Weight / Volume of Cooker

[Guest]

A catalog has the following item:

Cooking disc made from concave plow disc.

Constructed from heavy duty steel.

22" disc (don't know if circumference or diameter).

Thickness: 3.5 mm

Depth: 3 in.

Can the volume and weight be determined ?

[Guest]

no

[Guest]

What information is missing?

[Guest]

Cooking disc made from concave plow disc. .........Is this a concave dish rather than a disc. Part of a hollow sphere in shape.

Constructed from heavy duty steel........OK

22" disc (don't know if circumference or diameter). ........Is this the diameter of the rim.

Thickness: 3.5 mm .........Is the the thickness of the steel sheet it is formed from.

Depth: 3 in. ........Is this the overall depth of height from the edge of the rim to the outside of the curved concave surface.

[Guest]

"Is this a concave dish rather than a disc. Part of a hollow sphere in shape. " Constructed from disc. Similar to a wok.
" Is this the diameter of the rim." I am guessing it is.
" Is the the thickness of the steel sheet it is formed from. " Again I am guessing it is.
" Is this the overall depth of height from the edge of the rim to the outside of the curved concave surface. " Just indicates 3" deep.

[Guest]

OK, similar to a Wok.
A Wok generally has a flat base for sitting on a flat topped cooker, for good heat conduction....Does it have a flat base, if so, need to know the diameter of the flat base.
Do we assume uniform thickness throughout or has it a thicker base.
Take 22 inches as top rim outside diameter, with spherical curved sides.
Take 3 inches deep from bottom of flat base to top of rim.
You should be able to calculate now from this info if you decide the base flat or curved.

[Guest]

Not sure how to solve.

[Guest]

Does it have a flat base,????? if so, need to know the diameter of the flat base.....OR we could take it at half the rim diameter....that would be 11 inches.

Also take it as uniform thickness throughout....... 3.5 mm.

[Guest]

I still don't know how to solve.

[Guest]

Looks like you don't know if it has a flat base or not....so we will say no flat base, it actually makes it simpler....so spherical in shape, rounded bottom, 3 inches depth, 3.5 mm thick and 22 inches rim diameter.
So......
need to calculate the diameter of the whole sphere, outer diameter and inner diameter. use methods of intersecting chords or whatever.
Find outer and inner volumes
Find the fraction of the whole sphere that is the Wok.
Work out this fraction of volume for Wok volume.
Multiply by density of steel to get weight

[Guest]

I still can't solve.

[Guest]

If you just want an approx. answer, and it will be near enough in this case, Just work out the surface area of the Wok and multiply by the thickness.
That should be easier.

[Guest]

Surface area: 4 * pi * r^2 (sphere)

4 * 3.416 * 11 * 11 = 1520.53

1520.53 * .13370 (mm changed to in.) = 203.29 cu. in.

203.29 * .283 = 57.53 = 58 lbs.


I don't know.

[Guest]

You need to find radius or diameter first.
By intersecting chords.....

11 x 11 = 3 (D - 3)
121 = 3D - 9
130 = 3D
D = 43.333 OR R = 21.666

For part of a sphere

Surface Area of curved sphere bit = S = 2 x Pi x R x H
S = 2 x 3.142 x 21.66 x 3

S = 408.334 sq. ins

3.5 mm = 0.138 inches

volume of steel = 0.138 x 408.334 = 56.35 cu. ins

Density of steel = 0.284 lbs per cu.ins

Weight = 0.284 x 56.35 = 16 lbs.

[Guest]

Thanks.

A question: What are intersecting chords ?

[Guest]

A question: What are intersecting chords ?

An answer: Chords that intersect. Google the topic online
If you draw an X across a circle it cuts the circle in 4 bits. The 2 lines of the X are chords and they intersect. and are straight lines.
For line 1 of the X let A and B be the 2 bits of this line. And for the other line of the X let C and D be its 2 bits.

The intersecting chords theorem states that A x B = C x D.

For the question a line across the rim of the Wok is a chord of length the "rim" diameter. and a line (the depth) from the rim perpendicular to the centre point of the Wok's curved centre is another chord. That gives A = 11, B = 11. C = 3 and D = (sphere diameter - 3)

So 11 x 11 = 3( sphereDia - 3).........and work out the Dia and then the radius.

[Guest]

That is clear to me now. Thanks.

[leesajohnson]

I think that something is missing in the question which you have asked.

[Guest]

What do you think is missing? Can you be more specific please.
You expect questions to be asked, clearly, but your own statement is not very clear.