I have a question for an engineer please.

[Ben]

The weights I'm currently working with weigh exactly one pound according to my postal scale. They are cylindrical and measure 3/4" thick (or long) and have a 2 1/8" diameter (they're made from lead I poured into small Red Bull cans.). If I decrease the diameter to 1" how long will the cylinders be?
Thanks in advance for any help.

[Ben]

Oops- sorry- I found this link.
http://www.matweb.com/tools/weightcalculator.aspx
Looks like it would be 3.2 inches long (the internet never ceases to amaze me.)

[path_finder]

Dear Ben,
Unfortunately, you are too much confident in the computers. I guess your result was wrong.
On my own side I refuse to be the slave of any computer.
May be, I have been polluted by the HAL deconnection in '2001, a space odysse' from Stanley KUBRICK...
So far like Tarzan, I made the computation myself, see hereafter.
This kind of exercise is good against the Helzenheimer disease...
Many thank to you.


The volume of a cylinder is given by: V = pi x R² x L
where R is the radius, L the length, pi = 3.14

If you decrease the diameter of 1" the volume will decrease from V1 to V2 (V1=before, V2=after)
V1= 3.14 x (17/16) x (17/16) x (3/4)
V2= 3.14 x (9/16) x (9/16) x (3/4)"
V1 - V2 = 3.14 x ((289-81)/(16x16)) x 3/4 = 3.14 x (208/256) x 3/4
where (V1-V2) is the mass to add to the V2 cylinder, which will have a new length L3

The old volume V1 must be equal to the new volume
V1 = 3.14 x (17/16) x (17/16) x (3/4) = 3.14 x (9/16) x (9/16) x L3
then
L3 = (17/16) x (17/16) x (3/4) x (16/9) x (16/9)
L3 = (17 x 17 x 3 x 16 x 16) / (16 x 16 x 4 x 9 x 9) = (289 / 108)" = 2.6759" equal about to 2" 2/3


I can now testify the calculation of this computer is correct. See below the screen shots.

[Ben]

Thanks, PF, for your response and your time. Ungowa Tarzan!

[mickegg]

Path

Ben required the diameter to be reduced to 1 inch diameter....not by 1 inch diameter.

r1 = 1.0625 L1 = 0.75

r2 = 0.5

L2 = (r1 squared x L1) / r2 squared

L2 = 3.38671875

Regards

Mick

Edit.......pi cancels out

[Ben]

Yes, and I also noticed that no material was selected on the screen shots

[/img]

The actual weights are indeed 2 1/8 inches long. I drilled 1 1/4 holes into 4X4 blocks. The holes were about 2 1/2 inches deep. I then set the blocks on the scale and turned it on. It sets at zero. I then poured the molten lead into the hole until it read one pound. When the lead cooled, I split the wood with chisels.
So, the physical weight verifies the calculator- at least for lead.

Edited to add--

I did start with a 1" diameter in mind, but a three inch length was way too long for the model I'm working on, so I settled for a 1 1/4" diameter.