toroidal whistle, Video

[acoustics101]

Thanks. The earlier patent 4429656 actually provides for a two tone version.
http://www.delphion.com/cgi-bin/viewpat ... &MODE=fstv

A two tone (or more) version of my latter patent 4686928 would look somewhat like a gargantuan chime horn.
http://www.delphion.com/cgi-bin/viewpat ... &MODE=fstv

As for making a quintadena, the toroidal whistle is of such a large scale that it would probably totally overblow to the 3rd harmonic before getting a significant amount of 3rd harmonic in addition to its fundamental.

As to that fundamental; my 430 Hz Ultrawhistle prototype had a propagation loss of 6 dB per doubling the distance, due to the inverse square law, plus a low atmospheric absorption loss of about 0.8 dB/1000 feet. It was down by only 8 dB below the inverse square law loss at 10,000 feet! That explains the long echo on the video that was still audible over the running diesel generator from 500 feet.

A typical warning siren operating at 700 HZ also has the inverse square law loss of 6 dB per doubling the distance, plus an inverse square law loss of about 2 dB/1000 feet due to the higher frequency and its prominent high frequency harmonic multiples, which make it sound louder up close but actually detracts from it at a distance.

Welcome!

Like your hearing protection sign. It should be required that I post one next to my piano, and also display one when playing organ at any church.

Thank you for the details about your excellent efficient design.
Now to make a two-toned version! Minor third would be nice to produce a good low resultant!
Athough a major third would occur naturally in the harmonic series.

(I was able to voice a quintadena pipe with a VERY prominent 5th.
Was the most sour-sounding quintadena I had ever heard!)
Could you accomplish such an harmonic content with the toroidal whistle?

[Robert Gift]

...A typical warning siren operating at 700 HZ also has the inverse square law loss of 6 dB per doubling the distance, plus an inverse square law loss of about 2 dB/1000 feet due to the higher frequency and its prominent high frequency harmonic multiples, which make it sound louder up close but actually detracts from it at a distance.

I'm surprised there is that much greater loss at only 270 Hz higher frequency.
Is it an exponential or linear rise?
Do the high frequencies cause loss, for example through cancellation?
Or do they simply create an apparent higher dB output but are soon attenuated in the "ether"?
Thank you.

[acoustics101]

Atmospheric absorption loss is actually due to the higher frequencies transferring more of their energy as heat into the air than lower frequencies. It is much more apparent at the higher frequencies. The loss per given distance is a somewhat of a linear relationship with frequency.

A friend of mine built me a 1/20 scale working keyring version of my Ultrawhistle patent. The little whistle, no larger than an unscrewed plastic soda cap, produced 132 dB at 1 meter at a frequency of about 8.5 kHz at a pressure of 60 PSI.

By 100 feet (30 meters) it was down to 100 dB, so it was losing 2 dB/100 feet (20dB/1000 feet) in addition to the inverse square law. Although as loud as a vehicular siren, it would be inaudible past a kilometer. Those who run sound for large events are well aware of this loss. They EQ the higher frequencies to make up for it.

...A typical warning siren operating at 700 HZ also has the inverse square law loss of 6 dB per doubling the distance, plus an inverse square law loss of about 2 dB/1000 feet due to the higher frequency and its prominent high frequency harmonic multiples, which make it sound louder up close but actually detracts from it at a distance.

I'm surprised there is that much greater loss at only 270 Hz higher frequency.
Is it an exponential or linear rise?
Do the high frequencies cause loss, for example through cancellation?
Or do they simply create an apparent higher dB output but are soon attenuated in the "ether"?
Thank you.

[acoustics101]

As to the part about a source of higher frequency and more harmonic content sounding initially louder than a source of lower frequency and less harmonic content is simply due to the ear's sensitivity to various frequencies at various sound levels. This is known as the Fletcher-Munson Effect.
http://www.webervst.com/fm.htm

Past a distance of a kilometer or so, any advantages due to the Fletcher-Munson Effect will be outweighed by the high frequency losses due to atmospheric absorption, as losses per given distance (such as 1000 feet)are proportional to the frequency.

