Problem #6: Arguing Hoggs (Page 2): Then there’s the thought that I’m reading the argument and it makes no difference whether my phone is 4’ from me or 6’ from me. But I love all the...

Marg Loc: Canadian transplanted to NW Alabama

Then there’s the thought that I’m reading the argument and it makes no difference whether my phone is 4’ from me or 6’ from me.
But I love all the answers most especially Jim Bunk’s mathematical equation.

DirtFarmer Loc: Escaped from the NYC area, back to MA

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BigDen Loc: Alberta, Canada

Are they all wearing hearing aids? If not, it probably doesn’t matter.

SalvageDiver Loc: Huntington Beach CA

JimBunk wrote:

Assuming volume drops off with the square of the distance, and your standing in the middle...
@4’: 1/4’x4’ x 2 hhogs = 1/8 hhogs/ft2
@6’: 1/6’x6’ x 3 hhogs = 3/36 hhogs/ft2
So the 2hhogs will be louder

If you assumed volume drops off with the cube of the distance, the answer would be the same.

DocDav Loc: IN

Howard5252 wrote:

For this problem DO NOT send me Private Mail – just post on the thread. Post the answer / post a question / post a statement / post an opinion / post a correction / EVERYTHING is to be posted on the thread.

The Problem:

All other things being equal, which will cause the greater racket, two UHH’s arguing at a distance of four feet from you, or three UHH’s arguing at a distance of six feet from you?
I’ll be back in 48 hours to post the correct answer.

HAVE FUN!

At what decibel are they "arguing". is one or 2 or 3 women. Through what medium is the sound wave being carried. Are they talking politics or more importantly, GAS and whose lens is bigger? It's not the lens, it what you do with it. men with big lenses yell loudest I feel

Ollieboy

It's all fun until someone loses an eye.

DeanS Loc: Capital City area of North Carolina

Howard5252 wrote:

For this problem DO NOT send me Private Mail – just post on the thread. Post the answer / post a question / post a statement / post an opinion / post a correction / EVERYTHING is to be posted on the thread.

The Problem:

All other things being equal, which will cause the greater racket, two UHH’s arguing at a distance of four feet from you, or three UHH’s arguing at a distance of six feet from you?
I’ll be back in 48 hours to post the correct answer.

HAVE FUN!

It is highly unlikely, nee impossible, for this scenario to occur. We all are more than well aware that NO Hoggers on this forum have opinions of any sort, especially on camera gear, GAS, or politics. Therefore, I dismiss your postulate out-of-hand!

John from gpwmi Loc: Michigan

2 Hogs at 4 feet will be 50% loader than 3 at 6 feet. Sound drops off with the square of the distance. Except if they are arguing camera brands or RAW vs JPEQ. Then your ears will saturate at the level of pain and it won't matter.

DirtFarmer Loc: Escaped from the NYC area, back to MA

If the sound is loud enough you can produce intermodulation distortion by hitting the limits of your ear mechanism (a form of harmonic distortion caused by clipping).

Yoshicat

2/(4x4) > 3/(6x6)
Inverse square law

smf85 Loc: Freeport, IL

JimBunk wrote:

Assuming volume drops off with the square of the distance, and your standing in the middle...
@4’: 1/4’x4’ x 2 hhogs = 1/8 hhogs/ft2
@6’: 1/6’x6’ x 3 hhogs = 3/36 hhogs/ft2
So the 2hhogs will be louder

If you assumed volume drops off with the cube of the distance, the answer would be the same.

Its the square of the distance.

smf85 Loc: Freeport, IL

John from gpwmi wrote:

2 Hogs at 4 feet will be 50% loader than 3 at 6 feet. Sound drops off with the square of the distance. Except if they are arguing camera brands or RAW vs JPEQ. Then your ears will saturate at the level of pain and it won't matter.

TriX Loc: Raleigh, NC

JimBunk wrote:

Assuming volume drops off with the square of the distance, and your standing in the middle...
@4’: 1/4’x4’ x 2 hhogs = 1/8 hhogs/ft2
@6’: 1/6’x6’ x 3 hhogs = 3/36 hhogs/ft2
So the 2hhogs will be louder

If you assumed volume drops off with the cube of the distance, the answer would be the same.

Excellent. Since the speaker is effectively a point source and the sound propagates/expands in all three dimensions, I’d vote for the inverse cube relationship. This assumes that it occurs in free space (no walls or floors). If there are walls and/or floors present, it will depend on the distance from the surface(s) to the source and the frequencies of the voices. The question is whether the speaker is a point source or to what extent the voice of the speaker is directional which again is a function of the frequencies.

kymarto Loc: Portland OR and Milan Italy

[quote=Howard5252]For this problem DO NOT send me Private Mail – just post on the thread. Post the answer / post a question / post a statement / post an opinion / post a correction / EVERYTHING is to be posted on the thread.

The Problem:

All other things being equal, which will cause the greater racket, two UHH’s arguing at a distance of four feet from you, or three UHH’s arguing at a distance of six feet from you?
I’ll be back in 48 hours to post the correct answer.

HAVE FUN![/quote

Given the inverse square law and assuming equal steady state sound output, it would be two at the four foot distance from the listener, it seems to me without actually doing the math.

Howard5252 Loc: New York / Florida (now)

First a tip of the hat to JimBunk for being the first person to post the correct answer.

THE SOLUTION: Since the intensity of sound diminishes as the square of the distance from the source, the intensity of the sound made by the 3 hoggs six feet away may be represented by the number 3/36 and that of the 2 hoggs at a distance of four feet by the number 2/16. To compare the fractions, a common denominator must be used; in this case 144. The math looks like this:
2/16 = 18/144 while 3/36 = 12/144
18/144 is 1 ½ times as large as 12/144.
Thus the disturbance made by the 2 hoggs would be 1 ½ times as intense as that made by the 3 hoggs.