Laser 28 - Excellent example of this great design Hamble le rice |
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Rossiter Pintail Mortagne sur Gironde, near Bordeaux |
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Sam.Spoons ![]() Really should get out more ![]() Joined: 07 Mar 12 Location: Manchester UK Online Status: Offline Posts: 3400 |
![]() ![]() ![]() ![]() Posted: 14 Jan 19 at 3:59pm |
Thanks gents
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Spice 346 "Flat Broke"
Blaze 671 "supersonic soap dish" |
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Thanks for making a mock up! I'll use those pictures for the class if anyone asks!
Also, top makes to oatsandbeans, got it right in a matter of minutes. The system is what's used on most 800s. But class rules say 4:1. But the whole thing is stupid as it means having a moving cleat-block thing (trapeze adjuster) rubbing up against the mast. There's metal chafe pads but they're always being ripped off by the cleat moving around. It seems much more sensible to have a 4:1 cascade the traditional way up, and a cleat on the mast. Then you only have two very small blocks (or friction rings) moving around.
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pondlife1736 ![]() Posting king ![]() Joined: 17 Jan 14 Online Status: Offline Posts: 106 |
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Easy way to think about it is draw horizontal line through the cascade and sum the tensions. Above the middle pulley you get 3xblack + 1xgreen. Below you get 1xblack+2xgreen. Green =2xblack so in both cases you get 5:1 (assuming fall is vertical, take cosine of angle if not).
Don't think it matters which end is fixed.
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Sam.Spoons ![]() Really should get out more ![]() Joined: 07 Mar 12 Location: Manchester UK Online Status: Offline Posts: 3400 |
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What 'e said
![]() On my mock up the bottom was fixed and the top on a bungy. And you're welcome to use the pics.
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Spice 346 "Flat Broke"
Blaze 671 "supersonic soap dish" |
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I think the reason it matters is because if the top was fixed, then the 10kg you pull on the fall line would not count in the system.
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Do Different ![]() Really should get out more ![]() Joined: 26 Jan 12 Location: North Online Status: Offline Posts: 1312 |
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After all that I'm inclined to think the system purchase is 4:1, that is to say the tension being applied to the shrouds is four times the effort applied to the black line.
Surely: to be in balance forces are opposite and equal. Therefore your later diagrams showing 50kg of tension above the system and 40kg below the system cannot exist. |
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When you have hold of the tail, and are pulling it with 10kg of force, there is only 40kg of force on the fixed block at the bottom, but that is still 50kg of force pulling down on the cap shrouds. That 50kg is equalled by 50kg of tension in the cap shrouds.
You then let the line you are pulling slip back in to the cleat and the block at the bottom carries the extra load, so there is now no load on the black tail line, and 50kg on the fixed orange line, equalled to 50kg on the purple cap shroud line.
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423zero ![]() Really should get out more ![]() Joined: 08 Jan 15 Location: United Kingdom Online Status: Offline Posts: 3420 |
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I was always told number of strings not counting one in your hand and cascade doubles, so 4to1 under that system.
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Oatsandbeans ![]() Far too distracted from work ![]() Joined: 19 Sep 05 Location: United Kingdom Online Status: Offline Posts: 382 |
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It is quite simple if you think that the load is being taken by the various lines of rope. If you put a known force into one, by pulling the black rope, you can work out the loads in the others and add them up to get the total force produced by the system.
The loads in all the balck ropes must be the same. So that is x3 the applied load. The load in the green one is confusing some, but that must be double that of the black as it is conneced to the black by a simple 2:1 arrangement. So that makes 3+2. That is it |
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