How do you calculate this?

[fleabag]

You have a half sphere that is 3.75" deep and 11" across (diameter). So how do I figure out how many inches deep can an object that is 7.25" in diameter go? I'm sure there is some math involved but I have no idea how to approach this problem.

(I'm trying to put in a circular CFL into a light fixture and these are the dimensions I'm working with)

 

[SithSolo1]

Well if its 3.75" deep and 11" it definitely ain't spherical. More ovoid.

I'll get back to you on the math.

http://en.wikipedia.org/wiki/Ellipsoid

What you have described is an Oblate spheroid and your guess is as good as mine. Last geometry class I took was in high school 14 odd years ago.

 

[Paratus]

inflate it to .......... Oh nevermind

 

[PieIsAwesome]

I think. . .

Find 3-variable equation of ellipsoid with given info, substitute a point on the circle for the second thing, and you get the height at which the circle would fit.

General equation of ellispoid:
(x^2 / a^2) + (y^2 / b^2) + (z^2 / c^2) = 1
Equation of ellipsoid with given info:
(x^2 / 4.5^2) + (y^2 / 4.5^2) + (z^2 / 3.75^2) = 1
Substituting in point on circle (3.625, 0, z):
(3.625^2 / 4.5^2) + (0^2 / 4.5^2) + (z^2 / 3.75^2) = 1

Solve for z.

 

[her209]

800 psi

 

[guyver01]

(I'm trying to put in a circular CFL into a light fixture and these are the dimensions I'm working with)

please be sure to videotape your spectacular death when you try plugging a round light into a square socket, and electrocute yourself.

 

[guyver01]

So how do I figure out how many inches deep can an object that is 7.25" in diameter go?

If it's 7.25" in diameter...

then it's radius is 3.625" ... and it's circumference is 41.2812734"

you figure it out.

 

[Matthiasa]

please be sure to videotape your spectacular death when you try plugging a round light into a square socket, and electrocute yourself.

Is that like measuring the voltage on some wires with a voltage unknown to the person that asked. And then after the measurement you find out you would have died almost instantly with even the slightest of touch to it.

Now as for for the op...
Assuming it's any kind of standard light and the socket sticks out at all... it will go in to the depth that the socket allows.
But to really know... we need pics.

 

[jeanclaude]

You have a half sphere that is 3.75" deep and 11" across (diameter). So how do I figure out how many inches deep can an object that is 7.25" in diameter go? I'm sure there is some math involved but I have no idea how to approach this problem.

(I'm trying to put in a circular CFL into a light fixture and these are the dimensions I'm working with)

For an individual who has claimed to be smarter or at least more knowledgeable than a Korean automotive engineer you are surprisingly deficient in math skills.

Not to worry. You should cover this next year when you take senior math. Unless of course you elect to take home economic to further develop your sardines and pasta recipe repertoire.

 

[edro]

I would draw it in a free 2d CAD program.

 

[grrl]

I would draw it in a free 2d CAD program.

Or on a piece of paper. Or just test fit the actual parts. Talk about mole hills and mountains.

 

[deadlyapp]

You have a half sphere that is 3.75" deep and 11" across (diameter). So how do I figure out how many inches deep can an object that is 7.25" in diameter go? I'm sure there is some math involved but I have no idea how to approach this problem.

(I'm trying to put in a circular CFL into a light fixture and these are the dimensions I'm working with)

I'm fucking confused as to what you are actually asking.

The OPENING is 3.75" Deep and you want to know how DEEP a 7.25" OBJECT can go?

Maybe...3.75"?

I don't believe that the 7.25" diameter object will ever contact the sides of the other object, since the opening is much wider than the depth. You could probably do a simple graph on a calculator using parabolic curves to see if they intersect. Just take several points and plot a best fit line. I believe the equations up above will also do it.

 

[jeanclaude]

I'm fucking confused as to what you are actually asking.

The OPENING is 3.75" Deep and you want to know how DEEP a 7.25" OBJECT can go?

Maybe...3.75"?

I don't believe that the 7.25" diameter object will ever contact the sides of the other object, since the opening is much wider than the depth. You could probably do a simple graph on a calculator using parabolic curves to see if they intersect. Just take several points and plot a best fit line. I believe the equations up above will also do it.

Exactly. One again Fleabag provides limited information re his 'problem'.

The best advice was "Or just test fit the actual parts". Common sense. Problem solved.

But no. Failbag just impulsively posts whatever pops into his addled mind.
Just more evidence that he is a cognitively challenged loner with no one around to talk to.

Add lack of self control to his exhibited personality traits.

 

[grrl]

<snip> Add lack of self control to his exhibited personality traits.

Do you mean he chronically masturbates?

 

[PottedMeat]

Exactly. One again Fleabag provides limited information re his 'problem'.

The best advice was "Or just test fit the actual parts". Common sense. Problem solved.

But no. Failbag just impulsively posts whatever pops into his addled mind.
Just more evidence that he is a cognitively challenged loner with no one around to talk to.

Add lack of self control to his exhibited personality traits.

that's just so he can attack you later when you're forced to assume data for parts of the problem.

 

[grrl]

that's just so he can attack you later when you're forced to assume data for parts of the problem.

Oh, you are so pious!

 

[PottedMeat]

Oh, you are so pious!

:awe:

 

[Paperdoc]

My read of this post is that he does NOT have a "half sphere". What he has is less than half a sphere, but it is a slice off one side of a sphere. The particular part he has is 11" diameter at the periphery, and 3&#189;" deep. It would NOT be possible to calculate the real diameter of the sphere from these limited data. Without that, you cannot answer the question posed by calculation.

No doubt the best answer is, as many have said: fit the real parts together and figure it that way. There's a lot to recommend practicality over theory.

Wait, I take it back. It IS possible to calculate the radius of the sphere. The model: the sphere's center is x inches above the plane of the edge of the slice of sphere. We can set up two equations for the radius, r:
r = x + 3.75
r^2 = x^2 + 5.5^2 (Pythagorean theorem)
Substitute first into second, eliminate x^2 term and find that
(2 * 3.75) x + 3.75^ = 5.5^2
x = 2.15833
So, r = 5.90833 inches

Now, reverse the process. This question becomes:
We know the spherical radius, r
We know the new diameter of the slice, which is 7.25 inches (the lamp diameter)
We need to calculate the depth of the bowl from that diameter to the bottom of the bowl - in other words, this time the value we need is what replaces the 3.75 inches in the original problem. Let that be y and figure it out.