Double whip tackle

[Kneejerk Bombas]

I disagree, big mechanical advantage, the load moves twice as far/fast as a single rope.

It's not always about lowering the load, when you have too much power and not enough speed or distance, this is the ticket.

 

[rbtree]

Good discussion,guys, I edited my rather basic comments.

 

[TheTreeSpyder]

The mechanidal advantage is fractional i beleive; ie. 1/2 power therefore 2x speed; from same input.

This is because 1/2 the pull is exerted on anchor, and 1/2 on load; because the input is divided between these 2 by pulley on truck. As the truck pulls forward 1'; it shortens both legs into the pulley by 1' at same time; therefore truck will move forward 1' and load will move forward 2' (if line is already stretched); in the same amount of time as truck moves forward. This is where the speed comes in, as it is traded for power. So this could be used for shorter pulls than load movement and speed; when there is reduced run for input; but loads more power than needed; the opposite of having plenty of run; and not enough power.

Mechanical advantage is a multiplicative operation; whereby "0" advantage is more properly expressed as "1"; 2x power as 2/1; 2xspeed as 1/2. The resistance of any load is a set, singular number to be overcome by force; anytime the input moves a diffrent measured amount than the load; the ma is not none or "1"; it is either fractional or positive. If the input effort moves farther than load, that input energy is concentrated into a shorter distance; power increases; mechanical advantage is positive. If the input effort moves a shorter distance than load; the power of the input is 'diluted' over a longer distance; ma is fractional, power drops; but speed increases in trade. Since the load that must be overcome is a set number that = power x speed; as one of the 2 multipliers change; the other must change reciprocally to still equal same force/distance in time needed to move a given load. In other words; irrespective of friction, in simple (non-compound) machines; the change in power x change in speed will equal 1. ie. a 2/1 will move 1/2 as fast; so the total effort to move a given load will be the same; just manipulated bertween speed and power. Putting 2 mechanical advantage systems on a load parallel to each other will increase their power arithmetically; putting them in series (compound) will increase their power geometrically.

If you learn to see this daily in levers, screws, ramps, gears etc.; it will be easier to see and decode from our operations. For it is all the same; exactly the same but diffrent: trading back and forth speed and force to equal same product in an equation.

 

[TheTreeSpyder]

Joe and i were conversing on simple physics and he suggested that i post these 2 web sites:

http://www.sciencejoywagon.com/physicszone/lesson/05work/consofenergy.htm

http://www.sciencejoywagon.com/physicszone/lesson/05work/workpower.htm

Of course if you took the open system (control line off load); and put a climber on the lowering pulley; gave the climber the control line (closed system=control line attatched to load; doubling as another support leg) it would make the 2/1 DWT a 3/1. Write?

If you use a dropeye pulley (my favorite) on the anchor; with another pulley on the dropeye; you can redirect the terminating end from the dropeye back to the load. This would give 3 lines of support on load for an open system; 4 if a closed rig; for 3/1 or 4/1. Looking at single anchors here.

Raising from 2/1 to a 3/1 will put less pull on the anchor per same load in an open system. In a closed system the pull on anchor will be = to the load itself. In an open system the pull on anchor will always be greater; than the load. Looking at single anchors, no friction here, ground controlled.

The pull on anchor, open system, with pulleys i beleive could be expressed as = (load) + (load x (the inverse of the mechanical advantage)). That would be 100# hanging@ 'no' mechanical advantage would be=(100)+(100x(1/1))=200#. For 100#@2/1 that would be =(100)+(100x(1/2))=150#. For 100#@3/1; anchor pull would be = (100)+(100(1/3))=133etc. So another considerationfor using a mechanical advantage system; is manipulation of anchor load as well as direction,distance ,time (speed) and power.

 

[Joe]

You posted a reasonble reply KC. You're math seems to work for figuring overall rigging point loads when using pulleys for rigging. Another point in your post you didn't mention was the fact the rigging point sling needs to handle no more than the load instead of twice the load.

I've figured in the past a system using pulleys with 2" diameter sheaves(the small cmi pulley with steel cheek plates like at Sherrill) and 1/2" 16 strand line(breaking strength new=7000 lbf) to handle about 900 Lb•ft of work with many cycles to failure for the dwt. This means one can do quite a bit of lowering with this system without fear of the system failing.

The best part about the physics links you posted is they're very easy to understand.
The concepts jump out at the reader and there is no misunderstanding which concept is being discussed. If one is being introduced to physics for the 1st time, this is a good link for which to start learning. It's also a good and quick review for the people who've forgot the material.
Thanks for sharing them.

