Please solve a geometry problem for this mathematical idiot

Factual Questions
1

Okay, this is probably a very simple geometry problem, but I forgot all I knew about geometry roughly two hours after the final – and that was looooong ago.

Cross my heart and hope to die, this is NOT a homework problem. It’s for a book I’m writing, and if I ever manage to get it published, I’d really rather the first five hundred fans letters I get (hah!) don’t all point out that my math is crappy.

Here’s the situation: the story is set in a fifty year old colony on another planet. Humans are the settlers, and going home isn’t a possibility: they were sent out in hibernation on a slower than light ship. The computers were to locate a planet that fell into certain habitable ranges on various factors, land the ship, and revive the people. It was supposed to be a planet without sentient natives. Supposed to be. Heh.

Anyway, the colony is by and large underground due to weather reasons. The backbone of the colony was made by reusing the big cylinders that the space ship was mostly constructed from (with this reuse intended from the start.)
The cylinders have an internal diameter of 21 feet. Two parallel horizontal floors are built inside the cylinders, so that there is an 8 foot tall space between the two floors, and 6 foot spaces between the floors and the inside of the cylinders at the highest and lowest points. (The floors themselves use up 1 foot.) That is, you have three ‘levels’ with ‘ceilings’ of 6’, 8’, 6’ going from top to bottom.

The middle floor is the main one. The bottom floor has various mechanical systems and such running down it. The top floor is used for various things, mostly for storage but also living/playing space for children.

Now, obviously the height of the two outside floors varies, from six feet at the center, down to zero at the outside edges. Vertical walls, 4.5’ tall, have been built from the floor to ceiling along the lengths of the cylinders, dividing the top and bottom layers into central areas that are ‘mostly usable’ (with ceiling heights from 4.5’ to 6’) and outer areas that are too low to be traversed in any comfort. There are occasional doors leading into the ‘useless’ spaces, because the space is used for storage. Why, yes, those ‘useless’ spaces could provide handy access and out of sight passage to…anything…small enough to traverse them. :wink:
What I need to know is, where are these vertical walls located? How wide is the central passage, and how wide are the outer channels? As in, if the following really crude diagram survives, what are X and Y in feet? Assume the vertical walls have no width.

[cylinder wall]X_|Y|X[cylinder wall]
Thank you to anyone who will solve this for me. (I can also offer mention in my future ‘Author’s Note’ if you like.)

2

OK, there’s one problem with the question as put: 6 + 1 + 8 + 1 + 6 is not equal to 21. So you need to change a figure or two there. I’d change the 6 to 5.5, so the the ceiling and floor of the central part are equal.

Therefore, the floor is 6.5 feet from the bottom of the cyclinder, so it’s 4 feet from the centre of the cylinder. Construct a right-angled triangle with the right angle in the centre of the floor, the top angle in the centre of the cyliner, and the other angle where the floor meets the cylinder. Now the hypotenuse of this triangle is 10.5 feet (the radius of the cylinder), and the vertical sude is 4 feet. Therefore (using Pythagoras), the other side is the sgare root of 10.5[sup]2[/sup] minus 4[sup]2[/sup], i.e. about 9.38 feet. The total width of the floor is twice that, or 18.76 feet.

Does that help?

3

Wrong.

The floors take up 1 foot total and are thus each 6 inches thick.

Also the math is wrong. The hypotenuse is 10.5 ft, but the vertical side is 4.5 + .5 + 4 = 9 ft. The horizontal side is then SQRT(10.5^2 - 9^2) = 5.408, and the total width of the passage is 10.81665 ft.

.

4

Yes, we are calculating the widths of different floors/ceilings. The floor/ceiling of the central section is about 19.4 feet wide (i rechecked my calculations, and I’d put an error into them. The floor of the top section is about 10.8 feet wide, as jebert calculated.

5

Why not draw a diagram? Should take about 30sec, and if you change anything, you wouldn’t need to come back to the board.

6

I don’t believe it’s necessary to include this additional informationin your book.

Most readers after reading two sentences of this are likely to skip the rest of the chapter entitled, “Dimensions of the space cylinders.”

