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Post by jaydeehess on Jan 25, 2007 14:33:47 GMT -4
I see that BAUT is going to be closing down 9/11 threads and concentrating on astronomy and space instead.
There is one thread there that concerns terminal velocity of a piano dropped from the top of the WTC. Prior to 9/11 this would not have anything to do with conspiracies but the thread may disappear. ,,, and so I bring you all the math question For the terminal velocity of a piano I get;
Fg=ma Fd=0.5 r A Cd v2
Fg=Fd at terminal velocity
For a piano (grand piano style) use;
A=2 m2 m=200 Kg Cd=1.14(that of a prism/triangle) r(rho) for a lower part of the atmosphere = 1.2 Kg/m3 a=9.8m2
So ;
200(9.8)=0.5(1.2)(2)(1.14) v2 v=sqrt(1433) v=37.8 m/second = 136 kilometers per hour = 85 MPH
This assumes that the piano does not tumble but remains in a flat fall.
[/quote]
But then it gets interesting when one wants to find the time to reach terminal velocity
a(t) = g - (0.5*rho*S*Cd*v(t)^2)/m
a(t) =9.8- (0.5*1.2*2*1.14)v(t)^2)/200 a(t) = 9.8- (0.007)v(t)^2 Self check, substituting a=0 at terminal velocity 0 = 9.8 - (0.007)v(t)^2 v= sqrt(1400) =37.4 m/s (given rounding errors I think this is close enough to the above calc of terminal velocity)
So;
dv/dt = 9.8 - (0.007)(ds/dt)2
after that I get lost as it has been 30 years since calculus classes.
This bums me out since I realize this is a first year problem but I cannot seem to understand how I integrate this to find s(t) and t_ as Nicolas lays out
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Post by AstroSmurf on Jan 25, 2007 18:44:18 GMT -4
Unfortunately, this is a nonlinear differential equation, so I doubt you can solve it analytically. Time to start up your favourite math program and have it calculate it numerically.
Remember the boundary conditions s(0) = building height and v(0) = 0. That should give you a particular solution to work with.
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Post by jaydeehess on Jan 25, 2007 23:36:31 GMT -4
Unfortunately, this is a nonlinear differential equation, so I doubt you can solve it analytically. Time to start up your favourite math program and have it calculate it numerically. Remember the boundary conditions s(0) = building height and v(0) = 0. That should give you a particular solution to work with. Actually wouldn't s (final) be the building height if s = distance traveled? so s(0) = 0 We do have the condition that a = 0 when v = 37.4 m/s which is also when we want to know the distance and the time. After this time the velocity remains at 37.4 m/s (or perhaps more properly v approaches 37.4 asymptotically)
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Post by turbonium on Jan 26, 2007 1:02:09 GMT -4
200 kilos? Too light to be a grand piano - even a baby grand! Let's assume it's a "baby-baby" grand!
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Post by AstroSmurf on Jan 26, 2007 10:12:58 GMT -4
Quoth jaydeehess: Actually wouldn't s(final) be the building height if s = distance traveled? so s(0) = 0
Right, my bad - I used s as the height above ground, but that would have the signs wrong, and using s(0)=0 is more generic. The only difference is what value of s you use to get your final time. That too will have to be derived numerically. Joy.
We do have the condition that a = 0 when v = 37.4 m/s
That's not a boundary condition per se - it follows from the definition of a that's already been stated, so it's redundant. And it's not a unique time in this case, since terminal velocity will hold more or less constant once equilibrium between the forces has been reached. You might be interested in how long it takes to reach terminal velocity, but at best you can get an approximate time, since the equation isn't amenable to analytical solutions.
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Post by jaydeehess on Jan 26, 2007 12:11:44 GMT -4
Quoth jaydeehess: Actually wouldn't s(final) be the building height if s = distance traveled? so s(0) = 0Right, my bad - I used s as the height above ground, but that would have the signs wrong, and using s(0)=0 is more generic. The only difference is what value of s you use to get your final time. That too will have to be derived numerically. Joy. We do have the condition that a = 0 when v = 37.4 m/sThat's not a boundary condition per se - it follows from the definition of a that's already been stated, so it's redundant. And it's not a unique time in this case, since terminal velocity will hold more or less constant once equilibrium between the forces has been reached. You might be interested in how long it takes to reach terminal velocity, but at best you can get an approximate time, since the equation isn't amenable to analytical solutions. After I posted I tried working with the a=0 when v=37.4 and of course all i could get from that is that , yes, when v=37.4 that indeed a=0. As you say, redundant. In my now hopelessly obvious limited math, I think of this as just one point on a plot of the function a= dv/dt. There is the boundry of s=0 since there is no such thing in this case as a negative velocity and thus no negative distance from the origin. But technically the only upper boundry is infinity, as far as the math is concerned although reality is that the planet gets in the way.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Jan 26, 2007 12:19:31 GMT -4
When all else fails, perform a time-step simulation. I used jaydeehess' values for mass, area, and Cd. If dropped from 410 m (about the height of the WTC upper floor) to sea level elevation, it would take 13.5 seconds to reach the ground. Here are the results in tabular form.
