Island Time 1,235 Posted April 13, 2015 Author Share Posted April 13, 2015 The last block is no longer providing a 1:1 advantage - in this case not giving a 3:1 MA that it would have if the haul was parallel. Easy way to work it is that the last turn is Cos 45 (or whatever your angle is) - in this case COS 45 = .71, so the total purchase in this example is 2.71:1. The loads on the lines are 10/2.71 or around 3.7 KGs. Of course they would be 3 1/3 KG with the 3:1 tackle. Link to post Share on other sites
NevP 0 Posted April 13, 2015 Share Posted April 13, 2015 Ahh yes. I see that now. So if the angle goes all the way to 90 deg (cos 90 = 0) you get back to a 2:1 ratio. And if you go even further to 180 deg (cos 180 = -1) you have 1:1, because you are actually back at configuration 1. Link to post Share on other sites
ScottiE 174 Posted April 13, 2015 Share Posted April 13, 2015 Pretty close IT - here's my exact solution by hand calc - apologies for the hand writing - don't do this much any more. Was going to do a time-varying large-displacement geometric non-linear analysis to show you how it moves from point a to b but that would just be too nerdy and I'd have to introduce another assumption to cheat my analysis software (all that is long handed for can't be arsed!) 20150413170756.pdf Link to post Share on other sites
Island Time 1,235 Posted April 13, 2015 Author Share Posted April 13, 2015 Yep, you are right of course, if the load is free to move sideways. I should have drawn a better picture! Link to post Share on other sites
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