Heiwa's bathroom scale experiment
- Thread starter Heiwa
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defaultdotxbe
Drunken Shikigami
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yeah, lousy gravity's always keeping me down
Oops, you left out F=ma. This is your failure! You failed again due to lack of knowledge and practical application. Your lack of knowledge in physics is why you have vaporized joints?!
My weight was 175 on the ladder, zero during the fall, then it was a bunch! Gee my weight at impact was 1519 pounds, my scale is broken, my floor is marred; had to use a force sensor scale to get the 1519 pounds! This is a great scale, it works in real time sort of! Use knee braces to a leg damper, or break a leg.
Seeing is believing; my scale reads 175 on the ladder, 0 (zero) during the free fall, and 1519 at impact! Heiwa is proven wrong with my ladder experiment! Why is Heiwa unable to get physics right? What does he do in shipbuilding, interior design, or what?
Lennart Hyland
Muse
Jonnyclueless
Philosopher
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When I stepped on the Pizza, my feet got burned. Not sure if that proves or disproves Bazant....
ElMondoHummus
0.25 short of being half-witted
He's a marine safety engineer. The company mission as stated on the linked page says:
... and more information on what they do can be found here: http://heiwaco.tripod.com/prodservs.htm.
So anyway, anything he's involved in floats. I guess as long as you stay off ships who's owners have contracted with them, you'll be all right.
nicepants
Graduate Poster
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It's your experiment. What is your reason for not conducting it? Were you planning on performing it and reporting back with your results?
The only thing you've successfully debunked is any notion that you might have any basic level of physics knowledge.
If you are truly interested in learning this subject matter I would recommend you visit your nearest university (or even community college) and speak with a physics professor for about 5 minutes and see what he/she thinks of your experiments. Most of my college profs had the job because they loved teaching, and would have no problem giving up a few moments of their time to explain a thing or two.
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A W Smith
Philosopher
Heiwa
1) have someone jump of your shoulders from 3.7 meters
2) report back to us
1) have someone jump of your shoulders from 3.7 meters
2) report back to us
HawksFan
Muse
I still think he should try my crane/house experiment.
jaydeehess
Penultimate Amazing
Heiwa started out saying that the weight is always the same and it seemed , that jumping on a bathroom scale from 3.7 meters up would not do any harm to the scale.
I note that many JREF'ers missed the point where he changed this and now wants this to be a lesson that illustrates that the object that falls will suffer injury when it lands. His point of contention is that Bazant regards the upper block as rigid and does not take into account the energy transferred into the falling block which would break connections and bend steel in the portion of the upper block. Of course Bazant does this as a level of approximation whereas Heiwa requires every detail be accounted for.
In Heiwa's 'experiment' I strongly doubt that the scale would survive and I would expect that the falling subject would experience some injury.
where does his analogy break down?
First of all the floor of your house has a much greater capacity to absorb the impact it receives via its connection to the scale than did the floorspace of the impacted level of the WTC towers.
So I offer my own thought experiment(not even Heiwa expected anyone to try jumping from 10 feet onto a bathroom scale);
Take a typical three storey house, assume its sides to face the compass major points(N,S,E,W). The mass of the third storey is quite adequately handled by the vertical structural members at the second storey level and the mass of both the third and second by the structural load bearing members at the first floor level. Now imagine that the upper storey shifts 8 inches north and 8 inches west (almost a foot diagonally). The upper storey mass now falls but where does its mass impact? Does it hit the vertical, load bearing parts of the structure? No, it instead is impacting the second floor. Was the second floor designed to hold the mass of the third floor structure? No, it was designed to hold the load that could be expected to be on the floor space and transfer that load to the vertical structural components. So rather than be able to hold the mass of the top structure the floor itself will buckle under that mass let alone the impulse force of that mass dropping the approx 2.5 meter distance. Now the upper portion will suffer some damage certainly but the floor was not designed to hold that mass that will be on it whether rubbleized or intact.
If I started filling the second storey with sand and gravel as I took away an equivalent mass from the third storey( and this would be a much less violent act than what occured at the WTC towers), the second floor would fail before there was mass of gravel equivalent to that of the third storey walls and roof. Furthermore once the second floor failed the first floor would not be able to withstand the impact or the static mass it would subsequently be encountering.
It would be rather odd as well to have three stories of the outer frame standing all by itself with no lateral support normally supplied by the floor joists.
Bazant indeed has the mass of the falling upper block impacting the floor, not the vertical structural members and in the case of the WTC towers it is not just the mass of one storey but that of at least ten stories. Were the floor spaces of the towers ever designed to carry the mass of all the floors and walls above them? No, that is why the floor trusses are basically the same for each of the 110 stories whereas the columns were heavier farther down. The columns are designed to take vertical loads.
But the impacts were not directed at the columns, it was on the trusses which tore away from the columns. The columns however absolutly require lateral support in order to support even their own mass (long column instability as per Euler IIRC) That lateral support had been supplied to the perimeter and core column systems primarily via the trusses that connected them. No trusses, no lateral support.