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Post by jaydeehess on Jan 26, 2007 12:33:35 GMT -4
200 kilos? Too light to be a grand piano - even a baby grand! Let's assume it's a "baby-baby" grand! www.pianoplus.co.uk/yamaha-piano/grands.htmlLength: 161cm (5'3") Width: 149cm Height: 101cm Weight: 285kg Finished in Black Polyester You will note that terminal velocity and the time to reach terminal velocity will increase with greater mass, while the time to impact will decrease with greater mass, right? I seem to see that my surface area is almost twice that of this edxample too, further increasing the term. velocity. I used an approx of a triangle for the shape of a grand piano so surface area would then be (1.61m X 1.49 m)/2 = 1.19 sq. m. This thing gets closer to a vacuum free fall time the more one looks at it. At another forum a person with much greater math skills came up with a 13.7 seconds to impact using 417 meters drop and the specs I originally put out for a piano. Now if one uses 1.25 sq. m. for surface area and 285 Kg for mass that time will decrease. Further to this are other factors, first of all if the object tumbles then it will spill air from under it thus reducing fall time, second would be the effect of the material of the surface. In this case, a grand piano, the surface could be expected to be quite wind 'slippery'. The later can be ignored since of course the original reason this was brought up was to use a piano as analogous to a structural steel column which is not smooth. On the other hand steel columns are much denser than a piano and would therefore be less affected by air resistance. If the idea was to prove that air resistance would slow the collapse its a bust. A multitude of photos and videos show that the free debris falling away from the buildings outpaced the collapse. If a steel member took 13 seconds to fall through nothing but air and hit the ground then the videos show that the collapse took longer than that and if a steel member took 13 seconds to fall through nothing but air then that puts the error to the buildings falling in less time than that.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Jan 26, 2007 12:53:51 GMT -4
This thing gets closer to a vacuum free fall time the more one looks at it. At another forum a person with much greater math skills came up with a 13.7 seconds to impact using 417 meters drop and the specs I originally put out for a piano. Now if one uses 1.25 sq. m. for surface area and 285 Kg for mass that time will decrease. Using the revised numbers of 1.25 m 2, 285 kg, and 417 m drop height, I get a fall time of 11.2 seconds.
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Post by jaydeehess on Jan 26, 2007 14:09:57 GMT -4
This thing gets closer to a vacuum free fall time the more one looks at it. At another forum a person with much greater math skills came up with a 13.7 seconds to impact using 417 meters drop and the specs I originally put out for a piano. Now if one uses 1.25 sq. m. for surface area and 285 Kg for mass that time will decrease. Using the revised numbers of 1.25 m 2, 285 kg, and 417 m drop height, I get a fall time of 11.2 seconds. Then in order to make a claim of a 10 or 11 second collapse time the CT crowd now must come up with a way to explain how free debris falling outside the building not only out paced the collapse progression but how both the collapse and the free debris managed to hit the ground in such a short time. It would require some other downward acting force other than gravity. Perhaps [ct_mode]there were explosives on the roof that detonated as the top became shrouded in dust thus propelling the building down via the extra force on the hat truss[/ct_mode]
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Jan 26, 2007 17:34:16 GMT -4
How fast an object falls depends on its ballistic coefficient, which is a measure of a body’s ability to overcome air resistance. Ballistic coefficient combines all the factors relating to size, shape and mass into a single number. It is given by the equation, Cb = M/(A*Cd) Where M is mass, A is the cross-sectional area, and Cd is the drag coefficient. For the piano example, the ballistic coefficient is, Cb = 285/(1.25*1.14) = 200 All objects with the same ballistic coefficient will have the same terminal velocity. Below is the time it takes to fall from 417 m to 0 m mean sea level elevation for various ballistic coefficients: Cb Time(s) 25 22.1 50 16.6 100 13.2 150 11.9 200 11.2 250 10.8 300 10.5 400 10.2 500 10.0 600 9.9 700 9.8 800 9.7 1000 9.6 For comparison, the ballistic coefficient of a basketball is about 25 and that of a bowling ball is about 400. A cannonball can be upwards of 1,000.
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Post by Apollo Gnomon on Jan 26, 2007 17:51:20 GMT -4
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Post by turbonium on Jan 27, 2007 5:48:26 GMT -4
If the idea was to prove that air resistance would slow the collapse its a bust. A multitude of photos and videos show that the free debris falling away from the buildings outpaced the collapse. If a steel member took 13 seconds to fall through nothing but air and hit the ground then the videos show that the collapse took longer than that and if a steel member took 13 seconds to fall through nothing but air then that puts the error to the buildings falling in less time than that. The debris was forcefully ejected from the building - in every direction. Any debris ejected downward would be able to travel down faster than objects in a pure free-fall. You can't measure the speed of collapse for the tower by comparing it to the debris speed.
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Post by PhantomWolf on Jan 27, 2007 9:55:20 GMT -4
The debris was forcefully ejected from the building - in every direction. Any debris ejected downward would be able to travel down faster than objects in a pure free-fall. You can't measure the speed of collapse for the tower by comparing it to the debris speed.
However a number of parts can be shown to have started with an upwards component to their velocity and even with this, were still ahead of the collapse point. You can't say that they were travelling faster than free-fall.
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Post by jaydeehess on Jan 29, 2007 0:30:18 GMT -4
If the idea was to prove that air resistance would slow the collapse its a bust. A multitude of photos and videos show that the free debris falling away from the buildings outpaced the collapse. If a steel member took 13 seconds to fall through nothing but air and hit the ground then the videos show that the collapse took longer than that and if a steel member took 13 seconds to fall through nothing but air then that puts the error to the buildings falling in less time than that. The debris was forcefully ejected from the building - in every direction. Any debris ejected downward would be able to travel down faster than objects in a pure free-fall. You can't measure the speed of collapse for the tower by comparing it to the debris speed. Anything ejected with a downward component would by definition hit the rest of the building first.
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