Weight of Skillet

[Guest]

A skillet has the following description:

Made from cast iron.

Flat base.

Diameter = 15.5 in.

5.5 in. handle.

Inner base = 12.5 in.

Depth = 2.25 in.

Thickness = .25 in.

Cast iron weighs .26 lbs. per cu. in.

Determine weight of skillet.

How do I proceed ?

[Guest]

Do it the same method as the last one......you should be at least able to try it.
Find the surface area and multiply by the thickness
Are the sides straight, tapered as part of upside down truncated cone.???

Diameter = 15.5 in.........I assume this is outer rim diameter

5.5 in. handle.....this is only length...need crossectional dimensions

Inner base = 12.5 in.....OK this is the base diameter

Depth = 2.25 in......OK

Thickness = .25 in........OK

Cast iron weighs .26 lbs. per cu. in. .....OK

[Guest]

" Are the sides straight, tapered as part of upside down truncated cone.??? " Appears more straight.

5.5 in. handle..... "this is only length...need crossectional dimensions " Only length given.

[Guest]

" Are the sides straight, tapered as part of upside down truncated cone.??? " Appears more straight.

5.5 in. handle..... "this is only length...need crossectional dimensions " Only length given.
0
You can take an approx size for the handle....say it is 1 inch diameter.

Work it out as before for the Wok.....((surface area of the pan) x thickness) plus the handle.....It will give you a near enough approx answer.

[Guest]

My attempt:

7.75 * 7.75 = 2.5(D - 2.5)

60.0625 = 2.5D - 6.25

66.3125 = 2.5D

D = 26.525 R = 13.2625


2 * 3.1416 * 13.2625 * 2.5 = 208.327 sq. in.

208.327 * .25 = 52.08 cu .in.

52.08 * .26 = 13.54 = 14 lbs.

Where am I correct ?

[Guest]

Looks like you calculated as though it was a spherical shaped Wok.
You sai it had straight sides tapered as in part of a cone, and it had a flat base.
And you forgot the handle.

But the answer will not be all that much different if it is only an approx you want.

[Guest]

I wasn't sure.

[Guest]

Just curious. The skillet is listed in the same catalog as the cooker. I realize I could contact the company for the weight, but I wanted to determine it mathematically. The weight wasn't listed in the description.

[Guest]

Additional info:

2.5 in. lip.

Sorry for the omission.

[Guest]

You are the one with all the details, I am only advising of the the method to calculate the volume.

So far we have assumed the following...
Skillet is made as part of a hollow cone.
This inside dimensions are 15.5 dia at rim and 12.5 at inside base.
The outside dimensions are 16.0 dia at rim and 13.0 at outside base.
The inside depth is 2.25 and depth to outside of base is 2.5
Ignoring slight descrepancy due to how we measure thickness of the sloping sides.

Volume for truncated cone is V = 0.33 x Pi x H ( R^2 + r^2 + Rxr)
Work out the outer volume and subtract the inner volume
Vinner = 0.33 x Pi x 2.25(7.75^2 + 6.25^2 + 7.75x6.25)
Vinner = 347.73

Vouter = 0.33 x Pi x 2.5(8^2 + 6.5^2 + 8x6.5)
Vouter = 414.35

Difference = Volume of pan = 66.62 cu ins

Weight of pan = 66.62 x 0.26 = 17.32 lbs

You need to add on the handle.....5 inches long and 1 inch dia
Pix r^2 x L = Pi x 0.25 x 5 = 3.93 cu ins
3.93 x 0.26 = 1.02 lbs

total weight is = 18.35 lbs.

[Guest]

I haven't checked the figures calculated in last post so may be typo errors etc....you can check yourself

[Guest]

Thanks.

[Guest]

Thanks.