An off-the-wall question about sound output

[Daniel]

Forgive me if this is a stupid question, but I was thinking about this the other day. If there were two sirens (imagine they were Sentry 7V8's), identical in every way except that one had eight ports and the other had sixteen, which would be louder? The eight port siren running at full speed, or the sixteen port siren running at half speed, sounding the same pitch as the first siren? Would a sixteen blade rotor move as much air at half speed as an eight port rotor at full speed?

[Jim Z]

it's really hard to say. in theory, I'd expect the 16-port one running at 1/2 speed to be significantly quieter than the 8 port at full speed, since the rotor isn't pulling in and compressing as much air.

Would a sixteen blade rotor move as much air at half speed as an eight port rotor at full speed?

doubtful; the # of rotor ports doesn't really predict how much air flows through. that's far more dependent on the design of the impeller (the radial "blades" on the inside of the rotor.)

[SirenMadness]

Theoretically, no matter how many ports you have on a particular design of rotor, the opened area should be equal to the closed area, so at full speed either of the two sirens should be about the same in terms of sound output. At half-speed, I'd imagine you'd get much less compression, and so the output won't be that great. As said, the output would really depend on the arrangement of the blades moving the air if nothing else. Then again, more ports might also mean less time for compression before the air gets pushed out.

[Justin]

I'd be leaning towards the eight port 7V8 personally. You'd think the rotor would be designed to run at 100% duty cycle (DC), where the impellers can suck up more air since it would be working twice as hard (or presumably moving twice as fast) as 16-port 7V8 working at 50% DC. Thinking about this a bit more, it seems the 16-port 7V8 would be horribly inefficient at 50% DC as the air would be moving slower and possibly allowed to escape before passing through the vanes than at 100% DC...* I guess it depends on the rotor design and how well it can trap air through centrifugal (centripetal?) force. To be objective, the 16-port rotor would have more surface area on the vanes to force air, perhaps being as efficient as the 8-port 7V8 at 100% DC...

* I'm thinking this because 'self-supplying' sirens quiet down as the rotor slows. Blower fed (or fan-forced) sirens don't suffer this problem as the air supply is constant and redundant to the rotor speed (at least in theory).

If you were to factor in dual tone rotors, you would have to take into account losses where the ports don't completely line up on one tone but do on the other tone in a single rotation. Since this hypothetical is about a single tone rotor, this doesn't matter.

A stupid question this is not. It would be interesting to see the physics of this question in action. And somewhat difficult to answer in itself, depending on how you look at it.

[SirenMadness]

I guess it depends on the rotor design and how well it can trap air through centrifugal force. To be objective, the 16-port rotor would have more surface area on the vanes to force air, perhaps being as efficient as the 8-port 7V8 at 100% DC...

Theoretically, the rotor only traps air when the port is open during substantial speed and there is suction at that split-moment; that's how the air gets drawn in... hence why you have sound coming out of the intake as well, as the incoming air replaces the leaving air with an equal burst of pressure... also the reason why having both the solenoids closed on a running blower-siren would probably cause the bleed valve to open. I'm guessing that a siren with too many ports will not allow for maximum expansion of the air around the siren, as you have two shorter bursts of air instead of one large burst that would allow for more air to be pushed out without interruption. That's probably why sirens with a lower pitch propagate better than the higher sirens. Assuming identical overall dimensions and height of ports, both rotors should achieve approximately the same output in pressure but the lower one would be the better choice, simply due to its frequency. In other words, identical power output does not always mean identical performance.

[acoustics101]

Considering that nothing is changed but the number of ports and that the open to closed ratio and total radiating and flow area are the same, reducing the speed would reduce the pressure and thus the output on a siren with a common shaft for the blower and chopper.

In the case of a siren with an external blower, such as a Thunderbolt or Hurricane, output could be maintained when running the chopper at a slower speed if using a horn with a low enough cutoff frequency.

Since propagation is a function of both frequency and output, the Sentry 7V8 may or may not perform as good at a distance at the lower speed, but the Thunderbolt definitely would perform even better.

Forgive me if this is a stupid question, but I was thinking about this the other day. If there were two sirens (imagine they were Sentry 7V8's), identical in every way except that one had eight ports and the other had sixteen, which would be louder? The eight port siren running at full speed, or the sixteen port siren running at half speed, sounding the same pitch as the first siren? Would a sixteen blade rotor move as much air at half speed as an eight port rotor at full speed?

[ver tum]

The 16 port Sentry running at half speed would probably have half the sound output of the 8 port one, because the ports on the 16 port one would be smaller. Same as the air velocity on a 3 bladed fan, and a 6 bladed fan of the same size and blade pitch running at half the speed. In this situation, less ports and higher RPM = higher sound output, same as less blades and higher RPM = higher air velocity on a fan, given that the rotor is the same size, and the blades/ports have the same pitch. Someone please correct me if I'm wrong.

