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Last Post 12/16/2010 7:40 PM by  Tim_Johnson
Geometry Problems
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RJortberg
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08/31/2010 11:39 AM

A problem with using plans alone is that unless they are "as builts" you could easily miss important parts of the building. I have run into numerous occasions where original plans do not show new or modified construction. Think of a new or expanded garden center at a building like a Home Depot, Kmart or a Walmart. Most of the available plans are at building departments, and it is often easy to find the original permitting plans but difficult to find expansion plans.

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ChuckDeaton
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08/31/2010 2:13 PM
That is my point, exactly and the thought that if you hire a professional you should get a professional, not a "smart" adjuster.
"Prattling on and on about being an ass with experience doesn't make someone experienced. It just makes you an ass." Rod Buvens, Pilot grunt
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Medulus
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09/02/2010 3:23 PM
That's a false dichotomy, Chuck. For instance, I consider you to be both a professional and a smart adjuster.
Steve Ebner CPCU AIC AMIM

"With great power comes great responsibility." (Stanley Martin Lieber, Amazing Fantasy # 15 August 1962)
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Leland
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09/02/2010 8:17 PM

If we estimate the surface area of a flat roof of the same foot print and also the surface area of a half sphere with the same footprint we can establish a range within which our actual square footage will fall. This won't give us our exact answer, but I would like to do this because it will help me to check my math and I just feel comfortable doing it this way.

Using google earth I came up with a diameter of 715'. Using google earth may not be exactly accurate, but its a start. According to one website the diameter is 680, so I know I'm close. It's hard to tell from the satellite photo but it looks like it would be possible to walk around the bottom edge of the dome and measure the circumference. This could be done with one of those walking wheel things- perfectly accurate on a flat surface. Measuring the circumference will help to double check our diameter. The formula for that is:

diameter x Pi = circumference.

since Pi is 3.14159265 we can multiply our diameter (680) by 3.14159265 and we get 2136. If our wheel shows 2140 we know the diameter is correct at about 680.

One measurement confirms the other- the relationship between diameter and circumference is the same on any size circle.

Now, to establish a baseline minimum, a flat circle with a diameter of 680 would have an area of 363,168 square feet (680/2) * (680/2) * 3.14159265

We know this is not the right number because the dome is obviously not flat. But we now know for sure that the square footage is at least that much and not less.

Now let's consider a sphere- the surface area of a sphere is:

4 x pi x radius x radius

So it the superdome looked like the top half of a ball or globe the surface area would be:

(4 x 3.14159265 x 340 x 340 )2

This answer is 726,336 square feet.

Now we know this answer is also not correct because the dome has a much lower profile (flatter) than one half of a sphere or ball.

But know we have a range that will give us some confidence in our calculations- our final answer should be somewhere in between 363,168 and 726,336 square feet.

I know this is a big range but doing this calculation might help me avoid an embarrassing error.

The problem I see with this dome is that it is PART of a much larger sphere. It isn't the whole top half of a sphere, look how flat it is- it is just smaller portion of a much bigger imaginary sphere. So If I could know the size of the sphere that it is part of and now what fraction of that larger sphere it represents, I could possibly do the math. I think this is the correct way to calculate it. But I don't know how to do it that way.

By searching online, I discovered the superdome roof is a SPHERICAL CAP.

Here is the formula for the surface area:

2 x Pi ((n x n) + h x h)/2h)h

(I had to retype it as best as I could- the "n" stands for the flat radius and the "h" stands for the distance from the top of the dome to its imaginary base, see the picture on wikipedia)

So if the height of the dome (just the dome height, not counting the extra distance to the ground below it) is 100' then the surface area would be: 394,572 sq ft.

If I compare this answer with the area of a flat circle I notice that it is a bit more, but not too much more.

So I can have some confidence in my answer because it makes sense for a dome that is 680 feet across and only 100' high to have more surface area than a circle, but not too much more. My example is a fairly flat dome.

Obviously the actual dome may be higher than 100' and the surface area would be greater.

here is the wikipedia article: http://en.wikipedia.org/wiki/Spherical_cap


HOW TO GET THE MEASUREMENTS

will a laser pointed up to the inside top get the height from the ground? that number could be compared to the height of the base of the dome. The base could also be checked by laser, or a string could be dropped off the edge of the parapet wall to measure the lower half of the building below the base of the dome.