This is the very reason why you hear the crack of thunder from up close but only the rumble at a distance. It's also the reason why you hear the high pitched whine of a jet engine from up close, but only its roar at a distance.

Try making a good recording of a jet or train from up close and one from a couple miles away. If adjusted for equal volume during playback they will sound nothing alike!

Sounds not only attenuate with distance, but likewise mellow with distance. If this were not the case you would be hearing the crack or whine long after the rumble and roar instead of the other way around, due to the ear's greater sensitivity to higher frequencies.


[/quote]I'm surprised there is that much greater loss at only 270 Hz higher frequency.
Is it an exponential or linear rise?
Do the high frequencies cause loss, for example through cancellation?
Or do they simply create an apparent higher dB output but are soon attenuated in the "ether"?
Thank you.[/quote]

[AllSafe]

Good, but horribly inefficient.

[acoustics101]

A 100 HP motor driving a compressor delivering 15 PSIG at 1350 SCFM represents 74.6 kW. An output of 135 dB at 100 feet with a directivity index of 13 dB represents 20 kW acoustical energy. The efficiency is therefore 20/74.6 or about 27%. A typical 6" diameter locomotive steam whistle uses 150 PSIG at 600 SCFM (over 100 kW) to produce just 1 kW acoustical.

A loudspeaker of this efficiency would have a sensitivity rating of 107 dB/W/meter and would have to handle 75 kW to produce the same amount of sound.

Good, but horribly inefficient.

[acoustics101]

When it comes to atmospheric absorption loss, it's basically a linear rise with frequency. The 270 Hz discrepancy can be accounted for by the high amount of harmonic content in most sirens, as the effective frequency would have been higher.

Basically, you can figure a loss of about 1 dB/1000 feet for a relatively pure tone of a whistle of large scale. As the frequency gets lower, the loss is proportionally less. As the frequency gets higher, the loss is proportionally greater.

Sound propagation is always based on the inverse square law loss of 6 dB/doubling the distance, plus the frequency dependent atmospheric absorption loss. It definitely doesn't obey an arbitrary man made law such as the popular concept of losing 10 dB/doubling the distance.

I have found that you must first multiply a Federal Signal Thunderbolt's fundamental frequency by a correction factor of about 1.4 to determine the effective frequency to use in the calculation. For simplicity, 700 Hz multiplied by the correction factor of 1.4 effectively has the propagation value of a relatively pure tone of 980 Hz, so the effective difference in frequency would be closer to 550 Hz rather than 270 Hz. The ratio of 980 to 430 being about 2.28:1.

That is why a siren rated at 130 dB at 100 feet typically has a 70 dB radius of 6900 feet (1.3 miles) as opposed to a whistle rated at 125 dB at 100 feet achieving a measured 70 dB radius of 2.5 miles (13,200 feet). It's all a matter of the frequency spectrum of the signal and the frequency dependent effects of atmospheric absorption. You can't judge a sound signal's effectiveness simply by what it's rated at 100 feet.

For signals using more than one tone you need to first find the average frequency, then apply the correction factor if needed. For access to an interactive sound propagation calculator visit http://groups.yahoo.com/group/steam-whistles The Excel spreadsheet is available in our files for all members and takes the calculations out to 100,000 feet (about 20 miles). This calculator should be used in conjuction with the Fletcher-Munson Effect to determine audibility.
http://www.webervst.com/fm.htm

I'm surprised there is that much greater loss at only 270 Hz higher frequency.
Is it an exponential or linear rise?
Do the high frequencies cause loss, for example through cancellation?
Or do they simply create an apparent higher dB output but are soon attenuated in the "ether"?
Thank you.

[acoustics101]

I see I left out something very important-the frequency. The 1 dB/1000 foot loss is based on a whistle with a frequency of 500 Hz.


[quote="acoustics101"]
Basically, you can figure a loss of about 1 dB/1000 feet for a relatively pure tone of a whistle of large scale. As the frequency gets lower, the loss is proportionally less. As the frequency gets higher, the loss is proportionally greater.