John wrote... <<<With the optimistic 50% steel estimate, you have 1 * 0.065 * 0.5 = 0.0325 square inches of steel. Assume a load of 200 lbs. That results in stress in the steel of 200/0.0325 = 6153 psi. The tensile modulus of steel is 30,000,000 psi. So under 200 lbs load, the belt will stretch by 6153/30000000 = 0.0002" per inch. If you are using that belt for a 100" long linear axis, the stretch would be 0.020.
A ballscrew for an axis that long would most likely be over an inch in diameter. A 1" ballscrew probably has a root diameter of at least 0.8", resulting in 0.502 square inches of steel. That is 15 times more steel, or 15 times less stretch. So a 100" long, 1" diameter ballscrew loaded to 200 lbs would only stretch 0.0013" instead of 0.020".>>>> Interesting maths but what would the effect of changing temperature be - which would be the more stable, he belt or the ballscrew? Best wishes, Ian ------------------------- Ian W. Wright Sheffield UK ------------------------------------------------------------------------------ Crystal Reports - New Free Runtime and 30 Day Trial Check out the new simplified licensing option that enables unlimited royalty-free distribution of the report engine for externally facing server and web deployment. http://p.sf.net/sfu/businessobjects _______________________________________________ Emc-users mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/emc-users
