On 10/29/2014 12:04 AM, Danny D. wrote:
dpb wrote, on Tue, 28 Oct 2014 17:12:39 -0500:

Well, no, there's certainly been no mention of it and nothing in any of
the pictures I've seen is even _close_ to that...

There's almost no way you would have known how absolutely huge the
big redwood tree is downslope.

There were a few pictures of it, early on, in a different thread,
when we had laid the first few ladders and the cargo netting.

As you can tell from this diagram, there are three sets of trees
that matter, for our purposes (although many other trees exist):
http://i62.tinypic.com/ieeakx.gif

Here's a quick look downhill, from the path near the uphill path:
https://c3.staticflickr.com/3/2943/1...edf3ba0c_c.jpg

This is the uphill anchor point, on a small Monterey Pine tree:
https://c4.staticflickr.com/4/3870/1...f45d19e2_b.jpg

....

I don't need a picture of the big redwood; I'm perfectly willing to
allow as how they can and do get big; I've been through redwood country
a number of times. I was simply noting from the pictures posted near
that point in the thread there didn't seem to be much of any real size
and was more concerned of potential on the root system with the load
than whether the tree itself was sufficient presuming it was.

When the response to the question regarding angles for trying to
estimate tension needed to provide a given vertical force component
includes the justification that the angle will increase owing to the
tree flexure doesn't lend itself to thinking they're terribly big,
either. And, again, just "devil's advocate" position raising the
question...

From a diagram such as that with a few measurements one could get at
least a reasonable approximation using simple-enough analyses as
outlined in the following (beginning at 7-30ff)--

http://isdl.cau.ac.kr/education.data/statics/ch7.pdf

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