WeeDee

Not about to give up what, Hama?

For a minute there, I thought that they were playing it backwards! But it does have an hypnotic effect on the unwary listener, granted. Good post!

They are probably some of the many that I don't know!

What was the rope for? Or is it just some electrical line that came down with the tree?

I wish! I can identify at about the same speed as fungal growth!

That was SO easy! It has only taken me 20 minutes to sort out the ID. (Now for number 2, then number 3 ... and .... then....... Great idea - great fun! Thanks.

I don't use the art friction saver, but can it not be lowered with a throw line folowing the climbing line through, and then lowering the friction saver on the throwine?

I've not even heard of this hitch, so I'm as curious as you!

A mass of 1kg (sitting on the ground) exerts a force of (nearly) 10 Newtons due to the force of gravity. So when reading a pulley's rating as 22 kN - you could 'imagine' an approximate rating as 2.2 tonnes. Does this help at all, Hama?

Give that goldfish back!

The goldfish in the bag! I want the the goldfish!

An object will fall 4.9m in the first second - assuming air resistance is negligible and initial velocity is zero. Is this the answer to the quiz?

This is wrong - sorry! The answer should be 7448 N or 7.448 kN

A 12 stone man (76kg) falling for 1 second and stopping immediatetly (which wouldn't quite happen) will exert a force of nearly 745 Newtons on the rope. F = m X a Force = mass X acceleration due to gravity Force = 76 X 9.8 m per seconds (squared) = 744.8 Newtons I hope this helps. Dee

Ah yes - I see what you mean, Peter. Thanks.

If the Revolver were clipped straight to the Petzl OK, that would allow the blue line and dyneema sling to be left out. Would this still satisfy the purpose? It might not allow changeovers will suspended though.

That's a lot of oak! How long was spent on this job? Is there an estimated date for a re-pollard? Thanks, Dee.

Remember - that calculation assumes almost no stretch in the rope. To give a more useful result we could assume a stretch of 1m, thus giving a final force of just over 22 kN.

Hama, if it's of any use I've calculated some close figures to your imperial measurements. A 90kg mass falling 25m (approximate to your suggestion) with almost no travel after initial impact (the rope catching the falling mass) will exert a force of 315000 N on the system. So that's a force of 315 kN.

Not a good way to attract business. I wonder if Natwest require two backsides before sticking their policy up them? Productive shopping around, Dean.

Hama, let me ask you one question. WHO DO YOU THINK THIS COURSE IS DESIGNED FOR? You make it sound as if it's made for people who already HAVE a degree! But it isn't, of course. It's designed for applicants who have the ability to learn but don't have all the necessary skills and knowledge yet. I think that you're experiencing (possibly for the first time in a long time) the apprehension of venturing into a new situation without knowing all of the facts. In your everyday work you're probably on top of most situations - but this is 'beyond your control'. My advice is to trust the educational route that has been planned to get from the start of this course to the end, and have faith in your own abilities to learn something new in a situation that is not available to everyone. You'll end up one of the lucky ones - you just don't know that yet! Good luck and keep us posted with your progress. Dee

Oooooooooooooooooooooooh! Very up!

Yes - we understand each other. Thanks, Rod.

Let me ask - when retrieving the whole system, do you lower it out of the tree by the srt line with the 'ground anchor' end of the srt line eventually passing up and over the branch and then down onto the ground?

I might be misunderstanding you, Rod. But if we are talking about the same thing, Jeff Jepson describes this set-up as an Adjustable False Crotch.