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 evan-e-cent
   
Gender: Male
 |  | Calculating the heights of cranes masts buildings and this massive cooling tower
< on 9/26/2009 8:19 PM >
|  | | | I thought the discussion of how to estimate the height of a structure might be of interest to other explorers. Has anyone bothered to do this before? Do you have examples you could add or alternative methods you have used?
This discussion took place in an awesome thread about climbing a massive cooling tower at an abandoned nuclear power plant.
Check it out:
http://www.uer.ca/...d=72439&currpage=1
These photos might tempt you to look closer:
Photo by AnAppleSnail
Photo by AnAppleSnail
Photo from Google Earth
Posted by evan-e-cent
I have been trying to estimate how high that gigantic cooling tower is. From the photo of people inside the base of the tower, the open triangle base appears to be four times the height of a person or 25 feet high. The tower is about 16 times the height of the base. So that makes the whole tower about 400 feet tall. But that is a crude estimate. I wonder whether there any construction records. Anyone want to count the number of rungs on the ladder next time!
Posted by Air 33
400 feet is 37 stories. I think this is taller. It just could be optical illusion because there is nothing else around, but yeah.
Posted by evan-e-cent
I agree it looks higher, but I just checked the satellite pictures and the length of the shadow is 40 times the length of the shadow seen on typical surrounding buildings so about 40 stories high appears to be correct.
Posted by Rankin Cycle
The cooling tower for a unit this size would be around 350 to 425 feet tall depending on design conditions, and the amount of fill installed in it (this cooling tower never had the fill installed prior to the cancellation of the plant). This is a quite typical size cooling tower.
Posted by evan-e-cent
And here is a better method. The satellite pictures give a linear scale so you can accurately measure the diameter of the base as 320 feet. A distant view of the tower shows the height is approximately 1.25 times its base diameter. 320 x 1.25 = 400 feet ! The camera tilt may make a slight difference so it could be 425 feet.
Post by Rankin Cycle
Well remember, the tallest one in the world is in Germany at a coal fired plant, and its 600 feet tall, and it also doubles as the chimney....
Post by Rankin Cycle
it was one tower per 1,200 MW. I haven't had time to look into it further (outage at my plant...) but I would say that tower is indeed right around 500 feet.
Posted by evan-e-cent
CORRECTING FOR CAMERA TILT
I seem to be getting obsessed with estimating the height. Can't leave it alone! I enjoy the challenge of trying to solve problems like this. I hope a few others might find the discussion useful for other climbs. This should be the last step in the process.
The last estimate I did uses the height to width ratio measured from a distant view of the tower. This would be accurate if the camera was level with the midpoint of the tower (or the camera was positioned at infinity) so that the camera would not have to be tilted upwards to take the picture. Unfortunately the camera is at ground level and is tilted upwards and this has the effect of foreshortening the image (unless you have a fancy camera with a tilting film plane).
I can correct for this but we do not know how far the cameraman (or camerawoman) was from the tower. If I assume it was 1000 feet and the tower is about 500 feet high the sides of the triangle would be 1000, 500 and the hypotenuse 1500 according to Pythagorus. This triangle shows that the camera would be tilted up 30 degrees from horizontal. Skipping some of the geometry the ratio of shortening is 1000/1500 or 2/3.
So the image of the tower would look 2/3 its actual height. A bigger factor than I expected and the true height of the tower would be 400 x 3/2 or 600 feet. In actual fact the camera may be a little further away so I would just have to say it could be anywhere from 400 to 600 feet and most likely close to the higher end.
Before I drop this subject, can we use the technique to measure other objects using a camera. A surveyor would use a theodolite, an optical instrument used to measure angles.
MAKE YOUR OWN THEODOLITE
You can make your own crude theodolite which could be quite accurate. All you need is a drinking straw to act as the sights attached to a protractor and a weight on a string. I was going to explain how when I found plenty of articles about it by a Google search. Heres a great title:
The dangerous and daring blog for boys and girls!
http://dangerous-d...-high-is-that.html
You measure the distance you are from the object D. Then look through the straw to the top of the object. Measure the angle of the straw from horizontal. Plug them into the formula Height = tan of the angle. Simple.