Also trigonometry can be used to check height:

http://www.worsleyschool.net/scienc.../page.html

But I have to admit, I would probably trust the building plans. I think A high profile building like this, with so many architects and engineers working on it, would have fairly accurate measurements on the plans. It's not like the steel fabricator would make a bigger curve in the roof just to see if he could get away with it and not have anybody notice.

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Leland
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09/02/2010 8:35 PM
One problem with the "spherical cap" formula is that the superdome might not be a perfect spherical cap- examples:

A slice cut from a bowling ball: perfect spherical cap

a slice cut from the end of a long watermelon: not a perfect spherical cap, but close

a slice cut from the pointy end of an egg: not a spherical cap.
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ChuckDeaton
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09/02/2010 9:13 PM
Now we are getting somewhere!
"Prattling on and on about being an ass with experience doesn't make someone experienced. It just makes you an ass." Rod Buvens, Pilot grunt
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CatAdjusterX
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09/02/2010 9:50 PM
Posted By Leland on 02 Sep 2010 08:35 PM
One problem with the "spherical cap" formula is that the superdome might not be a perfect spherical cap- examples:

A slice cut from a bowling ball: perfect spherical cap

a slice cut from the end of a long watermelon: not a perfect spherical cap, but close

a slice cut from the pointy end of an egg: not a spherical cap.



 

Although the dome is not half of a sphere, it is essentially 1/3 of an oval ( Stewie from Family guys head) x.333

 

Base x height x ( I HAVE NO FREAKIN" IDEA)

I have to agree with Chuck in this thread is great for the mere ability to get us thinking and foraging for the right answer

 

"A good leader leads..... ..... but a great leader is followed !!" CatAdjusterX@gmail.com
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Goldust
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09/02/2010 10:22 PM

You got it Base x The height x 10 -15% waste plus steep. That oughta cover it.

 Remember man built it man can fix it.

JERRY TAYLOR
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Medulus
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09/03/2010 10:11 AM
I was wondering when you would weigh in, Leland. I think you nailed it. If you were actually assigned this claim, and therefore you were on site, you could use your laser measuring device to take a measurement from the center of the floor under the dome to the top of the interior ceiling. Then take a measurement from the floor to the edge of the dome. The difference would be the chord height that dacree was talking about. Using his formula you could then figure out what proportion of a hemisphere is constituted by the dome. The waste factor is probably 20% because it is a circle and because you are essentially measuring the inside of the dome, but the roofing is for the exterior. The exterior will be slightly larger than the interior. I believe that, for a project like this, the materials would be custom manufactured off site. Otherwise the waste factor might be even higher.
Steve Ebner CPCU AIC AMIM

"With great power comes great responsibility." (Stanley Martin Lieber, Amazing Fantasy # 15 August 1962)
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ChuckDeaton
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09/03/2010 11:41 AM
Has anybody heard from BobH, he hasn't posted in a while?
"Prattling on and on about being an ass with experience doesn't make someone experienced. It just makes you an ass." Rod Buvens, Pilot grunt
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ChuckDeaton
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09/03/2010 12:15 PM
In my opinion some of the comments, regarding handling, reflect limited, small claim influenced thinking. 15% waste on the amount required to reroof the Super Dome would be a large sum of money.

I agree that convention is to add waste to a relatively small residential roofing job, but on large commercial jobs adding a waste percentage needs to be considered and done with discretion.

"Prattling on and on about being an ass with experience doesn't make someone experienced. It just makes you an ass." Rod Buvens, Pilot grunt
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Medulus
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09/03/2010 1:02 PM
You're probably right about that, Chuck, and for the very reason I stated in my previous post. The roofing would probably be custom manufactured for such a job. If waste is applied, it would only be to the materials. And, in the real world, a job like this would be awarded by competitive bid or at labor and materials rather than on any square foot basis. The way in which it would be settled in real life is much more along the lines of your original suggestion. However, and also in the real world, carrier examiners, and even carrier management, is not the same as in the past. The superdome would still likely be given to a very experienced examiner if one were available. I know a few companies, though, who have taken a hatchett to their staff to such an extent that they have no one of real experience left. Consequently, a very experienced catadjuster may find that he or she is reporting to a carrier staff person with less than five years experience. The possibility that a catadjuster would be asked for information that might be more pertinent to smaller claims, or that is irrelevant is highly probable. The file might even be kicked back for that reason. I'm thinking of the examiner who once kicked a file back to me because I hadn't drawn a diagram of the bedroom. She called me to make a special point of it. I told her that it was a rectangle and the dimensions were on the estimate. She told me I wouldn't be paid until I gave her the diagram. I sketched a rectangle while we were on the phone, labelled the dimensions, scanned and emailed it to her while we were speaking. Some examiners are unable to think outside their box.
Steve Ebner CPCU AIC AMIM