7

You’ve got an arc of a circle of diameter 21 cut out by a line 6 from the parallel tangent. Construct a perpendicular to that line so that the distance from the line to the arc along the perpendicular is 4.5, right?

Consider the circle

x[sup]2[/sup] + (y+15)[sup]2[/sup] = 441

The x-axis corresponds to the floor of the upper space. Now, the question is the x-values where y is 4.5. Easy:

x = ± sqrt(216-30y-y[sup]2[/sup])

In particular, for y=4.5, we get x just under 7.8

That is, in your diagram the middle chamber is a little under 15.6 feet, while the outer chambers are a little under 6.5 feet (to keep six inch walls like the ceilings).

8

Say he includes some beastie of a certain size and says it fits into the outer chamber. You can’t see some pointy-headed geek actually doing this calculation and trying to ridicule him over some a trivial matter?

Frankly, I think it’s a great idea to come here and head this sort of thing off at the pass.

9

People in the future, even astronauts, still use feet and inches? Make it metric.

10

Not quite; metric might give the right ‘futuristic’ impression, but otoh, if you’re aiming at people who just can’t visulaise a m at all, then use imperial, and put a note at the start saying ‘translated from the metric’

11

Giles is getting closer, but the floor/ceiling of the middle section is not the same width as the floor/ceiling of the upper/lower sections because of the six inch floor difference. With that taken into account, you get the total upper/lower section floor/ceiling width of 18.97 feet (sqrt(10.5^2 - 4.5^2) * 2) and the individual section floor/ceiling widths as labeled would be x=4.08 feet (each of the outer sections) and y=10.8 feet (the center section).

12

The floor-width of each of the side sections is

X = sqrt(10.5^2 - 9^2) ft ≈ 5.4 ft.

The floor-width of the center section is

Y = 2*[ sqrt(10.5^2 - 4.5^2) - sqrt(10.5^2 - 9^2) ] ft ≈ 2*( 9.5 - 5.4) ft = 8.2 ft

13

/me sighs.

I’ve now seen two more quotes with no work shown. You people would fail one of my courses.

Read my first post and if you disagree with my results, point out where I made my mistake. It should be easy since I show my work.

14

Your first error was here:

This is not the correct equation for the circle you describe. For example, your circle has radius sqrt(441) = 21. However, the desired circle has diameter 21, and so radius 10.5.

15

I’m glad Tyrell could point out Mathochist’s problem, because I didn’t think the answer even made enough sense to address.

As for Tyrell’s answer, what you’ve calculated for X is actually half of Y; that is, the horizontal component of the triangle with hypotenuse 10.5 and height of 9 is the length from the center of the cylinder to the wall, not from the wall to the edge of the cylinder.

16

You’re correct. Two points out of ten off for an error, though I recover partial credit for the twchnique. Let’s tweak this then…

You’ve got an arc of a circle of diameter 21 cut out by a line 6 from the parallel tangent. Construct a perpendicular to that line so that the distance from the line to the arc along the perpendicular is 4.5, right?

Consider the circle

x[sup]2[/sup] + (y+4.5)[sup]2[/sup] = 110.25

The x-axis corresponds to the floor of the upper space. Now, the question is the x-values where y is 4.5. Easy:

x = ± sqrt(110.25-(y+4.5)[sup]2[/sup])

In particular, for y=4.5, we get x just over 5.4

That is, in your diagram the middle chamber is a little over 10.8 feet, while the outer chambers are a little under 3.6 feet (to keep six inch walls like the ceilings).
*

Now, this is still off from your result of 5.4 feet for the side sections.

17

Yes, you’re right.

18

Yes, I was wrong.

19

FWIW, Mathochist, I used the same technique you did (looking at the equation for the circle and solving for x), but, once I found it, I forgot to subtract x from where the wall meets the floor at sqrt(10.5^2 - 4.5^2) to get the desired quantity.

20

I think I might also be a bit ranty from finals week. Now that this is settled, StarvingButStrong needs to get our mailing addresses for those royalty checks.

I’m serious. Stipends pay worth sh*t.