[bonanova]

I just bought a circular glass top for my coffee table and I wanted to measure its precise size.

I grabbed my measuring tape; but alas, with use all its numbers had rubbed off, and it could not be read.

I also have a calibrated square whose side is finely adjustable from arbitrarily small up to 1 foot.

With the square, I determined the diameter was greater - by several multiples - than 2^1/2 feet: the square's largest diagonal length.

I thought for a moment then, on a hunch, I speed dialed Morty's and described my problem to Alex.

Well, he said, do ya still have the box the thing came in?

Yes, I replied, and that would give me its diameter; but if I could measure the box, I could certainly ...

Ahem! I heard over the phone, as if I was not supposed to speak while Alex was thinking.

What I was going to say was ...

Following Alex's advice, I made a single measurement with my square ruler and found the radius was exactly 3 feet!

What was the reading on my [square] ruler?

Edited December 3, 2008 by bonanova
clarifying that I used my square ruler

[Guest]

there was no reading on ur ruler cos it was blank!

[Guest]

I just bought a circular glass top for my coffee table and I wanted to measure its precise size.

I grabbed my measuring tape; but alas, with use all its numbers had rubbed off, and it could not be read.

I also have a calibrated square whose side is finely adjustable from arbitrarily small up to 1 foot.

With the square, I determined the diameter was greater - by several multiples - than 2^1/2 feet: the square's largest diagonal length.

I thought for a moment then, on a hunch, I speed dialed Morty's and described my problem to Alex.

Well, he said, do ya still have the box the thing came in?

Yes, I replied, and that would give me its diameter; but if I could measure the box, I could certainly ...

Ahem! I heard over the phone, as if I was not supposed to speak while Alex was thinking.

What I was going to say was ...

Following Alex's advice, I made a single measurement with my square ruler and found the radius was exactly 3 feet!

What was the reading on my ruler?

Is the calibrated square only 1ft x 1 ft or bigger than that?

[bonanova]

there was no reading on ur ruler cos it was blank!

Following Alex's advice, I made a single measurement with my square ruler and found the radius was exactly 3 feet!

Editing OP to make that clear.

[bonanova]

Is the calibrated square only 1ft x 1 ft or bigger than that?

No bigger than 1ft x 1ft.

[Guest]

No bigger than 1ft x 1ft.

There is some folding actions to be done to the box?

[bonanova]

There is some folding actions to be done to the box?

no folding - or spindling - or mutilating

[Guest]

I'm not sure if I understood, but:

3(2^1/2-1)

[Guest]

I just bought a circular glass top for my coffee table and I wanted to measure its precise size.

I grabbed my measuring tape; but alas, with use all its numbers had rubbed off, and it could not be read.

I also have a calibrated square whose side is finely adjustable from arbitrarily small up to 1 foot.

With the square, I determined the diameter was greater - by several multiples - than 2^1/2 feet: the square's largest diagonal length.

I thought for a moment then, on a hunch, I speed dialed Morty's and described my problem to Alex.

Well, he said, do ya still have the box the thing came in?

Yes, I replied, and that would give me its diameter; but if I could measure the box, I could certainly ...

Ahem! I heard over the phone, as if I was not supposed to speak while Alex was thinking.

What I was going to say was ...

Following Alex's advice, I made a single measurement with my square ruler and found the radius was exactly 3 feet!

What was the reading on my [square] ruler?

Did you... Segment the square in some way... should be able to do this exactly by marking off the four mid points of the edges of the squares, and connecting them to each other and the four corners and the centre point... and whatever other points you produce in the process.

Somewhere in this drawing lines procedure you'll end up with a length that is small enough to measure using the 1ft square, and a bit more calculation will let you determine the total size of the square.

_____

|\/|\/|

|/\|/\|

|\/|\/|

|/\|/\|

------

Then the distance from the middle of the smaller x's to the edge of the square will be exactl 3/4 feet, and can be acurately measured using your square.

[Guest]

I'm not sure if I understood, but:

3(2^1/2-1)

I think nobody is there, but there might be a step missing:

ie the box it came in is a square with sides of 3 ft:

the diagonal of the calibrated square would be measured as 3.(2^1/2 -1) meaning the sides of the ruler are 3.(1 - 2^-1/2)

[Guest]

put the table in the box, put the square in the corner of the box, the square's diagonal touches the table when fully extended.

