Your Homework This Weekend

[pope]

OK, geeks, here's a question that can be answered by an average student of high school physics:

 

You're rigging a rappel off of two questionable bolts, which are separated laterally by a distance of X and which are level with each other vertically. You have only one 24-inch sewn sling. You want to connect to these bolts in a way that will minimize the force on each one.

 

1. Is the American Triangle always better than using a load-equalizing "V"? Does the answer depend on the separation X?

 

2. Suppose that for some separation X, the force on a bolt in the American triangle is F1, but the force on a bolt in the "V" system is F2. For what value of X is the ratio of F1 to F2 highest? What is the ratio in this case?

[allthumbs]

Bong-up dude!

[Necronomicon]

Force is a vector quantity, and as such, requires a direction. Please restate question 2 so that it makes sense.

Try this for help...

http://www.physicsclassroom.com/Class/vectors/U3L3b.html

[pope]

quote:

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Originally posted by trask:

Bong-up dude!

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One more outburst like that and you'll be explaining your insolence and piss-poor attitude to the Headmaster!

[pope]

quote:

---

Originally posted by Necronomicon:

Force is a vector quantity, and as such, requires a direction. Please restate question 2 so that it makes sense.

Try this for help...

http://www.physicsclassroom.com/Class/vectors/U3L3b.html

---

OK, wiseacre, we're looking for the value of X (bolt separation) that makes the ratio of MAGNITUDES of F1 to F2 highest. That is, F1, it turns out, always has a magnitude higher than F2 (which answers the first question), making the American Triangle a poor choice for all values of X.

 

In question 2, you'll need to write F1's magnitude as a function of X. Do the same with F2, then form the ratio of magnitudes F1/F2 as a function of X. Study this function for a maximum, then report.

 

In essence, there is a bolt separation X for which choosing the American Triangle is not only the wrong choice, the degree to which it is a poorer choice is highest. Want a hint? OK: 0<X<24.

 

No more hints.

[allthumbs]

quote:

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Originally posted by pope:

quote:

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Originally posted by trask:

Bong-up dude!

---

One more outburst like that and you'll be explaining your insolence and piss-poor attitude to the Headmaster!

---

If she's that tight-assed Headmistress with the bodacious funbags in the French-maid outfit, she can spank my bare ass any time. I look forward to her flogging!

 

[ 06-08-2002, 03:13 PM: Message edited by: trask ]

[Necronomicon]

Could you provide a free-body diagram? Are we to use the metric system? Can you provide a load value? Are you talking about the MAGNITUDE of the RESULTANT force on each bolt, the HORIZONTAL vector component, or the VERTICAL vector component?

 

[ 06-08-2002, 03:21 PM: Message edited by: Necronomicon ]

[allthumbs]

Of the Headmistress or Pope's lesson?

[Terminal_Gravity]

Pope you are such a pompous Dork. You have also really presented an unworkable or incomplete problem.

 

Firstly, if all you really have is one "sewn" sling your options are very limited. Girth hitch one bolt and then what??

 

I assume that you have a way to connect the sling to the bolts...a couple of carabiners, maybe? So "x" really becomes the distance between connection points.

 

Also, I assume that you would always tie at least a slider not in the V method which will cost about an inch of sling. That very point invalidates your later statement that the AT is never stronger than a V. (yes, for any real case senario but you are talking in absolutes)

 

Furthermore, niether system is the best for most reasonable cases. An AT clove hitched at both connections puts the least load on both bolts...you know it, and I know it.

 

I will compare the three systems not just your simple and innapropriate two.

 

For the sake of analysis I will make a couple of assumptions.

 

First, You have a 'biner or split ring to hang at the bottom of the sling (powerpoint).

 

Second, a slider knot at the bottom of the V requires one inch of sling.

 

Third, 2 clove hitches at the connection points require 4 inches of sling.

 

Fourth, the spacing "x" is from the connection point not the bolts themselves and those connection points don't affect the calculations by moving...not true in the real world.

 

And fifth, the load is perpendicular to a line drawn between the connection points...also not always true.

 

Weight of Rappeler = Lr

Load on each bolt = Lb

 

For the American Triangle the formula is:

 

Lb=1/Cosine(90-(Arccosine((x/2)/((24-x)/2))/2))*Lr/2

 

For the V sling the formula is:

 

Lb=1/Cosine(90-(Arccosine((x/2)/((24-1)/4)))*Lr/2

 

And, not surprisingly, the formula for a clove hitched AT is:

 

Lb=1/cosine(90-Arccosine((x/2)/((24-4-x)/2)))*Lr/2

 

Note: I kept extra parenthasis and did not reduce anything so that even you could follow.

 

If you solve for x, you quickly see that the AT and V are equivalent at slightly over 11 inches and the AT is superior above that ... refuteing your low brain statement.

 

Also, the clove hitched AT is stronger ( by a bit ) then the "V" up to an x of 8.5 inches

 

So, in summary the strongest is the CH-AT to 8.5 then the V to 11.2 and then the AT.

 

Below is a chart

 

Load to bolts (#'s) with 1000# system load

Spacing------AT-------V------CH/AT

---1--------723-----502-----501

--1.5-------732-----504-----502

---2--------742-----508-----503

--2.5-------752-----512-----505

---3--------764-----518-----508

--3.5-------776-----525-----512

---4--------791-----533-----516

--4.5-------806-----543-----523

---5--------824-----555-----530

--5.5-------844-----569-----540

---6--------866-----586-----553

--6.5-------892-----606-----570

---7--------922-----630-----593

--7.5-------957-----660-----625

---8-------1000-----696-----671

--8.5------1052-----742-----742

---9-------1118-----803-----870

--9.5------1204-----887----1174

--10-------1323----1013-----N/A

-10.5------1500----1226

--11-------1803----1714

-11.5------2500-----N/A

 

Clearly, the AT is not the best choice. If the spacing is ever more than 9.5 inches you had better figure out something else like hacking off a piece of your rope.

