Doppler Cents Puzzler

[Mark Schecter]

Hi, Vladan.

Thanks for your thorough explanation. This puzzler continues to 
fascinate, in part because I feel close, but no cigar, to having enough
math to figure it out myself. I do understand your point about 2^1/12 
and 2^1/1200. Out of context, I knew that. Where I get into trouble is 
connecting a mathematical representation to a "real-world" problem. I 
appreciate your (and Amy Z's and John D's) abilities in that regard. 
Sorry it took me so long to reply.

-Mark Schecter

V T wrote:
> Hi Mark,
> 
> Here is my commentary:
> 
>> The fact that the tone goes flat 10 cents when going
>> away merely confirms that the difference between the
>> stopped truck and the moving cycle produces a 10 cent
>> differential in pitch.
> 
> True.  He didn't even slow down!  Incredible.
> 
>> So I considered the pitch coming from the stopped
>> truck to be 1, and the sound to be travelling at 1100
>> feet per second. In order to reach a pitch of 2, the 
>> cycle would have to be moving at the speed of sound
>> toward the truck, to achieve a total of 2200 feet per
>> second closing speed.
> 
> Correct.
> 
>> With that thought in mind, I just calculated that a
> 10 >cent increase in pitch equalled 10/1200 of the
> speed >of sound, so:
> 
> This is where the trouble started.  It is true that
> the octave consists of 1200 cents, but the frequency
> increments (in Hz) between those 1200 tones are not
> linearly divided into equal amounts.  Note the
> following:
> 
> - One semi-tone is equal to a frequency ratio of
> 1.05946 times, but
> 
> - One cent is equal to frequency ratio of 1.0005778
> times.
> 
> Note that 1.0005778 is not equal to 1.05946/100, just
> as one semi-tone is not equal to a pitch increment of
> 1+(1/12)=1.08333 times.
> 
> You have to multiply 1.0005778 one hundred times _with
> itself_ (raise it to the hundredth power) to get
> 1.05946.
> 
> So, the principle is that each higher pitch is equal
> to the previous pitch multiplied by a constant.  That
> constant will depend on the pitch increment you
> choose.  If you want to step in semi-tone increments,
> the constant that defines two adjacent steps will be
> 2^(1/12).  If you want to step in one cent increments,
> that constant will be 2^(1/1200).  If you want to step
> in octave increments, the constant will be 2^(1/1). I
> am using "^" as the symbol for "raised to the the
> power of ...".
> 
> Some examples:
> 
> One cent sharp from A440 is:
> 
> 440*1.0005778 = 440.254228Hz
> 
> Note that the 1 cent increment up from 440 Hz is equal
> to 0.254228 Hz
> 
> Two cents sharp from A440 is equal to:
> 
> 440*1.0005778*1.0005778 = 440.508602 Hz
> 
> The difference in pitch between the last two tones is
> 0.2543745 Hz, which is slightly different from
> 0.254228.  It doesn't seem like much, but if you make
> that small error 1200 times in a row, it turns into an
> ice cream disaster.
> 
>> 10 cents higher than nominal pitch =
>> 10/1200 * (speed of sound in air)
>> or 1/120 * (1100 ft/sec) = 9.1666 ft/sec (speed of
>> bicycle)
>> 9.166 ft/sec * 3600 secs/hour = 33,000 ft/hour
>> 33,000 / 5280 (ft/mi) = 6.25 mph
> 
>> However, you arrived at 2 meters/second, which equals
>> 7200 meters/hour, 
>> which translates to 4.47 miles per hour. So would you
>> tell me how you got there? Thanks!
>>
>> -Mark Schecter
> 
> The other small difference which is not really
> relevant is that I used 345m/sec for the velocity of
> sound which is slightly off from 1100 ft/sec.
> 
> Vladan
> 
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