"With great power comes great responsibility." (Stanley Martin Lieber, Amazing Fantasy # 15 August 1962)
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jdacree
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09/03/2010 2:00 PM
Getting away from the superdome, the formulae for a radius when you have a chord, and a chord height will work. Also getting away from the superdome utilizing the laser method from the inside, if you put the laser on a tripod, you could then take three or more measurements on the inside FROM the same point if they are ALL the same then you have proven that the surface is a true radius. If the measurements are differenct then you have a parobilic arc, and a formulae for that is also availiable.

Other methods listed were to get on top and drop a weighted rope, then measure the length of the rope, this would work BUT it would not tell you if you had a radius or a parobilic arc. In general there were some very good comments on this subject. If I were to run into a geodisic(sp) dome house, I would have more options from which to choose to figure the number of squares of shingles to use. Upon contacting my supervisor I would probably figure my waste utilizing a 15-20% due to the facets of a dome containing so many triangles, or trapozides being like a lot of hips on the roof, just no valleys.
Jim Acree Stupidity is the art of not trying to learn Ignorance is the lack of opportunity to learn I am ignorant
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ChuckDeaton
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09/03/2010 2:22 PM
How about a helium balloon on a string.
"Prattling on and on about being an ass with experience doesn't make someone experienced. It just makes you an ass." Rod Buvens, Pilot grunt
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jdacree
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09/03/2010 2:50 PM
Chuck, surley you jest, a helium ballon on a string would just go straight up. If you are referring to my comment about a weighted rope, that is used to measure excavations in the grond to determine depth. A weighted rope or string has also been used by builders to measure a difficult slope.
Jim Acree Stupidity is the art of not trying to learn Ignorance is the lack of opportunity to learn I am ignorant
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Leland
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09/03/2010 3:48 PM
If the dome is not a "spherical cap" then we can break the surface up into triangles. I think someone mentioned this above.

Imagine two fly fishermen standing on the very center top. Each casts a lead weighted line out to the edge of the dome. The two weights fall to the edge and lay 10 feet apart. The "triangle" is the area between the two lines.

We know how to measure the area of a triangle, if the circumference is 2010 feet then there would be 201 such triangles. Because of the each triangle is not perfectly flat, there would be a small error. Instead of 201 triangles with a 10' base, you could calculate 402 triangles with 5' bases, that would be a bit more accurate.

It is theoretically possible to measure that way in the field. You would need to get on the top, identify the center, and have a string 400' to 500' feet long. You could use a fishing line because it won't stretch as much.

I would like to estimate the length of the string from the same numbers I used before.

We can use the pythagorean theorem to get close.

We could imagine a hook inserted in the top of the ceiling of the dome and a trapeeze artist suspended by a 100' rope below the hook. The trapeeze artist points a laser at the inside base of the dome.

I only give this visual description so I can describe the geometry without an actual picture.

The rope is one leg of a right triangle, and the laser is the other leg. It is a "right triangle" because the angle between the two lines is 90 degrees.

The third side of this imaginary triangle is the hypotenuse. The "diagonal" of a right triangle is called a hypotenuse.

The hypotenuse runs from the hook in the ceiling to the spot where the laser is pointing. It runs in a straight line underneath the curved surface of the dome.

The formula for determining the hypotenuse is : (side x side) + (side x side) = square of hypotenuse.

This is an extremely useful formula for determining the length of a roof slope, many adjusters know about this formula. Parden me if I am making this too simple or too hard, everyone has a different math background.

So let's plug in the numbers:

(100 x 100) + (340 x 340) = the square of the hypotenuse

if we do the math we get:

125,600 = square of hypotenuse

so now we need to fine the square root of 125,600

a really easy way is to go to google and just type "square root of 125,600" in the search box- the answer is 354, a bit longer than the 340 as it should be.

So now we know how long a straight line under the curved roof would be. The actual exterior surface length of the fly fisherman's line would of course be longer.

But I would like to try this number just as a reference.

I earlier calculated a circumference of 2136.

If we think of flat triangles with 354 on each side and bases of 5 feet, each triangle would have a surface area of very close to 885 sq feet. (This will be very slightly off, I can explain why if anyone really needs to know)

With a circumference of 2136 you could place 472 and 2/10ths of triangles on the dome and cover it completely.

