Hensley Arrow: How does it REALLY work?

Milt, in your first image you made the statement,

Stressor wrote:
Hooking a come along to between point B and a Willow tree would result only in a lateral load being applied at point C.

This is not correct. Since you can't move C without trying to shorten or lengthen segment AC, there is also a vertical component to the load at C.

The case presented in your second illustration is not clear to me, but if "there is no place for the force to go" for a force applied to A or B, how can there be a lateral load at C if there is no impetus for movement? Are you fixing point C or any of the lengths AD, DC, BE, or EC? Fixing either the points or the segment lengths is redundant and simply results in a truss structure.

In your third illustration you try to convert your truss structure to a Hensley analogy, but it doesn't work that way. First you say

Stressor wrote:
distance C-E and C-D must remain the same

Since there is no physical point C, what exactly can possibly constrain those segments to be equal?

You also said

Stressor wrote:
C is what has been called the Virtual Pivot Point

This is incorrect. Up to date in this thread, the Virtual Pivot Point has always been described as the imaginary point at which segments AD and BE would intersect if they were extended to their point of intersection. In theory, this is also termed the Instantaneous Center of Rotation. Instantaneous because no two angles between AD and BE ever intersect at the same point.

When you say

Stressor wrote:
In the Hensley, we do not actually extend the legs to point C, because we hold distance D-E and A-B constant

it is incongruent since making segments DE and AB fixed lengths has nothing to do with enabling the omission of fixed-length segments to a point C.

Point C as you describe doesn't ever exist on a Hensley application nor is it useful for any purpose.

Tim