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Jug vs eD A vs RE XXX
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<blockquote data-quote="guitar maestro" data-source="post: 300806" data-attributes="member: 548874"><p>I'm a physics major as well, and had thought about this before</p><p></p><p>BumpinDoug you are incorrect.</p><p></p><p>the actual surface area of the cone does not matter</p><p></p><p>the projected (or radiating surface area) is what matters</p><p></p><p>Proof once and for all:</p><p></p><p>(for sake of simplicity, this is in 2-d, but can be generalized to 3-d)</p><p></p><p>assume a flat cone, 10" in diameter, that is to move 1/2" in excursion</p><p></p><p>now, take a horizontal line on the x-y axis to represent the profile of the cone</p><p></p><p>so y1(x)=0 from x=0 to x=10</p><p></p><p>at 1/2" excursion, the can be defined as another piece-wise function, y2(x)=1/2 from x=0 to x=10</p><p></p><p>now we find the area that it has displaced by moving from y=0 to y=1/2":</p><p></p><p>integral(y2-y1,x,0,10)=integral(1/2-0,x,0,10)= (1/2)(10)-(1/2)(0)= <strong>5 in^2</strong></p><p></p><p>now lets use a conical profile:</p><p></p><p>the piecewise function shall be outrageously deep (5") for the sake of emphasis,</p><p></p><p>y1(x)= -x+5 from x=0 to x=5, and x-5 from x=5 to x=10</p><p></p><p>so y2(x)= -x+5+1/2, x-5+1/2</p><p></p><p>we find the area this conical profile displaces from y=0 to y=1/2"</p><p></p><p>1st-integral[(-x+5+1/2)-(-x+5),x,0,5]</p><p></p><p>the function to be integrated simplifies to simply '1/2' after you subtract in the expression</p><p></p><p>therefore 1st-integral now becomes int(1/2,x,0,5)</p><p></p><p>2nd-integral[(x-5+1/2)-(x-5),x,5,10]</p><p></p><p>the function to be integrated also simplifies to 1/2 after you subtract in the expression</p><p></p><p>so the 2nd-integral now becomes int(1/2,x,5,10)</p><p></p><p>so anyone who knows their math and physics, as bumpindoug claims, can see that we can combine these integrals into one of the form:</p><p></p><p>int(1/2,x,0,10).........(the same function integrated from 0 to 5, then from 5 to 10, is the same as 0 to 10) = <strong>5 in^2</strong></p><p></p><p>which is exactly what we integrated with a flat cone......so it doesn't matter if an outrageously deep sub has more surface area, because the expressions that account for the slope, or profile, always cancel out in the integral, and the only thing left is the distance the cone moves</p><p></p><p>that is why manufacturers (knowledgeable ones anyways) never define the parameter "Sd" as 'surface area' in their parameter sheets....they always define it as "Sd (radiating area)= xxx.xx in^2 or xxx.xxx cm^2"</p><p></p><p>if anyone actually thinks they have a valid rebuttle for my argument for doing it in R^2 instead of R^3, or some similar non-sense, think about this.....when you subtract the original function from the one shifted up in the volume integral, what do you end up with? the functional expressions cancel out and all you're left with is the distance shifted upwards (or downwards), in this case which is 1/2"</p><p></p><p>class is over</p></blockquote><p></p>
[QUOTE="guitar maestro, post: 300806, member: 548874"] I'm a physics major as well, and had thought about this before BumpinDoug you are incorrect. the actual surface area of the cone does not matter the projected (or radiating surface area) is what matters Proof once and for all: (for sake of simplicity, this is in 2-d, but can be generalized to 3-d) assume a flat cone, 10" in diameter, that is to move 1/2" in excursion now, take a horizontal line on the x-y axis to represent the profile of the cone so y1(x)=0 from x=0 to x=10 at 1/2" excursion, the can be defined as another piece-wise function, y2(x)=1/2 from x=0 to x=10 now we find the area that it has displaced by moving from y=0 to y=1/2": integral(y2-y1,x,0,10)=integral(1/2-0,x,0,10)= (1/2)(10)-(1/2)(0)= [B]5 in^2[/B] now lets use a conical profile: the piecewise function shall be outrageously deep (5") for the sake of emphasis, y1(x)= -x+5 from x=0 to x=5, and x-5 from x=5 to x=10 so y2(x)= -x+5+1/2, x-5+1/2 we find the area this conical profile displaces from y=0 to y=1/2" 1st-integral[(-x+5+1/2)-(-x+5),x,0,5] the function to be integrated simplifies to simply '1/2' after you subtract in the expression therefore 1st-integral now becomes int(1/2,x,0,5) 2nd-integral[(x-5+1/2)-(x-5),x,5,10] the function to be integrated also simplifies to 1/2 after you subtract in the expression so the 2nd-integral now becomes int(1/2,x,5,10) so anyone who knows their math and physics, as bumpindoug claims, can see that we can combine these integrals into one of the form: int(1/2,x,0,10).........(the same function integrated from 0 to 5, then from 5 to 10, is the same as 0 to 10) = [B]5 in^2[/B] which is exactly what we integrated with a flat cone......so it doesn't matter if an outrageously deep sub has more surface area, because the expressions that account for the slope, or profile, always cancel out in the integral, and the only thing left is the distance the cone moves that is why manufacturers (knowledgeable ones anyways) never define the parameter "Sd" as 'surface area' in their parameter sheets....they always define it as "Sd (radiating area)= xxx.xx in^2 or xxx.xxx cm^2" if anyone actually thinks they have a valid rebuttle for my argument for doing it in R^2 instead of R^3, or some similar non-sense, think about this.....when you subtract the original function from the one shifted up in the volume integral, what do you end up with? the functional expressions cancel out and all you're left with is the distance shifted upwards (or downwards), in this case which is 1/2" class is over [/QUOTE]
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