Unless the initial collapse was halted by impact with the next floor the falling mass would be moving even faster at the impact of the next floor down. Unless that floor is sufficient to halt the collapse it will be even faster at the next level, and so on. As the velocity increases, the available energy increases proportional to the square of the increase in velocity. The building stood no chance after the initial failure.
I note that many JREF'ers missed the point where he changed this and now wants this to be a lesson that illustrates that the object that falls will suffer injury when it lands. His point of contention is that Bazant regards the upper block as rigid and does not take into account the energy transferred into the falling block which would break connections and bend steel in the portion of the upper block. Of course Bazant does this as a level of approximation whereas Heiwa requires every detail be accounted for.
In Heiwa's 'experiment' I strongly doubt that the scale would survive and I would expect that the falling subject would experience some injury.
where does his analogy break down?
First of all the floor of your house has a much greater capacity to absorb the impact it receives via its connection to the scale than did the floorspace of the impacted level of the WTC towers.
So I offer my own thought experiment(not even Heiwa expected anyone to try jumping from 10 feet onto a bathroom scale);
Take a typical three storey house, assume its sides to face the compass major points(N,S,E,W). The mass of the third storey is quite adequately handled by the vertical structural members at the second storey level and the mass of both the third and second by the structural load bearing members at the first floor level. Now imagine that the upper storey shifts 8 inches north and 8 inches west (almost a foot diagonally). The upper storey mass now falls but where does its mass impact? Does it hit the vertical, load bearing parts of the structure? No, it instead is impacting the second floor. Was the second floor designed to hold the mass of the third floor structure? No, it was designed to hold the load that could be expected to be on the floor space and transfer that load to the vertical structural components. So rather than be able to hold the mass of the top structure the floor itself will buckle under that mass let alone the impulse force of that mass dropping the approx 2.5 meter distance. Now the upper portion will suffer some damage certainly but the floor was not designed to hold that mass that will be on it whether rubbleized or intact.
If I started filling the second storey with sand and gravel as I took away an equivalent mass from the third storey( and this would be a much less violent act than what occured at the WTC towers), the second floor would fail before there was mass of gravel equivalent to that of the third storey walls and roof. Furthermore once the second floor failed the first floor would not be able to withstand the impact or the static mass it would subsequently be encountering.
It would be rather odd as well to have three stories of the outer frame standing all by itself with no lateral support normally supplied by the floor joists.
Bazant indeed has the mass of the falling upper block impacting the floor, not the vertical structural members and in the case of the WTC towers it is not just the mass of one storey but that of at least ten stories. Were the floor spaces of the towers ever designed to carry the mass of all the floors and walls above them? No, that is why the floor trusses are basically the same for each of the 110 stories whereas the columns were heavier farther down. The columns are designed to take vertical loads.
But the impacts were not directed at the columns, it was on the trusses which tore away from the columns. The columns however absolutly require lateral support in order to support even their own mass (long column instability as per Euler IIRC) That lateral support had been supplied to the perimeter and core column systems primarily via the trusses that connected them. No trusses, no lateral support.
Unless the initial collapse was halted by impact with the next floor the falling mass would be moving even faster at the impact of the next floor down. Unless that floor is sufficient to halt the collapse it will be even faster at the next level, and so on. As the velocity increases, the available energy increases proportional to the square of the increase in velocity. The building stood no chance after the initial failure.
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roger
Penultimate Amazing
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I followed this to the logical conclusion.
I dropped a pizza box onto the scale.
And, you dumb debunkers, even though I held it right over the scale, it missed it completely! A gust of wind sent it down the driveway.
Videos prove there was wind on 9/11. The top of the towers should have blown away.
Silly debunkers.
I dropped a pizza box onto the scale.
And, you dumb debunkers, even though I held it right over the scale, it missed it completely! A gust of wind sent it down the driveway.
Videos prove there was wind on 9/11. The top of the towers should have blown away.
Silly debunkers.
But I have! In scale 1/100 of course, i.e. I just jumped from 3.7 cms height above the scale. My weight remained the same before and after impact. Only an idiot would jump from 3.7 m! But the result would be the same. The weight does not change. But probably the body. Not rigid?
But do not blame me. Blame Bazant. He suggests that your weight increases while you penetrate the scale - unhurt. You must be rigid - and the scale will be added to your mass, etc. And then you drop through the floor!
Don't forget to say hello to your neighbours on the floors below when you drop by!
ElMondoHummus
0.25 short of being half-witted
lmao!!!
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WildCat
NWO Master Conspirator
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Lennart Hyland
Muse
Still you demand people to perform the experiment in the proper scale it was ment to be tested:
http://www.internationalskeptics.com/forums/showpost.php?p=4155238&postcount=7
Since you already knew this wasnt necessary (according to your comment above), you failed to meet his argument.
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alexi_drago
Graduate Poster
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- Oct 2, 2006
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I've not done this stuff for years so there's probably some mistakes but anyone with any involvement in engineering should have some idea about this.
