# Measuring height of building with barometer

I have a problem where one reading is 743mm at the top of the building and 760mm at the bottom of building. In order to find the height of the building I tried getting the difference of the two readings and multiply by the density of the mercury that it contains. I should come up with 179 m. but don't.

Now I know this can open up a can of worms, but I referenced this subject on the net, and found tons of jokes relating to this subject, but none with the solution I need. :) Granted all of the comments or jokes were valid, however it doesn't help.

Thanks

## Answers and Replies

hmm i come up with 191m.

first you need to calculate the pressure values that correspond to the given mercury lengths via p =1.36*10³*10*0.76 and 1.36*10³*10*0.743, I take g = 10 and 1.36*10³ is the mercury-density. Then use the law : p_bottom - p_top = 1.21*10*h and 1.21kg/m³ is the air-density, h is the height. I get then h = 191 meters...

marlon

Astronuc
Staff Emeritus
Basically, one needs to understand the equivalence in pressure, or differential pressure, and elevation.

Air pressure arises from the fact that a mass of air extends from the point of interest (measurement) to the top of the atmosphere. If one ascends through the atmosphere, the amount of air decreases because 1) the height of the atmosphere above is decreasing, and 2) the density of the atmosphere is decreasing.

But for this problem, let's assume that the density of air is constant.

A barometer works because the air pressure is pushing on some reference area (say a piston, diaphragm, or surface of liquid - like mercury). So the air pressure acts on that surface. Acting in the opposite direction, in the case of a mercury barometer, is the liquid mercury. For static equilibrium, the air pressure acting on the surface must be balance by the pressure exerted by the column of mercury in the barometer. That pressure ( or force acting on the area) is given simply by mgh, where m is the mass of mercury, g is accel of gravity, and h is the displacement from some reference (equilibrium) point. So when the pressure changes the height of mercury (or other fluid) changes. The denser the fluid (like Hg, density = 13.6 g/cm^3), the smaller the displacement.

So how does this relate to the air.

Well, one needs to find the height of a column of air $$h_{air}$$ acting on the reference surface in the barometer in opposition to the mercury.

It all comes down to find a balance of equivalence of force or weight which is proportional to mass, which is proportional to density and volume. But volume is area x height, so the problem simplies find to find an equivalence:

$$\rho_{air}h_{air} = \rho_{Hg}h_{Hg}$$

knowing the density of Hg = 13.6 g/cc, and the height is 17 mm, then one can find the height of the air, which is the height at which one measures the air pressure.

Try density of air at ~0.00129 g/cc,

and the answer for the height should be ~ 179 m or 179,000 mm.

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or you could take the barometer to the architect, and offer him that nice pretty barometer if he'll tell you how tall the building is. :tongue2:

marlon said:
hmm i come up with 191m.

first you need to calculate the pressure values that correspond to the given mercury lengths via p =1.36*10³*10*0.76 and 1.36*10³*10*0.743, I take g = 10 and 1.36*10³ is the mercury-density. Then use the law : p_bottom - p_top = 1.21*10*h and 1.21kg/m³ is the air-density, h is the height. I get then h = 191 meters...

marlon

yes the answer is indeed 179m.

I used 1.21kg/m³ for the air density while it should be 1.29kg/m³

marlon

Clausius2
Gold Member
marlon said:
yes the answer is indeed 179m.

I used 1.21kg/m³ for the air density while it should be 1.29kg/m³

marlon

Sorry, but the jokes are unavoidable here.

There's another form of measuring the height with a barometer. Go up to the roof of the building and throw away the barometer. Then measure the time till it crashes into the floor with your chronometer, then the height will be:

$$h=gt^2$$

:rofl: :rofl: :rofl: Marvellous...

EDIT: I read once that it was the test that Rutherford gave to Bohr (an excellent alumn at that time) for solving. Then Bohr went to the top of the building as I have described and threw away the barometer. Rutherford was impressed at first sight, but lately he kicked the Bohr ass for breaking his barometer. After some time, Rutherford recognized Bohr as his best pupil.

BTW: nobody has said it here, but the pressure of the air (and any gas) is given by the gas hydrostatics or the Boltzmann Barometric Formula:

$$P(z)=P(0)e^{-\frac{gz}{R_g T}}$$

With two pressures you can obtain any difference of heights.

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nice post clausius...

marlon

BobG
Homework Helper
Clausius2 said:
Sorry, but the jokes are unavoidable here.

There's another form of measuring the height with a barometer. Go up to the roof of the building and throw away the barometer. Then measure the time till it crashes into the floor with your chronometer, then the height will be:

$$h=gt^2$$

:rofl: :rofl: :rofl: Marvellous...

EDIT: I read once that it was the test that Rutherford gave to Bohr (an excellent alumn at that time) for solving. Then Bohr went to the top of the building as I have described and threw away the barometer. Rutherford was impressed at first sight, but lately he kicked the Bohr ass for breaking his barometer. After some time, Rutherford recognized Bohr as his best pupil.

BTW: nobody has said it here, but the pressure of the air (and any gas) is given by the gas hydrostatics or the Boltzmann Barometric Formula:

$$P(z)=P(0)e^{-\frac{gz}{R_g T}}$$

With two pressures you can obtain any difference of heights.

That's just one common variation of the story. It's doubtful that this really involved Rutherford and Bohr, since the first person known to use this story was Dr Alexander Calandra. But it is purported to be a true story of Calandra and one of his Physics students in the 1950s.

http://www.landiss.com/teaching/calandra.htm

Also, Neils Bohr is not the only Dane to win the Nobel Prize. His son, Aage Bohr and another Dane, Benjamin Mottelson, also won the Nobel Prize (1975).