Edited December 3, 2008 by mickjc75

[Guest]

I think nobody is there, but there might be a step missing:

ie the box it came in is a square with sides of 3 ft:

the diagonal of the calibrated square would be measured as 3.(2^1/2 -1) meaning the sides of the ruler are 3.(1 - 2^-1/2)

I've never seen such a square ruler, so I couldn't imagine it throughly. But logically, as you say, it must be scaled by its sides, rather than diagonal.

[Guest]

I've never seen such a square ruler, so I couldn't imagine it throughly. But logically, as you say, it must be scaled by its sides, rather than diagonal.

The sides are not 3 feet, they are 6 feet.

The table has a radius of 3 feet.

If the largest diagonal on the square is 2.5ft, then we know the other sides are 1.75ft

When the square touches the table, the diagonal forms a section of an octagon.

The box side is then the length of 1 octagon segment (2.5ft) + 2 of the square sides (3.5ft)

[Guest]

I haven't figured out how to insert a picture yet.

First, using your blank tape mneasure as a makeshift straight edge, make lines from corner to corner inside the box. Now place the glass circle inside the box, marking the points where the glass makes contact with the box. Use your tape measure to again connect the dots, forming a smaller square inside the first (half the size of the original). This will create four points where the lines intersect.

Now, draw lines through those points, vertically and horizontally (four new lines), creating a new square, 1/4 the dimensions of the first, in the middle of the box, and a 1/8th dimension square in each corner of the box.

Divide one of the corner boxes in half using the 2 unconnected points. The resulting line, perpindicular from the center of the line you just drew to the corner of the original box, will be exactly 1/8th the overall diagonal dimension of the original box.

If that is too great a distance to measure at this point, you should be able to use the square to bisect that corner square once again, making the resulting measurement 1/16th the overall length. Rinse and repeat until you end up with a measureable length.

Use that to find the dimensions of the box, which will give you the diameter of the circle. The rest is a piece of Pi.

But to answer your question, your measurement was either .75ft or 9 inches.

Of course, this only works if you have a pencil/pen/crayon, etc.

[Prof. Templeton]

1.2426 feet. Place the table top centered on the box, and measure from the corner of the box to the outside of the table top.

[Guest]

1.2426 feet. Place the table top centered on the box, and measure from the corner of the box to the outside of the table top.

Sure, if "simple" and "elegant" are what you're looking for.

My solution has the added benefit of being complicated and hard to follow (a sign of true genius, or so I've heard.)

[Guest]

1.2426 feet. Place the table top centered on the box, and measure from the corner of the box to the outside of the table top.

I see one flaw..

...can you be sure that you have the shortest distance? even being off by a couple radians (degrees) will significantly alter your accuracy. This may be able to be remedied by using the tape measurer to act as a straight edge from corner to corner of the box and then measure along edge of tape to edge of circle?

Or maybe I am being to nit-picky

[Guest]

I haven't figured out how to insert a picture yet.

First, using your blank tape mneasure as a makeshift straight edge, make lines from corner to corner inside the box. Now place the glass circle inside the box, marking the points where the glass makes contact with the box. Use your tape measure to again connect the dots, forming a smaller square inside the first (half the size of the original). This will create four points where the lines intersect.

Now, draw lines through those points, vertically and horizontally (four new lines), creating a new square, 1/4 the dimensions of the first, in the middle of the box, and a 1/8th dimension square in each corner of the box.

You make an interesting case, but (I think) are lucky in your analysis that your mistake at the start (in red) is then fixed in your next step!

Let's call the original Box A, your "1/2" box B, your "1/4" box C and your "1/8" box in the corner D

A has sides of length x

B therefore has sides = sqrt( 2 . (x/2)² ) = x/√2. It has an area which is half of the original, but not half the sides

C has sides = sqrt( 2. (x/2√2)² ) = x/2. It has an area which is 1/4 of A and half the sides of the original A.

D takes up the space in the corners of C in A (with C exactly in the centre) and has sides = x/4 and an area which is 1/16 of A.