 

Pope on a rope...sounds like a bath soap.

 

[ 06-09-2002, 01:24 PM: Message edited by: Terminal Gravity ]

[pope]

quote:

---

Originally posted by Necronomicon:

Could you provide a free-body diagram? Are we to use the metric system? Can you provide a load value? Are you talking about the MAGNITUDE of the RESULTANT force on each bolt, the HORIZONTAL vector component, or the VERTICAL vector component?

---

Let F1 symbolize the magnitude of the resultant force on either bolt in the American Triangle. The solution is independent of the type of units you use for force, and it is independent of the load applied at the power point, although if you wish, you may assume a 500 pound load is applied. BTW, the vertical component of force on each bolt is always half the load at the power point, in either system. Let me see what I can do about a diagram.

 

Trask, that's some find video...you are trying to dislodge a cocaine booger?

[Terminal_Gravity]

Sorry about the poor formating on the chart. I'd rather wiz on the computer than become a wiz on it.

 

I also made some full colour graphs but they would not transfer at all.

 

I'm dissapointed in you, Pope. That eloquent troll I posted above and no comeback?

[pope]

Terminal,

 

Not to be an asshole, but before you try to find "the problem with the problem" (suggesting the question is invalid since I foggot to account for the amount of webbing used in wrapping half way around a 'biner, etc.), you might want to check your mathematics. For example, the V-sling equation you've given could be written as Lb/(Lr/2)=[(24-x)/2]/(x/2), when you recall that cos(Arccos(y))=y, etc. Examine this proportion and then refer to the similar triangles in your diagram. I think you'll see that you're hot on the trail of the solution, but not quite there. Don't quit (quitting becomes a habit that is difficult to quit).

 

The American Triangle is several degrees more complicated. For simplicity sake, just assume that the webbing used in wrapping around a biner is negligible and that you're slipping the webbing over cold shuts, etc. Don't get bogged down in the messy details until you get the big picture.

 

My solution shows the American Triangle always applying slightly higher forces on each bolt (although never as high as 200% of what the V-sling system does for a given value of X).

 

Also, I'm thinking of a 24-inch sling as being a loop that is 24 inches when you stretch it between your fingers (would be 48 inches if you cut the loop and stretched it out).

 

For the V-system, I came up with f=(0.5F)(12)/sqrt[144-(x/2)^2)], where f=the magnitude of the force applied to the bolt and F=weight of your fat girlfriend.

 

I'll release the American Triangle calculation after you scream, "Uncle!"

[pope]

BTW, what is this "American Triangle cloved hitched at both connection points"? Anything I can visualize would seem to shorten the sling so dramatically that horizontal forces would become pretty high.

 

Since you enjoy extrapolating the hypothesis, imagine if you just cut ten feet off of your rope and used that for a sling!

[pope]

OK, guys, I'm going to the neighbors to watch the big fight. Tomorrow I'll be in Leavenworth. Solutions presented after 10 pm Sunday night will receive only partial credit. Good luck, and may the force be with you (yuckity yuckity yuck).

[Terminal_Gravity]

Yes, indeed - cos(arccos(y))=y but my formula is cos(90-arccos(y)).

 

Simplified as you wish, I believe the force on the AT is Sqrt(2) times the V when x = 0 and reaches a maximum of 1.5 times the load on the V (using your sling) when x = 12 inches.

 

I also believe that I do understand the big picture and my formula for the AT is correct (if you substitute 48 for 24), but not reduced. Check it out.

 

But you win, my brain is getting all mushy and shit.

 

BTW, she's not that fat.

 

If you carefully clove hitch each cold shut so that the horizontal run of an AT is not tight, the force vector is not 1/2 of the angle of the lower legs of the triangle; it is the angle just like in a V system. In cases where the extra loops don't use a significant amount of the sling (most) you get a better angle and less resultant load than the V. That being said, the difference may not be worth the extra complication and if a CH slips you could be back to the AT.

[pope]

quote:

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Originally posted by Terminal Gravity:

Yes, indeed - cos(arccos(y))=y but my formula is cos(90-arccos(y)).

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My mistake...I was reading your clove-hitched AT formula.

 

quote:

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Originally posted by Terminal Gravity:

Simplified as you wish, I believe the force on the AT is Sqrt(2) times the V when x = 0 and reaches a maximum of 1.5 times the load on the V (using your sling) when x = 12 inches.

 

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Nice work, this is what I was looking for! BTW, Iron Mike Tyson took a beating. Yawn.

[Terminal_Gravity]

Thank you, Pope. I feel all warm & fuzzy inside. The trig review I had to put myself through was valuable. I have to design a tower to support a cantalivered malt bin this month... so that I can make more beer.

 

I'm sorry I called you a pompous dork...I take it back.

[pope]

And I take back what I said about your girlfriend. Now tell her to stop calling me!

[Peter_Puget]

I too can ateest to the usefulness of Trig. Just last week I was straining to stay awake in a meeting that was seemingly endless when all of the sudden I was inspired to design some garage gyms. Trig was essential in calulating the dimensions. Too bad I didn't have a trig functions on my calc and had to resort to quick approximations. But everyone thought ole Peter was hard at work taking notes and checking the numbers that were being spewed forth. My homework ought to be doing simple sums by hand as without a calc. I am worthless!