Doppler puzzle

[Mark Schecter]

Hi, Amy.

Thank you! I appreciate your math abilities, and that you understand 
what I'm trying to articulate. You got it exactly, and now I'm on to 
John's reply to try to convince him! Thanks again.

-Mark Schecter

Amy Zilk wrote:
> Try dividing any two frequencies that differ by 10 cents.  It's a 
> constant.  Ten cents difference anywhere in the spectrum is a constant 
> frequency ratio of ~1.0579.  The doppler shift is found in terms of this 
> ratio and you can convert the 10 cents difference into the ratio of 
> frequencies.  It's counterintuitive for me (but not for Mark), that it 
> doesn't matter which pitch you are measuring but that's the way it works.
> az
> 
> 
> John Delmore wrote:
>> Hi Mark:
>> You're not all wet --perhaps just a little damp!!  
>>
>> It's not your speed that changes if you measure different pitches--it's the
>> size of the cents around that pitch.  So, if you're working with a
>> difference in cents from a certain frequency, you'll get a different result.
>> If the question had been phrased in Hz, you wouldn't have this problem.
>>
>> For example:
>>
>> A440+10 cents = 442.5489 Hz, a difference of 2.5489 Hz
>> A440-10 cents = 437.4658 Hz, a difference of 2.5342 Hz
>> A220+10 cents = 221.2744 Hz, a difference of 1.2744 Hz
>> And A220-10 cents = 218.7329 Hz, a difference of 1.2671 Hz.
>>
>> (I hope I got these right, I'm using the "2^1/1200:1" definition that I
>> learned thanks to this discussion)
>>
>> I'm working on the general formula to determine "cents difference" from a
>> pitch for any frequency, but that's a little more math than I have time for
>> right now...it involves the dreaded logarithmus naturalis, I think.
>>   
> Log base two, not natural logs (base e).
> az
>> I hope these minutia aren't too far afield...they're a lot more "on-topic"
>> than "that other thread"...
>>
>> John
>>
>>
>>
>>