Joe

 

[TheTreeSpyder]

Hmmmmmmm thanx Joe................

But, i would think that the sling on the anchor would carry the same load as the anchor. Actually, that the anchor, sling, biners, pulley etc.; would be taken as a 'redirect' unit, ie. all carry the same load? So, that in a simple redirect, no friction, 1/1, they all would be carrying 2xload?

Personally, i prefer the 2.37" pulleys from Sherrill. For as the numbers expressed by Samson; for a 1/2" braided line; the percentage of maintained line strength is up 20-25% over the 2" pulley. Also, 5/8 can be run through them; and they are rated at a higher capacity characteristically. i think the dropeye is the most flexible; and use it as a standard pulley, on my belt.

Still don't understand your previous statement that a redirect with friction puts more load on anchor than redirect without friction (pulley). It seems to me that the total load on anchor would be sum total of all lines (support and control) of pull. On a pulley that would match load, with friction, brake force on control line would be less i think. Wereby if total stopping or braking were employed at anchor; control line load would reduce to '0', and anchor would only bear force of load, as there would be no brakeforce redirecting to ground. As, less friction was allowed on anchor, control line would have to make that up to maintain enough brake force to control load, as this force is on contrlo line pulling on anchor, load on anchor increases; to my feeble way of thinking...............


i also have quite a lot of success with choking a sling and pulley on a vertical spar and and using as a back up anchor for delivery point anchor; place over lwering zone. In this way i can adjust somewhat how much each anchor carries by sliding system up and down on the spar, adjusting for diffrent loads and primary anchors around the tree from this central point. So an anchor point that is excellent for where it lowers too, but questionable for load; can be used with this backup support! Also, in this way, rigging line runs in upside down "U" rather than "V"; keeping groundies from being pulled into load.

 

[Joe]

Originally posted by TheTreeSpyder
Hmmmmmmm thanx Joe................

But, i would think that the sling on the anchor would carry the same load as the anchor. Actually, that the anchor, sling, biners, pulley etc.; would be taken as a 'redirect' unit, ie. all carry the same load? So, that in a simple redirect, no friction, 1/1, they all would be carrying 2xload?

<i>I should have thought this statement through a little better. As long as the working end of line is anchored separate of the rigging point sling, like with the dwt,
then the load on the rigging point anchor sling will = the load <b>+</b> or <b>-</b> rope angles.</i>

Personally, i prefer the 2.37" pulleys from Sherrill. For as the numbers expressed by Samson; for a 1/2" braided line; the percentage of maintained line strength is up 20-25% over the 2" pulley. Also, 5/8 can be run through them; and they are rated at a higher capacity characteristically. i think the dropeye is the most flexible; and use it as a standard pulley, on my belt.

<i>The rescue pulley which has a 2.37" diameter sheave has a 2" tread diameter. This gives an acceptable 4:1 bend ratio. The actual pulley is pictured in the March Sherrill advertisement at this site. I don't think I'd use a 5/8" line with this pulley.</i>

Joe

 

[TheTreeSpyder]

Thanx, for correcting me on that Joe!

i just assumed that the rated diameter of the sheave would be of the necessary info for calculating line percentage strength! Especially when rated that way by a manufacturer of both line and device! But as you pointed out i missed that in "Arborist Equip.". And it is no small diffrence to me for i still state this is the part of the graph were the line strength is decreasing quicker per drop in sheave diameter! So in my mind; putting a 1"pulley on a 1/" line to load is really pushing it, especially with human cargo!

As far as not using a 5/8 line in these pulleys; it kinda goes both ways i think! i wouldn't load them with 5/8 singlely because of the small bight per rope diameter. But, in an application with 2 anchors; like super strong high anchor, sending force straight into ground all day; then 'floating' one of these pulleys around with the 5/8 in it; placing it to run off another anchor too; in this way loads are shared, and the line doesn't make such a sharp bight, so is safe i believe. This is a very strong, flexible strategy for working diffrent loads to diffrent positions, with less swing all day.

One interesting way i've used it is to make the 'primary' anchor (the one the load lowers directly from?) a springy limb for rigging slight lift. Pulling tight gives 2/1 on tightening this spring; till it got 'bow' tight of course. But it can help pull around a hinging load horizontally by taking up line slack! In doing so the 'primary' anchor bends down some; the lower it bends the more the 'other' anchor takes the load for it is now higher in comparison! The 'other' anchor can also be backup in case of total failure. i find these to be very interesting features to manipulate! And once again as i can understand the flow of the math; can make informed decisions.

See ya; gonna go play!