{"id":524,"date":"2012-04-02T03:36:57","date_gmt":"2012-04-02T03:36:57","guid":{"rendered":"http:\/\/wirerope.net\/2015\/2012\/04\/02\/convers-chart\/"},"modified":"2020-01-14T21:57:02","modified_gmt":"2020-01-14T21:57:02","slug":"convers-chart","status":"publish","type":"post","link":"https:\/\/wirerope.net\/azwr\/convers-chart\/","title":{"rendered":"Conversion Chart"},"content":{"rendered":"

CALCULATION OF DRUM CAPACITIES <\/strong><\/h2>\n

CAPACITIES IN FEET OF ROPE (A + B) ACM <\/strong><\/h3>\n

Example <\/strong>
\nHow much 1\/4″ diameter rope can be spooled on a drum having a 36″ diameter flange, 24″ diameter drum and 30″ width between flanges?<\/p>\n

Using the above formula the number of feet equals (6 + 24) x 6 x 30 x .152 Answer \u2014 821 feet. The value of “M” is taken from the previously calculated table of values shown below. For 1\/4″ diameter this value “M” is .152<\/p>\n

Values for M <\/strong><\/p>\n<\/div>\n\n\n\n\n\n\n\n\n\n\n
Size Rope Inches<\/span><\/td>\nM<\/span><\/td>\nSize Rope Inches<\/span><\/td>\nM<\/span><\/td>\nSize Rope Inches<\/span><\/td>\nM<\/span><\/td>\n<\/tr>\n
1\/4<\/td>\n3.29<\/td>\n7\/8<\/td>\n.308<\/td>\n1 3\/4<\/td>\n.0770<\/td>\n<\/tr>\n
3\/8<\/td>\n1.58<\/td>\n1<\/td>\n.239<\/td>\n1 7\/8<\/td>\n.0675<\/td>\n<\/tr>\n
7\/16<\/td>\n1.19<\/td>\n1 1\/8<\/td>\n.191<\/td>\n2<\/td>\n.0597<\/td>\n<\/tr>\n
1\/2<\/td>\n.925<\/td>\n1 1\/4<\/td>\n.152<\/td>\n2 1\/8<\/td>\n.0532<\/td>\n<\/tr>\n
3\/16<\/td>\n.741<\/td>\n1 3\/8<\/td>\n.127<\/td>\n2 1\/4<\/td>\n.0476<\/td>\n<\/tr>\n
5\/8<\/td>\n.607<\/td>\n1 1\/2<\/td>\n.107<\/td>\n2 3\/8<\/td>\n.0419<\/td>\n<\/tr>\n
3\/4<\/td>\n.428<\/td>\n1 5\/8<\/td>\n.0886<\/td>\n2 1\/2<\/td>\n.0380<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
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