So the surface would be 472.20 x 885 = 417,897 sq feet.

But remember, this is based on a straight line and not the actual roof that bulges out and away from our imaginary line. The real number would have to be a bit bigger. This 417,897 would actually be the area of a pointy cone the same width and height of the dome.

Notice how close this number is to 397,492 I calculated earlier. It is about 5% bigger, fairly close.But it would have made more sense to me if this number had come out smaller, because my original number was based on the textbook formula allowing for the curvature. But the real number should be perhaps 10% bigger due to the curvature, so I am disappointed the numbers didn't come out closer.

If this was real life I could double check all the numbers more carefully and make sure it didn't have any rounding errors and extend all the decimals out several places to make sure my math is as precise as possible.

But since this is just an exercise I am going to go next door and get a frozen yogurt with lots of fresh fruit and come back and work on some claims and forget about this problem.






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Leland
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09/03/2010 3:59 PM

Just in case you didn't know the ancient Egyptians used a string with knots and the pythagorean thereom to make square corners on the pyramids.

If you take any length, like a wood stick or the distance from your king's elbow to the tip of his finger you can also lay out the corner of a pyramid.

Using some length that you like, make 13 evenly spaced knots in a string (so that you have 12 equal sections of string)

Lay it out on the ground as a triangle with 3 sides: one side is 3 sections, one side is 4, and the third side is 5.

Get two other adjusters and each person grabs a corner and stretches tight.

the angle made by the 3 section side and the 4 section side is a 90 degree angle. (you have made a "right triangle')

pound your stakes into the ground and start laying your blocks.

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ChuckDeaton
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09/03/2010 4:15 PM
A square + B square = C square (Pythagorean theorem) Some laser measuring devices will calculate the length of a wall using this formula.
"Prattling on and on about being an ass with experience doesn't make someone experienced. It just makes you an ass." Rod Buvens, Pilot grunt
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jdacree
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09/03/2010 6:11 PM
Carpenters use the 3-4-5 or 6-8-10 rule for laying out 2 walls square to each other. If you snap a line 3' long and a line 4' long intersecting where the 3 and 4 lines cross, then open or close these lines until you have 5' at the end of the 3 and 4' lines then you have a perfect 90 degree wall. This is based on the formulae
Square root of a sq (3*3+9)+b sq (4*4+16) (9+15+25)= c (sq root of 25 + 5)

As far as trying to calculate a spherical radius by triangulation, it will not be accurate. As noted above you could take a length and get the drop and use the formulae for a right triangle BUT you still have a circular surface that rises above the hypotunse (long side of a right triangle). The fewer triangles used the greater the error, lots of little triangles WILL smoth out the error but that takes a lot of measurements.

Jim Acree Stupidity is the art of not trying to learn Ignorance is the lack of opportunity to learn I am ignorant
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DASavage
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09/03/2010 11:58 PM

I think Leland’s spherical cap post yesterday was dead on. I’ve been waiting for someone to include the height of the cap into their formula and not simply the radius of the cap.

The problem with most formulas related to spheres is they require the knowledge of the radius of the sphere in order to work correctly. Unless you are dealing with a perfect hemisphere, the radius of the dome perimeter is not the radius of the roof/ceiling. In the case of the Superdome, the center point of the radius of the roof/ceiling is located several hundred feet underground. This is not a measurement that can be directly obtained physically. It must be found mathematically.

Radius of roof/ceiling = r = (h2 + rb2) / (2 x h)

Where h is the height of the dome cap and rb is the radius of the dome cap base.

The formula Leland located on wikipedia for calculating the surface area of a spherical cap has the formula for determining the radius of the roof/ceiling imbedded within it.

Α = 2πrh
or
Α = 2π (h2 + rb2) / (2 x h) h

This can be simplified into a more user-friendly form.

Α = π (h2 + rb2)

This formula can be used to calculate the surface area of a dome regardless of height.



For future convenience, I’ve included both of them in MS Excel format.

=2*PI()*(((CAP_BASE_RADIUS*CAP_BASE_RADIUS)+(CAP_HEIGHT*CAP_HEIGHT))/(2*CAP_HEIGHT))*CAP_HEIGHT
or
=PI()*(POWER(CAP_BASE_RADIUS,2)+POWER(CAP_HEIGHT,2))


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David A. Savage

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