USE A SHADOW TO ESTIMATE HEIGHT
Here is another middle school project you might remember and it is even simpler. Push a stick into the ground so that it is vertical. Measure its height and the length of the shadow. Calculate the ratio of height over shadow length. Measure the length of the shadow of the tower or crane. Multiply by the ratio and you get the height of the object. This works fine unless you always explore on moonless nights!
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MindHacker
 
Location: Suburbs of DC
Gender: Male
If you spot a terrorist arrow, pin it to the wall with your shoulder.
| | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 1 on 9/27/2009 2:02 AM >
| | | | My favorite method is "count the seconds"
.5*9.8*(seconds)^2 = height in meters. But with terminal velocity the theodolite would be more accurate.
Oh, and thanks for the crosspost, I almost missed this!
"That's just my opinion. I would, however, advocate for explosive breaching, since speed and looking cool are both concerns in my job."-Wilkinshire |
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Trap
 
Location: Dayton, Ohio
Gender: Male
Son, I am disappoint
| | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 2 on 9/30/2009 6:40 PM >
| | | | Man, I really need to get out and explore with Apps sometime!
Posted by Send4Help:
ITS EIGHT FUCKING THIRTY!!
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AnAppleSnail
Location: Charlotte, NC
Gender: Male
ALL the flashlights!
| |  |  | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 3 on 9/30/2009 6:52 PM >
| | | | Posted by Shutchatrap
Man, I really need to get out and explore with Apps sometime!
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Of course you do. When are you free? We're doing tall things tonight.
Achievement Unlocked
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Shutter
Location: Raleigh NC
Gender: Male
| | | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 4 on 9/30/2009 6:57 PM >
| | | | you could do a rough estimate on site using yourself, your shadow, and the structure's shadow using simple geometry.
As your measurement of the shadow for a tall structure would be iffy (unless you have perfect pacing) it would have a significant deviation, but its good enough for bragging rights.
Edit: wow...I completely missed that you said that in the original post. Carry on. Carry on.
[last edit 9/30/2009 6:59 PM by Shutter - edited 1 times]
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uLiveAndYouBurn
Location: Beyond
Anarchocommunist
| | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 5 on 10/5/2009 1:57 PM >
| | | | If its a permanent structure, the FAA has already done the hard work for you. Paper Sectional charts are about 5 bucks each and usually cover thousands of square miles. These charts delineate every single structure over 150 feet above ground level with a triangle like icon. Near the bottom of the icon is the height, written first in Mean Sea Level, and in actual height above ground in parentheses.
So if you have a cooling tower like the one in question you'd see something like this:
........................
............/\..........
.............1070.......
.............(540). .....
........................
Online version is available at http://skyvector.com/
"Aint nothin' to it but to do it"
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Bettner12
Location: Pittsburgh, PA
Gender: Male
I Win!
| | | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 6 on 10/15/2009 3:42 AM >
| | | | looks like the concrete was laid in 8' rings judging from the portrait in the other thread. just like a tree, count the rings?
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MindHacker
Location: Suburbs of DC
Gender: Male
If you spot a terrorist arrow, pin it to the wall with your shoulder.
| | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 7 on 10/15/2009 4:21 AM >
| | | | Posted by Bettner12
looks like the concrete was laid in 8' rings judging from the portrait in the other thread. just like a tree, count the rings?
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Concrete was a continuous pour for strength, the rings are reinforcement... but otherwise I think your theory is sound.
"That's just my opinion. I would, however, advocate for explosive breaching, since speed and looking cool are both concerns in my job."-Wilkinshire |
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evan-e-cent
Gender: Male
| | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 8 on 11/3/2009 3:36 PM >
| | | | Photo #1 shows someone at the top of the tower. DJ Craig thought it was him, but now everyone who climbed the tower believes the photo is of them. We need iris pattern recognition to figure this one out.
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makman
Location: Rochester, NY
Gender: Male
I live alone with a criminal
| | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 10 on 11/3/2009 5:51 PM >
| | | | Were you there in person, you could try some of these methods:
http://www.gomilpitas.com/humor/119.htm
Be careful, not safe.
"Urbex- so much fun that it should be illegal." |
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\/adder
Location: DunkarooLand
Gender: Male
I'm the worst of the best but I'm in this race.
| | | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 11 on 11/3/2009 8:56 PM >
| | | | Posted by evan-e-cent
Photo #1 shows someone at the top of the tower. DJ Craig thought it was him, but now everyone who climbed the tower believes the photo is of them. We need iris pattern recognition to figure this one out.
