Formula for Determining Size of Upset to Create Correct Tenon?

[glilley]

Making a replacement mailbox stand with post being 1" square solid and the cross bar (what mailbox will sit on) 1" x 1/2", all A36 using mortise/tenon construction. I have already punched/drifted the post 1/2" square to form the mortise and I intend to upset the 1" X 1/2" cross bar where it will join the post to create a good shoulder and butcher/swage part of it down to a 1/2" square tenon.

Now, I know the tenon needs to be 1.5 times the length of the mortise (1" post so 1.5" in this case) but how do I determine what the dimensions of my upset need to be in order to provide enough material to draw out a 1/2" square tenon 1.5" long and still leave enough of a shoulder to provide support against the post? I feel confident I can eyeball it or get some clay and work it out but is there a formula one can use to determine dimensions of upset for size/length of tenon needed? Thanks!

[ThomasPowers]

Cubic inches of steel = cubic inches of steel, fudge a bit for scale losses.

[Don A]

If I'm following your question, and considering Thomas's reply, you would need 3/4" of .5 x 1" to render a .5 x .5 x 1.5" tenon.

.5 X .5 X 1.5 = .375 (target)

1 X .5 X .75 = .375 (beginning dimension)

Correct?

[glilley]

Thomas and DonPowers - This looks like what I'm looking for. Looks like you determine cubic inches of desired result (.5 * .5 * 1.5) and reverse engineer where you maintain the width of the stock and plug in appropriate values for the other dimensions. Thanks!

[glilley]

Sorry - ...maintain the width and height of stock and plug in appropriate depth values.