Integral
Staff Emeritus
Gold Member
Clausius2 said:
Sorry, but the jokes are unavoidable here.

There's another form of measuring the height with a barometer. Go up to the roof of the building and throw away the barometer. Then measure the time till it crashes into the floor with your chronometer, then the height will be:

$$h=gt^2$$

:rofl: :rofl: :rofl: Marvellous...

EDIT: I read once that it was the test that Rutherford gave to Bohr (an excellent alumn at that time) for solving. Then Bohr went to the top of the building as I have described and threw away the barometer. Rutherford was impressed at first sight, but lately he kicked the Bohr ass for breaking his barometer. After some time, Rutherford recognized Bohr as his best pupil.

BTW: nobody has said it here, but the pressure of the air (and any gas) is given by the gas hydrostatics or the Boltzmann Barometric Formula:

$$P(z)=P(0)e^{-\frac{gz}{R_g T}}$$

With two pressures you can obtain any difference of heights.
Throwing the barometer is good, but you must take into account the time required for the sound of the crashing barometer to return to the top of the building.

Perhaps the most accurate number is had by the other popular joke solution. You take the barometer to the architect of the building and say "I will give you this beautiful barometer if you tell me how tall the building is"

Clausius2
Gold Member
Integral said:
Throwing the barometer is good, but you must take into account the time required for the sound of the crashing barometer to return to the top of the building.

Perhaps the most accurate number is had by the other popular joke solution. You take the barometer to the architect of the building and say "I will give you this beautiful barometer if you tell me how tall the building is"

Also, you must employ relativistic dynamics to have the exact movement of the barometer, and the returning of the light ray to know if the barometer has crashed yet or not is governed by general relativity :!!)

Maybe the next step of the story is hearing the architect answer:

"What the hell is a barometer?? and why are you bribing me with that?? Go away !!!"

But there's another possibility:

Take the barometer while you are going upstairs the building to measure progressively the height of the building. I mean, if the barometer is $$L$$ meters long, then by Euclides' postulates of the Geometry, the height of the building is:

$$h=k\cdot L$$ where k>1 is some number you can figure out while you are going upstairs. :rofl: :rofl:

HallsofIvy
Homework Helper
Another "standard" method for using a barometer to measure the height of a building (less destructive to the barometer) is this: go to the top of the building, tie a long string to the barometer and lower the barometer until it is ALMOST touching the ground. Now start it swinging and measure the period of the swing. You can then calculate the length of the string from its period and thus the height of the building.

ZapperZ
Staff Emeritus
.. and yet ANOTHER method would be to wait for a sunny day, measure the length of the building's shadow on the ground, and then measure the length of the barometer's shadow. Knowing the height of the barometer, one can use similar triangles to find the find the height of the building.

Of course, one should do this close to the middle of the day or else the shadow of the building will be too long for it to be practical to measure. :)

Zz.

BobG
Homework Helper
Of course, in the real world, your boss would only ask you to do this in the middle of the night. Being preoccupied by thinking up new adjectives for your boss you'll reach the roof only to realize your barometer's missing. Of course, when you sweep the flashlight over the ground in the vicinity of where you think you were standing, it will be obvious what happened to your barometer. It's still on the ground where you placed it in order to get a truly accurate reading of the pressure at ground level.

Hence, the true ingeniousness of ZapperZ's method. Tie your flashlight to the top of the roof, being careful to aim the beam so it's center (the little filament impression) barely clears the top of the barometer and then measure the angle of the flashlight relative to the side of the building.

Of course, by time you've made it down the stairs in the dark, only falling down three flights of steps in the process, you'll find a raccoon has made off with your barometer.

No problem. You can still see the image of the filaments on the ground, so you can just measure the distance from the filaments to the side of the building. Knowing two angles (provided you're not measuring the Leaning Tower of Pisa) and one side, you can find the height of the building ..... all thanks to your trusty barometer.

Clausius2
Gold Member
That's out of the problem. The problem stated originally you and a barometer. Who usually have a filament or a string 20 m long in his pocket?. BUT it is probably you'll have a chronometer in your watch. So let's think in realistic solutions or at least solutions according to the problem statement: a barometer, you and your usual possesions (it's not valid a thermometer, a laser, a 10m string, a filament, and that sorts of things that nobody except a crazy guy will carry on in his pocket).

Hmmm.......Surely there's another way..... If I get down my trousers....

I suggest a game here. As for now, the last post here suggesting a valid solution is the winner. Who wants to play?

BobG
Homework Helper
Clausius2 said:
That's out of the problem. The problem stated originally you and a barometer. Who usually have a filament or a string 20 m long in his pocket?. BUT it is probably you'll have a chronometer in your watch. So let's think in realistic solutions or at least solutions according to the problem statement: a barometer, you and your usual possesions (it's not valid a thermometer, a laser, a 10m string, a filament, and that sorts of things that nobody except a crazy guy will carry on in his pocket).

Hmmm.......Surely there's another way..... If I get down my trousers....

I suggest a game here. As for now, the last post here suggesting a valid solution is the winner. Who wants to play?

The filament is in the light bulb in the flashlight, not in your pocket. Check here for an explanation: http://www.bulbcollector.com/ubb/Forum10/HTML/000415.html [Broken]

And, for your sake, I hope this building isn't so tall that the roof level is above the electrical power lines. :surprised

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