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It is of course, none of them.
But in fact the pissed off ghost of J. Robert Oppenheimer. Angry because even though we have managed to come up with a positive use for his work, bureaucrats still fuck it up.
"No risk, no reward, no fun."
"Go all the way or walk away"
escensi omnis... |
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\/adder
Location: DunkarooLand
Gender: Male
I'm the worst of the best but I'm in this race.
| | | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 12 on 11/3/2009 9:35 PM >
| | | | If you know the height, and the structure has aviation (orange & white) painting on it you can estimate how high up you are, as you are climbing it.
For example let's say I'm on a 111m tower
111m ~ 365'
365 falls under the 1/7th category. It will have seven bands and band should be 1/7th of the height.
365/7 ~ 52' per band. If you were to climb up to the start of the fourth band (the top of the 3rd) You would have climbed 158'.
Another important thing to note is that according to the FCC, the band colors must start and end in orange.
A seven band tower would have the following paint scheme.
Orange, White, Orange, White, Orange, White. Orange.
It is also important to note that these are not exact numbers, according to the FCC each band should be approximately 1/7th of the tower height.
But it is useful for measuring how high you have climbed, how much is remaining to the top.
It is also useful because most cell towers have "safe working heights for RF exposure" if you wanted to stay below that height
196 foot tower, safe rf level height is say 162 feet.
196/7=28
Each band is 28 feet tall. 168/28 = 6. So you could climb to the top of the sixth band (the last white band) before your risk of rf exposure would be potentially hazardous to your health.
Of course these are just approximations, but it's a good thing to know.
edit: my source is here
https://oeaaa.faa....t/AC70_7460_1K.pdf
[last edit 11/3/2009 9:44 PM by \/adder - edited 1 times]
"No risk, no reward, no fun."
"Go all the way or walk away"
escensi omnis... |
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Alx_xlA
Location: Edmonton, Alberta
Gender: Male
| | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 13 on 11/12/2009 4:16 AM >
|  | | | Hooray for Trig! Move a known distance from the base of the tower and measure the angle from the ground to the top. Then calculate as follows: Tan*(Angle in degrees)*(Distance)
Knowledge is power. Power corrupts. Study hard and be evil. |
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evan-e-cent
Gender: Male
| | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 14 on 11/27/2009 6:25 AM >
| | | | You could of course measure the height of a tower by dropping an object from the top and measuring how many seconds it takes to reach the bottom. Or you could jump off the top and start counting, a bit like the you do when the anesthesiologist puts you to sleep for surgery. A recent conversation with DJ Craig by text messages while he was climbing may be relevant to this:
DJ Craig asked
How many seconds does it take to fall 650 ft?
That will take me a while to calculate the formula by calculus. Actually you should be able to do it. Rate of acceleration is
dV/dt = g
g is 32 feet per second per second.
Integrate this formula twice and you get the equation.
first integration gives
V = g.t
Second integration gives
X = g.t.t /2
Where X is distance fallen
rearrange the formula to extract t
t = SQRT( 2 . X / g)
6.37 sec to fall 650 feet
5.8 sec to fall 540 feet
The first equation can be used to calculate how fast you are falling after a given number of seconds.
V = g . T
If t is 5.8 seconds after falling 540 feet then your speed is
32 x 5.8 = 185.6
We are working in units of feet and seconds so velocity would be in feet per second. The terminal velocity of the human body is 120 miles per hour so I want to convert to MPH. There are 5280 feet in a mile.
1 MPH is 5280 feet per hour or
5280 / 60 / 60
1.46666 feet per second
So 185 feet per second after falling 540 feet converts to
126 MPH
This is almost exactly equal to the terminal velocity of 120 MPH !
Conversely how many seconds or feet does it take to reach terminal velocity which is;
120 MPH
or
176 feet per second
The first equation is
V = g.t
Or
t = V / g
t = 176 / 32 = 5.5 seconds
How far do you fall in 5.5 sec.?
The second equation is
X = g.t.t /2
X = 32 x 5.5 x 5.5 / 2
= 484 feet
So you reach terminal velocity of 120 MPH after 5.5 seconds and by that stage you have fallen 484 feet.