[acoustics101]

It would definitely be less than half the output running it at half the speed. That would decrease the output by only 3 dB. Generally it takes about four times the power to run a machine with wind resistance at twice the speed. This would indicate a loss of at least 6 dB. If the frequency dropped below the cutoff frequency of the horn(s), the loss would be even greater!

Just look at how much more HP it takes to run a race car at twice the speed! In the case of a siren, not only would the pressure to the chopper drop, but also the flow in CFM. A way to tell for sure how much difference in power is involved would be to measure the motor current vs the shaft RPM, but this might by difficult to do without a frequency controller.
http://en.wikipedia.org/wiki/Variable-frequency_drive

Remember that I originally said "Considering that nothing is changed but the number of ports and that the open to closed ratio and total radiating and flow area are the same, reducing the speed would reduce the pressure and thus the output on a siren with a common shaft for the blower and chopper."

The 16 port Sentry running at half speed would probably have half the sound output of the 8 port one, because the ports on the 16 port one would be smaller. Same as the air velocity on a 3 bladed fan, and a 6 bladed fan of the same size and blade pitch running at half the speed. In this situation, less ports and higher RPM = higher sound output, same as less blades and higher RPM = higher air velocity on a fan, given that the rotor is the same size, and the blades/ports have the same pitch. Someone please correct me if I'm wrong.

[SirenMadness]

The power output of a siren in terms of air movement depends on the amount of trapped air and the pressure a rotor attains during the closed cycle versus the time the air has to escape during the open cycle, which is why the ports are always equal in width to the walls of the rotor - too big a port, and you'll lose compression; too small a port, and you'll lose maximum attainable air output. However, since the supposed rotor with sixteen ports has half the compression time between bursts it only loses about half the air in return per port, displacing the air at about the same output speed per chop as the eight-port rotor, assuming both are running at the same speed, considering both rotors will achieve approximately the same negative pressure on the inside.
Yes, there has to be a certain minimum number of vanes for efficient suction, but I don't think eight-versus-sixteen would make too much of a difference in suction, as long as the volume between the vanes matches the rate of chopping.
Going back to my second post as well as Richard's first post, talking about propagation, would it be possible for a humongous sixteen-port rotor to produce nearly the same propagation as a normal eight-port rotor half its size due to the fact that you will have nearly the same amount of air exiting per chop? I'm thinking that since the high rotor will have shorter chops regardless, its double size would make up for the loss of output time. It may be extremely difficult and inefficient but would be fun to achieve better penetration with a high pitch by ridiculously increasing the physical size of the chops, though there really would be no point to it in the end.

Here's a more practical analogy as to what I mean in general. Let's say you have two identical plastic bowls; you place one on boiling water and the other on a working hotplate. Both the water and the hotplate are set to the exact same temperature, so both the water molecules and the iron atoms in the hotplate have the same relative average power acting on them individually. At about boiling point, you'll notice the bowl floating on the the water start to move up and down while the bowl on the hotplate is just standing there. Both the bowls are receiving about the same amount of power from their own sources of heat, yet one bowl is moving. Why? Because the water particles are allowed to move longer distances even at the same temperature as the iron and therefore move the bowl. The iron just vibrates randomly about a few nanometers back and forth and so does not move anything. You could double the temperature on the hotplate and you'll still move nothing. The hotplate represents the high rotor, while the water represents the low rotor.

[acoustics101]

In regard to the performance, power and output are all in the RPM. It's basically running the same centrifugal compressor at a different speed. The blower effect is where the power that is converted to acoustical energy at the ports of the chopper comes from. Even though the frequency would be the same at half the speed with twice the number of ports, the pressure and flow in CFM to the ports would not, considering that everything else is the same. Since the frequency is the same, the louder siren with fewer ports will carry further.

In regard to propagation, lower frequencies lose fewer dB with distance than higher frequencies. Propagation is a combination of the inverse square law loss of 6 dB/doubling the distance + a frequency dependent loss of x dB/1000 feet, along with the initial output. A pure tone of 500 Hz has a loss of about 1 dB/1000 feet. Since sirens do not produce pure tones, the real world loss is higher than this and increases with frequency. For this reason a higher dB rated siren with a higher frequency tone will decay faster with distance than a lower dB rated siren with a lower frequency tone.

The carrying power or propagation of two similar sirens of the same frequency, such as the proposed 8 port and the 16 port running at half speed, depend only on their initial output. The 8 port running at full speed would produce the higher initial output, due to it producing more blower power. I don't believe I can put it any simpler than this.