Divide one of the corner boxes in half using the 2 unconnected points. The resulting line, perpindicular from the center of the line you just drew to the corner of the original box, will be exactly 1/8th the overall diagonal dimension of the original box.

Actually, you were correct up to this point, but the diagonal of box D will be x/2√2 (or 3/√2 feet, given x = 6ft).

Either way, your biggest problem would be that the calibrated ruler/square doesn't reach across the box, so there's no way to draw your lines accurately in the box, I'm afraid.

The diagonal of the calibrated square is (diagonal of box A - radius of table) = 3√2 - 3 = 1.2426 feet and the sides are 3.(1 - 1/√2) = 0.8787 feet.

edit: to HokieKen, that's why the measurement tool is a calibrated square not just a ruler!

Edited December 3, 2008 by foolonthehill

[Guest]

I think I got it!

Maybe some of you have figure out but sincerely I figure out this one myself.

the reading on the square ruler shows 0.828 ft or 2 (2 ^1/2 - 1) ft.

and below is the explanation:

I hope I am correct this time.

[Guest]

okay, ive got it

you read the outside of the box

it probably says on it

[Prof. Templeton]

I think I got it!

Maybe some of you have figure out but sincerely I figure out this one myself.

the reading on the square ruler shows 0.828 ft or 2 (2 ^1/2 - 1) ft.

and below is the explanation:

I hope I am correct this time.

is 3 x 3. And it's diagonal (from corner to corner) is sqrt(3² + 3²) or 4.2426, then take away the three for the table top and it leaves 1.2426

[Guest]

is 3 x 3. And it's diagonal (from corner to corner) is sqrt(3² + 3²) or 4.2426, then take away the three for the table top and it leaves 1.2426

Prof, Bonanova is asking on what the square ruler reads, not the difference of the diagonal length to the radius.

[Prof. Templeton]

Prof, Bonanova is asking on what the square ruler reads, not the difference of the diagonal length to the radius.

O.K. I finally understand the "square ruler" concept. So, Yes, your absolutely correct.

[Guest]

O.K. I finally understand the "square ruler" concept. So, Yes, your absolutely correct.

I think that both the Prof and Woon have good ideas, but...

The answer is in fact 0.78822

Prof T didn't give a complete solution, 'cause the square ruler can't measure anything over a foot, hence 1.2426 feet can't be measured.

Woon had the right idea to put the square in the corner.

If you look at my crudely modified version of Woons drawing, you can place the square as shown.

From Prof T, AE = 3root2 = 4.2426

We know that BE = 3

Therefore, AB = 1.2426

That is the hypotenuse of a triangle formed with 2 sides of length BZ, which we'll call L

L**2 + L**2 = 1.2426

L**2 = 0.6213

L = 0.78822

So you could place your square, make sure the length on both sides is the same, and either measure 0.78822, or 0.21178 from the edge of the box to the other end of the square.

I think W is incorrect; he asserted that:

line segment BC = line segment CD = 1 - X = 0.172 feet.

You can tell from the diagram that in fact CD > BC

AB = 1.2426

AC = 1.4142

So AB = 1.4142 - 1.2426 = 0.1716

0.1716 + 0.828 = 0.9996; this is close to, but not equal to 1

Edited December 4, 2008 by xucam

[Guest]

You can tell from the diagram that in fact CD > BC

You can't really see the true dimension from the diagram. It is just a sketch I draw in a hurry, not a fully scaled drawing.

If you really figure out that the diagonal of the square ruler is co-linear with the radius, then you will sure agree CD = BC.

[Guest]

So AB = 1.4142 - 1.2426 = 0.1716

0.1716 + 0.828 = 0.9996; this is close to, but not equal to 1

It is not tally because the first one you used 4 decimal and the second part you are using 3 decimal.

The right one is

AB= 3 (2 ^1/2) - 3 = 3 (2 ^1/2 -1) .... or 1.2426

AC = 2 ^1/2 ........ or 1.4142

BC = AC - AB

= 3 - 2 (2 ^1/2) ....... 0.1716

BC + CD = 3 - 2 (2 ^1/2) + 2 ( 2 ^1/2 - 1)....... or 0.1716 + 0.8284

= 3 - 2

= 1.

dang!

Proved!