AIR RESISTANCE:
Actually to do this more accurately you should take account of the air resistance all the way down. This formula assumes no air resistance.
Resistance increases with the square of the velocity. (That is the formula I derived for the rocket program. For more complicated cases you could use the same numerical simulation but I think we can derive this using calculus.)
You need the coefficient of air resistance R for the human body and that could be calculated from the known terminal velocity. Terminal velocity occurs when the acceleration due to gravity is exactly equal to the deceleration caused by air resistance.
Air resistance creates a force against the body which tends to make it decelerate. This can be subtracted from the acceleration of gravity g in the above formulas.
This negative acceleration comes from the formula
F = m . a
Or force = mass times acceleration .
Rearrange to get
a = F/m
Where m is the mass or weight of a human and F is the force caused by air resistance.
Now introduce the formula for air resistance:
F = R . V*2
Where V*2 is velocity squared.
Now just combine these 3 formulas and integrate twice like we did before and you should be able to produce an equation for seconds flying time corrected for air resistance.
The coefficient of air resistance R is calculated from the last formula but at terminal velocity of 176 feet/sec.
a = - g so it becomes
F = m . a = R . V*2
- g . m = R . V*2
So. R = - g . m / V*2
The ideal male weighs 70 Kg or
m = 154 pounds.
g=32 feet per second per second
V = 176 feet per second
So R = - 0.159
The units of R are not really needed but it is
Foot.pounds per square meter.
Quite Easily Done QED as they say.
lol all i wanted to know was 6.3  we're down now
Oh ... I see ... OK then ... I hope it took you longer than 6.3 seconds. Just the right time for me to calculate it!
What were you climbing?
google The FUBAR in FUBAR
[edited by DJ Craig because uhh...definitely don't want this location posted]
Cool building!
Sent from my iPhone
[last edit 11/27/2009 8:57 AM by evan-e-cent - edited 1 times]
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RailGuy88
This member has been banned. See the banlist for more information.
Location: Where you're not...
Gender: Male
| | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 15 on 11/27/2009 9:04 PM >
| | | | Posted by evan-e-cent
You could of course measure the height of a tower by dropping an object from the top and measuring how many seconds it takes to reach the bottom. Or you could jump off the top and start counting, a bit like the you do when the anesthesiologist puts you to sleep for surgery. A recent conversation with DJ Craig by text messages while he was climbing may be relevant to this:
...
Sent from my iPhone
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Seriously, someone has waaayyyy too much time on their hands! 
Going where others can't... |
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shadowbot
Location: Massachusetts
Gender: Male
| | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 16 on 12/9/2009 4:50 AM >
| | | |
I'm guessing some of you have handheld GPS units, most of which will tell you your height above sea level. Look at how high you are before you climb, then after and subtract.
Another slightly different way to do it, but possibly easier than finding the angle might be:
1. Stand a measured distance from the building.
2. Use a rangefinder and point it at the very top of the tower.
3. Since you know the angle at the base is 90 degrees and the tower isnt crooked. (assuming the ground is flat-ish) You can assume you have a triangle with angles of 45 45 and 90 degrees.
4. If you know the angles and the length of 2 of the sides, the rest is literally just a high school math problem.
[last edit 12/9/2009 4:52 AM by shadowbot - edited 1 times]
My explorations documented on flickr: http://www.flickr..../8528952@N02/sets/ |
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Dick Winter
Location: Richmond
Gender: Male
| | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 17 on 12/21/2009 1:46 AM >
| | | | The tower is 540 feet. I looked it up on the maps uLiveAndYouBurn provided.
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gretschdiva
Location: Johnson City, Tennessee
Gender: Female
I'm just a chick, playing a chick, disguised as another chick
| | | Re: Calculating the heights of cranes masts buildings and this massive cooling tower
<Reply # 18 on 12/28/2009 10:08 AM >
| | | | Posted by TheVicariousVadder
It is of course, none of them.
But in fact the pissed off ghost of J. Robert Oppenheimer. Angry because even though we have managed to come up with a positive use for his work, bureaucrats still fuck it up.
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Vadder, I believe you are spot on.
Kudos to everyone who calculated the height of the cooling tower. Soon, I am going on a trip to visit said location with DJ Craig and whoever else would like to come along. I would be very interested to know the number of uncaged ladder rungs I am going to scale when I take on this beast.
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