If you place a 1Kg mass onto a set of scales the mass will exert a force of 9.8N and there will be the equal reactive force exerted in the opposite direction from the scales.
Let's drop the 1Kg mass from 3.7m onto scales.
The time taken to fall 3.7m due to gravity can be found with:
s= u*t + 0.5*a*t*t
we know:
distance traveled (s) is 3.7
initial velocity (u) is 0.0 so we can get rid of u*t
accelearation due to gravity (a) is 9.8
3.7=0.5*9.8*t*t
or
t*t=3.7/(0.5*9.8)=0.7551
t=sqr(0.7551)=0.869 seconds for an initaially stationary object to fall 3.7m
The velocity of the object after that time is:
v=u+at
we know:
initial velocity (u) is 0.0
acceleration (a) is 9.8
time (t) is 0.869
v=9.8*0.869=8.516 m/s
Lets say that when the mass hits the scales it takes 0.1 secs for the mass to actually come to rest (all energy is absorbed through deformation and the mass doesn't bounce off) what is the acceleration of the mass duing that time?
we know:
time (t) is 0.1 secs
initial velocity before impact (u) is 8.516
final velocity after impact (v) is 0.0
a=(v-u)/t
a=(0.0-8.516)/0.1=85.16 m/s/s
And the force required to give that acceleration is found with F=m*a
we know:
mass (m) is 1Kg
acceleration (a) is 85.16
F=1.0*85.15=85.16N
So while a stationary 1Kg mass placed on a scale will exert a force of 9.8N , a similar mass dropped from 3.7m will exert an average force of 85.16N for 0.1 secs.
The time of 0.1 secs for such a collision is probably very generous and should be much lower and give a shorter acceleration time and higher force.
If you place a 1Kg mass onto a set of scales the mass will exert a force of 9.8N and there will be the equal reactive force exerted in the opposite direction from the scales.
Let's drop the 1Kg mass from 3.7m onto scales.
The time taken to fall 3.7m due to gravity can be found with:
s= u*t + 0.5*a*t*t
we know:
distance traveled (s) is 3.7
initial velocity (u) is 0.0 so we can get rid of u*t
accelearation due to gravity (a) is 9.8
3.7=0.5*9.8*t*t
or
t*t=3.7/(0.5*9.8)=0.7551
t=sqr(0.7551)=0.869 seconds for an initaially stationary object to fall 3.7m
The velocity of the object after that time is:
v=u+at
we know:
initial velocity (u) is 0.0
acceleration (a) is 9.8
time (t) is 0.869
v=9.8*0.869=8.516 m/s
Lets say that when the mass hits the scales it takes 0.1 secs for the mass to actually come to rest (all energy is absorbed through deformation and the mass doesn't bounce off) what is the acceleration of the mass duing that time?
we know:
time (t) is 0.1 secs
initial velocity before impact (u) is 8.516
final velocity after impact (v) is 0.0
a=(v-u)/t
a=(0.0-8.516)/0.1=85.16 m/s/s
And the force required to give that acceleration is found with F=m*a
we know:
mass (m) is 1Kg
acceleration (a) is 85.16
F=1.0*85.15=85.16N
So while a stationary 1Kg mass placed on a scale will exert a force of 9.8N , a similar mass dropped from 3.7m will exert an average force of 85.16N for 0.1 secs.
The time of 0.1 secs for such a collision is probably very generous and should be much lower and give a shorter acceleration time and higher force.
Grizzly Bear
このマスクに&#
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Heiwa... you're blatantly IGNORING that there's a dynamic load involved when you fall 'X' distance at the acceleration of gravity. This requires ZERO math to understand....
The weight (mg) is a force, and this force that gravity exerts on an object at rest is mg. Drop that same object from several feet in the air and you add a net acceleration and velocity, this means that the force an object exerts upon hitting the ground plane will momentarily be greater than it's weight (mg)
If someone can word this better than me go ahead BTW... (and this doesn't mean you Heiwa)
Let's take an extreme example for you. Throw a bullet at something as hard as you can with your hands. Now fire that same bullet out of a gun and compare the results ....
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jaydeehess
Penultimate Amazing
See post 110
BTW mass and weight are not the same thing.
Force and weight are.
http://en.wikipedia.org/wiki/Weight
Thus the spike on the scale when one lands on it (even from 3.7 cm) because bathroom scales measure force and assume that the acelleration is that due to gravity on Earth. In fact they aren't even true 'scales' since they use a spring. A true scale uses a known counter mass and one is comparing the force of one object against that on a known one.
A spring 'scale' will measure differently on the Moon than it does on Earth whereas a balance scale will indicate the same. Use a balance on the Moon and if you have a mass of 100 kilos and put 100 kilos on the other side it will balance. Step on a your Earthian bathroom spring scale and you will see it reports you as being 16.7 Kilos.
,, and so given that the force of an impact can greatly exceed the force due to gravity it may well be the Heiwa assumes that Bazant is saying that the weight increases. Illustrates how well(not) Heiwa understands the concepts.
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