Grant Posted January 7, 2009 Share Posted January 7, 2009 Good morning all I hope everyone had pleasant holidays and has recovered from same. I am designing a shelf bracket (using the term designing loosely) and need help with bar length. If one wanted to end up with a 1/2" round bar with 5/8" knobs on each end and a total length of 7", how long would the 1/2" bar be in the beginning and what would the cube measure. I expect there is some approximation that goes If you want a 5/8 ball one needs a ?" cube and to get a ?" cube from 1/2 round you need to upset ??" of 1/2 round bar. tks grant Quote Link to comment Share on other sites More sharing options...
Rich Hale Posted January 8, 2009 Share Posted January 8, 2009 No one can tell you how much stock you will take to make a shape on the end of a piece of stock..And how much it takes you now will be different than after you have done a hundred pieces. So for now..take a piece of scrap bar and write down the length, bump it up and make the shape you want then measure again. that will tell you how much you need to allow for this porcess. Quote Link to comment Share on other sites More sharing options...
philip in china Posted January 8, 2009 Share Posted January 8, 2009 Or make the first upset on an overlength piece of stock. You know how much material that one took so you know how much to allow when you cut it down to length. Quote Link to comment Share on other sites More sharing options...
dablacksmith Posted January 8, 2009 Share Posted January 8, 2009 if i was makeing 5/8 knob on a 1/2 in rod i would take some 1/2 by1/4 strap wrap it around the 1/2 round and forge weld it ... as you weld it you can change it to 5/8 ball or cube ... the leingth wont change much then also...... Quote Link to comment Share on other sites More sharing options...
Mark Aspery Posted January 8, 2009 Share Posted January 8, 2009 (edited) You are going to need a bit more of a neck under the sphere to make the sphere. Trying to make a 5/8 sphere from 1/2 is going to be very frustrating. Using pi x (R squared) x h to calculate the volume of a cylinder and 4/3 x pi x (r cubed) to calculate the volume of a sphere a 5/8 diameter sphere =0.128 cubic inches and a 0.651 ( 11/16) of 1/2 inch diameter round = 0.128 cubic inches About the same So by the time you knock the corners of the 1/2 inch round and forge it to a sphere you will need 11/16 of 1/2 round per ball and 5 3/4 of 1/2 inch round between the balls for a total length of 7 1/2 inches. BUT...If you make a neck under the sphere (and I think you will have to) that measurement will have to change. I have attached some photos of a ball formed from 1/2 inch diameter round with no upset. Edited January 8, 2009 by Mark Aspery Quote Link to comment Share on other sites More sharing options...
Grant Posted January 8, 2009 Author Share Posted January 8, 2009 Good Morning All Thank you for your thoughtful responses. I realize that we are not rendezvousing in space, however I am trying to get a general sense of things and my current sense is that every time the material is worked some is lost. You can not get it all off your shoe. The way I get to a sphere is from a cube; divide the cube into upper and lower hemispheres then knock the quarters to eights, eights to sixtenths....... If one started with a 5/8 cube (.244 cubic inch) you would have twice the material needed for a 5/8 sphere (.128 cubic inch) in the arithmetic sense. I see the process as squaring the 1/2 round, upsetting and forging this square to a cube and forging the cube into a sphere. Should I figure a loss factor in these steps? Dablacksmitdh, adding additional material is a piratical idea. Mark, are you suggesting that more material needs to be added to the transition from bar to sphere. -grant Quote Link to comment Share on other sites More sharing options...
Grant Posted January 8, 2009 Author Share Posted January 8, 2009 (edited) Mark, thanks for the pictures. I'm trying not to neck the transition. The bracket is a typical L shaped shelf bracket with a 4 1/2" id circle in the corner out of 5/8 round. Two 1/2 round elements (the ones in question) tie the ends of the bracket to center circle. I want the 5/8 knobs to integrate the 1/2" pieces with the circle. Thanks for all the thought you have put into this. -grant Edited January 8, 2009 by Grant clarify Quote Link to comment Share on other sites More sharing options...
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