so tiny as to be negligible for for really long range measurements could be a pretty big factor (or if the car was approaching the speed of light)

33. Originally Posted by mayflow
I am not sure if you want to add the time for the radar beam to get to the car, or get back to the car or return from the car or for the full round trip.
We already established (see question 1) that the car was at distance when the radar beam hit it.
Where with the car be when the radar beam came back to the radar gun?

34. I thought we had already factored in both ways of the radar beam in cT/2. If it were the emitted beam only, would it not be just cT?

35. Originally Posted by mayflow
If it were the emitted beam only, would it not be just cT?
Nope. When the radar beam hits the car, at the time , the car is at distance from the gun.
What is the car-gun distance at , when the radar beam completes the roundrip to the radar gun?

36. I can't buy that we would have divided cT by 2 if it did not have to travel d*2. If a car goes 60mph and we go for one hour in one direction we go 60 miles. The same equation holds true for light. That is D=tV - We would not divide that by 2 to get that answer. It is simply V*T. We divided V*T (D ) by 2 because the T was doubled because the D was doubled. The V (in this case C remained constant). In your scenario, the light had to travel two ways, so we multiplied the distance it had to travel by 2. There would be no reason to do that if it did not have to travel fro and back before we could meaure the distance it traveled, so the return of the beam is already factored in in cT/2 or it would be just D=cT

I found this online which I think corroborates what I am saying.

D= V x T
V= D/T
T= D/V

37. Originally Posted by mayflow
I can't buy that we would have divided cT by 2 if it did not have to travel d*2.
You are reverting to your old habits. Not good.

If a car goes 60mph and we go for one hour in one direction we go 60 miles. The same equation holds true for light. That is D=tV - We would not divide that by 2 to get that answer.
You are regressing rather rapidly. The roundtrip of the radar beam took the total time . Therefore, when the radar beam "hit" the car, the car was at distance. You , yourself managed to solve this at point 1. Why are you going backwards?

I found this online which I think corroborates what I am saying.

D= V x T
V= D/T
T= D/V
You need to make the effort to think for yourself. Copying and pasting stuff off the internet doesn't help you.

38. I already had thought it out for myself. That is why I decided that D is simply vT, not vT/2. The internet equations seem to confirm this to me and also go back and look when I said the the radar beams had to travel double the distance before being received. The internet equations I found and posted make sense to me according to what my own reasonings already had deduced from the various relationships and units involved, and thinking it out for myself. I only looked for equations interralating T, V and D, and these coincided with what I had reasoned out from the discussion for myself.

It is 100% unintersting to me if you think I am regressing or reverting or not making an effort. They have nothing to do with the topic.

39. Originally Posted by mayflow
I already had thought it out for myself. That is why I decided that D is simply vT, not vT/2. The internet equations seem to confirm this to me and also go back and look when I said the the radar beams had to travel double the distance before being received. The internet equations I found and posted make sense to me according to what my own reasonings already had deduced from the various relationships and units involved, and thinking it out for myself. I only looked for equations interralating T, V and D, and these coincided with what I had reasoned out from the discussion for myself.

It is 100% unintersting to me if you think I am regressing or reverting or not making an effort. They have nothing to do with the topic.
It is good that you reasoned for yourself. It is bad that your reasoning is wrong. What is worse, is that you are happy with your incorrect reasoning.

Here is the correct reasoning, for your education:

The distance car-radar gun varies with time according to the equation:

where is the distance car-radar gun when the radar beam has completed its roundtrip.
Therefore so:

See if you can figure out why . Bye.

40. If you did D = cT to begin with this may make sense, but you divided by 2 to account for the return beam.

I gave you the answer in How about X=Xo-VT. There was no need to apply the returned signal twice, and it also makes it incorrect. It may have been correct had not you not included the return signal in the original equation then tried to include it a second time.

The radar beam was not affected by the v of the vehicle. It hit at at speed C at the separation distance and returned at speed C as well. The vehicle in this time frame traveled a super teenie distance and we accounted for this by recognizing that the beam traveled double the distance already in your original formula - it is not accurate to factor the return beam a second time.

41. Originally Posted by mayflow
If you did D = cT to begin with this may make sense, but you divided by 2 to account for the return beam.

...and I explained to you why it is a wrong answer.

There was no need to apply the returned signal twice, and it also makes it incorrect.
The problem with you is not only that you cannot think right, is that you insist that your wrong thinking is right. You are truly hopeless. I tried to teach you how to think. I am sorry that the experiment failed.

42. Always make a sketch and define your variables on that sketch. This is an 8th grade algebra problem.

43. Not a problem. In your mind you were right and I was wrong, and I will give you one thing. You are correct that I have nothing more to learn from you in light of your ways of thinking here. I learned some cool stuff for a short while due to reasoning out the equations but I did come to the conclusion that you jumped around from one equation to another, and ended up being mixed up. The fact still remains that if you factored in the return beam in the first place, you can not factor it in again. The thing also is it is not a good example to use the speed of of a radar beam going to and being reflected by a moving car as really anything meaningful in comparison with the speed of a car. How fast can that car go with respect to 670 616 629 miles per hour? A radar speed screen does not likely even extend beyound whole digits. How many micro or even nano miles would that car have traveled during the radar beam round trip (or in your example as of last, 3 trips) since you think the beam has for some reason to return twice?

44. Originally Posted by pikpobedy
Always make a sketch and define your variables on that sketch. This is an 8th grade algebra problem.
I suck at geometry. When I try to discuss certain math problems with the Engineers at work, they do often draw vector diagrams and stuff to try to figure things out though. That said, what is this 8th grade algebra problem and what is the answer to it?

45. Originally Posted by mayflow
Not a problem. In your mind you were right and I was wrong, and I will give you one thing. You are correct that I have nothing more to learn from you in light of your ways of thinking here. I learned some cool stuff for a short while due to reasoning out the equations but I did come to the conclusion that you jumped around from one equation to another, and ended up being mixed up. The fact still remains that if you factored in the return beam in the first place, you can not factor it in again. The thing also is it is not a good example to use the speed of of a radar beam going to and being reflected by a moving car as really anything meaningful in comparison with the speed of a car. How fast can that car go with respect to 670 616 629 miles per hour? A radar speed screen does not likely even extend beyound whole digits. How many micro or even nano miles would that car have traveled during the radar beam round trip (or in your example as of last, 3 trips) since you think the beam has for some reason to return twice?
The problem with you is that you are one of those students who is unteachable. Not only that you are wrong but you also argue. I tried to teach you but you cannot be taught. It is your problem and since you are quite advanced in age, it is no longer fixable. Take care.

46. That kind of thing does not belong on this forum, xOx. It is a physics forum, not a name calling forum.

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