Calculating anchor loads

[TheTreeSpyder]

What is greatest Anchor Load?

 

[murphy4trees]

Reminds me of the kind of math question I loved back in high school.
OK.
Count the # of legs coming off each wieght and divide 1500 lbs by that # to get lbs on each leg.
A=1500/3 legs= 500# each
B=1500/1=1500
C=1500/3=500
D=1500/4=375 given rope is tied off to wieght as per diagram, and it is not tied off to other anchor on ground
Now to get wieght on each anchor it is necessary to add lbs of "free" end of rope pulling down on anchor, and add this to the 1500 being held up, on A,B, and C. D has no free end.
So for A,B and C the wight being held by free end= wieght being held by each leg.
A=1500 +500=2000
B=1500+1500=3000
C=1500+500=2000
D=375x4=1500
So B clearly anchors the most wieght
The final factor to consider is the effect that lifting, lowering, and holding have on anchor. I think that lifting would add wieght, holding stays same and lowering lessens wieght on anchor. Anyone know the math here?
God Bless All,
Daniel

 

[Kneejerk Bombas]

It variable, depending on how fast you lift and lower.

 

[TheTreeSpyder]

Very good, very important IMHO!

But, i think your math is off on 1 of'em.......

But still, i think it shows maximum load will be with single pulley with no mom (MA) is greatest load on anchor commonly. The control line would be like an anchor, the load a pulling force, which takes 2/1 advantage on the anchor!

Minimum load will be the load itself, as expressed in scenarios where all 'legs' of line terminate totally at anchor and load.

All lines of pull on anchor must be added, including any control lines seperate or doubling as load line. The control line would add the weight of pull it took to control load as calculated to fight the last previous leg's pull.

A load line goes from anchor to load, each straight length of line is a 'leg' (of support, especially betwixt load and anchor). The control line would be the final leg (running end) that gives final control to load's suspension; sometimes doubling as a load line, such as in typical tree climber's retrievable loop (DDRT).

Sometimes this must all be taken into consideration for making a sound rigging strategy. Whether for load or life lines.

Lowering will be less, especially expressed with smooth motion or run, as the system isn't holding the whole weight. Through a smooth running pulley system, any lifting pressure should be immediately spread through other legs, load would give up and raise as they did, limiting the amount of pressure from escalating. For the load would always be that 'mechanical fuse' giving up and moving as it is overcome.

 

[TheTreeSpyder]

Applied on life line.

Anchor Load will be the sum of all lines of pull on it.

If a control line doubles as a load line, mechanical advantage is increased.

By understanding how these things werk, you can take the same materials and assemble systems tailored with diffrent strengths and properties.

 

[murphy4trees]

Spidy,
Got that.
Check "climber lift". Doesn't work for me.
Rechecking the math I saw something I missed though it was right in front of my nose. I didn't notice the pulley vs. friction (natural crotch) legend.

wieght on anchor A=2000 as above
weight on anchor C is less than 2000 due to friction reducing weight required to hold free leg.
So on C, wieght on anchor= 1500+[(1500-wieght of friction) / 3] which will be less than 2000.
God Bless,
Daniel

 

[TheTreeSpyder]

having troubles............. just caught your post.........trying again..........

Yes that is the way i saw the math; that with single support, non frictional redirect, OAload will be max @ 2xLoad(B).

You can't beat the weight of the load itself, it will be constant, so you have LOAD + CONTROL LINE = OA LOAD. So the load will always be there, all we can do is beat on the control line pull after the load's pull to lower the pull on the OA.

A single pulley makes you match the load's pull, to keep it afloat, so you have CONTROL LINE=LOAD (B) Always, this is the outer, higher constraint i believe. i think that without imapact that the max. OA load is 2x as in this example.

Any friction, helps take load off the control line, so you have CONTROL LINE = LOAD - FRICTION (C or D or typical line over limb); will always be less than 2x, how much less depends on how much friction, but that is lowering, if lifting that friction fights rather than helps, so must be overcome, increasign the OA LOAD.

Any more pulleys on OA that redirect line to load, lowers the CONTROL LINE PULL, because it only has to match the pull of the final leg of the support. Whereby, CONTROL LINE=LOAD/LEGS OF SUPPORT ON LOAD(A). Whether that is less than the previous example depends on how many lines of support and how much friction, always looking to lessen the control line pull, as i can't beat the weight of the Load itself

If all lines end at the LOAD and OA; ie. no pull on control line below load, then the CONTROL LINE is also a leg of support; and there is no other pulls on OA that don't end at LOAD, so no more pull is on OA than the LOAD (D or DdRTor single line hanging between laod and anchor etc.). In all cases, i beleive this is the minimal constraint, whereby you can't beat the weight of the laod itself, without running it, sharing supports, or slanting lines.

So i think, in all cases (B) that doubles the LOAD'S pull on anchor will be the answer, it will be 2x LOAD hanging.

Similiarily (D) will be least at 1x.

All others (that don't lift against friction) would be between 1x LOAD and 2x LOAD.

 

[TheTreeSpyder]

Safety Tip

Some of the 2/1 that i apply for safety is for setting/testing the lifeline.

Instead, of grabbing both lines to set lifeline that is thrown into tree, bouncing up and down. Anchor one end, and jump on the other leg. That puts 2x the pull (-friction in parallel lines) to set/test. The same amount of line, setup, etc. just used diffrently and better, by applying these things.

 

[TheTreeSpyder]

All this is real simple, once ya see it, respect it and feel it a few times.

 

[Kevin]

This is an interesting link for those that haven`t seen it...
http://www.howstuffworks.com/pulley.htm

 

[murphy4trees]

This math becomes very relevant when using an adjustable false crotch with a 3:1 MA sytem to lift climber. That is line end is secured above pulley on AFC... line comes down to pulley on saddle... back up through pulley on AFC.... then back down to climber. Rides like an elevator when used with a french prussic and a little help from the ground.
That extra leg in sytem making it 3:1 instead of 2:1 significantly reduces the added wieght from control line. The wieght reduction to OA will be double that reduced to control line (effected by 3:1 over 2:1) due to AFC.
Good to know.
God Bless All,
Daniel