Highway: Deriving speed from Doppler shift

[DaveC426913]

TL;DR Summary

Can I compute the speed of a passing motorcycle from the Doppler shift of its motor?

I passed a motorcycle on the highway going the opposite direction. I know I was doing 125/km/h. I estimated that the frequency of his motor dropped by an entire octave, so that's a doubling of the wavelength.

My intuition is telling me that's extremely unlikely. I can't actually calculate how fast he was going with just that information, can I?

It seems to me, I have to know the absolute frequency of one of those tones, either shifted up or down or unshifted, yes?

I tried to mimic the tone. The up-shifted tone was about as high as I can hum without scrunching up my face. Using an online pitch analyzer I (subsequently) estimate that was about 440Hz (and the lower tone then being 220Hz).

 

[berkeman]

Remember that you have to add your relative velocities, since you were both moving...

 

[DaveC426913]

Yah, no. I'm not sure how to do this. For starters, I need to figure out what the un-shifted freq is (since that's what the equation starts with). It's logarithmically halfway between 220 and 440. which looks like it's about ... 311?

This is what happens if I plug it into a Doppler shift calculator:

If my numbers were right, I should get a consistent source velocity between the two, just with an inverted sign, correct?

So I have to ask myself where I made the (biggest) error. Either it wasn't a full octave shift I heard, or it wasn't 440Hz when it was oncoming.

Lets assume I was right about the octave and wrong about the up-shifted frequency.

My question then becomes: what would the relative velocity have to be to produce a doubling of wavelength from up-shifted to down-shifted Doppler pitches?

 

[Vanadium 50]

I need to figure out what the un-shifted fre

Yes, and you won't solve one equation in two unknowns. You need at least two frequency measurements at two angles, and more information is better. Then you can do a fit.

 

[hutchphd]

Actually I think one needs to calculate each velocity relative to the still air. I don't think you need absolute frequency
Let $\alpha$ be your Mach number and $\beta$ be that of the motorcycle then $$f_{\pm}=f_0 \frac {1\pm \alpha} {1\mp \beta}$$

 

[DaveC426913]

Yes, and you won't solve one equation in two unknowns. You need at least two frequency measurements at two angles, and more information is better. Then you can do a fit.

Right so I fix one unknown in terms of the other:

...what would the relative velocity have to be to produce a doubling of wavelength from up-shifted to down-shifted Doppler pitches?

So I set up-shifted = 2 * down-shifted.

 

[DaveC426913]

OK, so I want to set

$$\alpha = 2* \beta$$

 

[hutchphd]

I sent that prematurely. Butt keyboarding...... let me patch it up

 

[hutchphd]

Actually I think one needs to calculate each velocity relative to the still air. I don't think you need absolute frequency
Let $\alpha$ be your Mach number and $\beta$ be that of the motorcycle then $$f_{\pm}=f_0 \frac {1\pm \alpha} {1\mp \beta}$$

where $$\Delta f =f_+-f_-$$ and you want $$\frac {\Delta f} {f_0}= \frac 1 2$$

Suppose for simplicity you are both moving the same speed $\alpha=\beta$. I get $$\alpha =\frac 1 {\sqrt 3} $$
Might be correct.....

 

[hmmm27]

Yah, no. I'm not sure how to do this. For starters, I need to figure out what the un-shifted freq is (since that's what the equation starts with). It's logarithmically halfway between 220 and 440. which is about ... 305? Let me check a pitch chart....

I don't think you need to calculate the actual engine frequency, though it takes ephemeral part in the equations.

Simplifying to a stationary observer and applying reasonable assumptions :

Coming in at half the speed of sound would be up an octave : doubling frequency. $f_{in}=\frac c {c-v}f$

Leaving at half the speed of sound would be 2/3 frequency. $f_{out}=\frac c {c+v} f$

so we can get to...

$$f_{in} \times \frac {c-v} {c} = f_{out} \times \frac {c+v} {c} $$ $$\Rightarrow$$ $$\frac {f_{in}} {f_{out}} = \frac {c+v} {c-v} $$

 

[DaveC426913]

Yah. My intuition was telling me there's no way you can get a full octave change at any plausible highway speeds.

I mean he was goin' fast, but he wasn't goin' fighter jet fast.

 

[hmmm27]

Yah. My intuition was telling me there's no way you can get a full octave change at any plausible highway speeds.

I mean he was goin' fast, but he wasn't goin' fighter jet fast.

Well... for a stationary observer, that'd be ~410km/h, but for a moving one ? If I'm putting the offset in the right place, that's 285km/h = 177mph ; not inconceivable.

 

[DaveC426913]

Well... for a stationary observer, that'd be ~410km/h, but for a moving one ? If I'm putting the offset in the right place, that's 285km/h = 177mph ; not inconceivable.

Really!? Huh.

 

[hmmm27]

Really!? Huh.

The 410's solid for a stationary observer : referring to previous equation : $\frac {f_{in}} {f_{out}} = \frac {c+v} {c-v} \Rightarrow 2 = \frac {1230 + x} {1230 - x} \Rightarrow x=1230/3=410$. Pretty sure that - based on puttering around with the doppler effect a few years ago, probably here somewheres - offsetting by your 125 is that simple as long as nobody hits Mach 1 in any way, shape or form.

 

[DaveC426913]

The 410's solid for a stationary observer : referring to previous equation : $\frac {f_{in}} {f_{out}} = \frac {c+v} {c-v} \Rightarrow 2 = \frac {1230 + x} {1230 - x} \Rightarrow x=1230/3=410$. Pretty sure that - based on puttering around with the doppler effect a few years ago, probably here somewheres - offsetting by your 125 is that simple as long as nobody hits Mach 1 in any way, shape or form.

So, to be clear, 285km/h (absolute) would have produced the perceived effect.

 

[hmmm27]

I pulled the (1d) formula out of a handy orifice [edit: it jives with Wp:Doppler Effect], but the numbers add up. Your relative velocity is the biker's speed plus yours : a (constant, subsonic) relative wind velocity doesn't affect pitch shifting, so subtracting your groundspeed should give the biker's groundspeed.

 

[A.T.]

you want $$\frac {\Delta f} {f_0}= \frac 1 2$$

Suppose for simplicity you are both moving the same speed $\alpha=\beta$. I get $$\alpha =\frac 1 {\sqrt 3} $$

If both are at the same speed there is no frequency shift between them.

 

[hutchphd]

Same speed. Opposite velocity. Jeez.

 

[A.T.]

Same speed. Opposite velocity. Jeez.

Sorry I misread the OP.

 

[DaveC426913]

I pulled the (1d) formula out of a handy orifice

I saw what you did there...