Advanced quiz question for you all....

[Yoda]

Yoda - Is there still fog in the swamp? I'm just trying to avoid hatching one of these.



Thanks,

Bagger

Hey, you're the guy who 'couldn't help yourself.' [Post #16 (of 91!)]

So...Leave me out of this!

I'm perfectly at home watching the 'braintrust' work from the sidelines. Plus, Mathew clued me in before he launched this little TGM soiree. So, I claim disqualification by reason of competitive advantage. Or insanity. Take your pick!

And so it goes...

From the arena floor:

"We who are about to die salute you."

-- Gladiator's Salute

And from the grandstand:

[Mathew]

I want to proceed further mainly to get these never talked about relationships down and ill post a picture at some point to clarify things.

Now the next question is how do we plug the hinge action accumulator3 motion into our previous equation relating to the plane I discussed here.

You need to reference the angle of the left arm relative to the inclined plane. If we reference this plane from the inclined plane itself and not the ground like we do for hinge action, it will be directly vertical to the inclined plane through the angle of the left arm. We can then work out the angle that the left arm is above the inclined plane. This can be the same as the angled hinge action (although unrelated) if you are talking about the angled plane that goes through the inclined plane vertically and not one of its other infinite possibilities. This plane will be the same as Jens plane when accumulator 3 has turned directly towards the inclined plane.

Key

a = is the angle of the relationship between the where the plane I quoted above from when it directly goes into the ground plane (low point) vs where it actually is in degrees - relative to the angle this makes on the inclined plane.

b = This angle is the inclined plane going into the ground plane (this isn't exactly true but to elaborate would require more complication) at an angle. If you create a plane that vertically goes through both the ground and the inclined plane at 90 degrees of its intersection which is the plane line, you can find the angle on this plane

c = The hinge action plane off from the horizontal eg - Horizontal hinge action = 0 degrees and Vertical Hinge Action = 90 degrees

The equation is - ill write this like your doing it on a calculator

a divided by 90 times b minus c

Congrats you now know everything there is to know about the left arm in its own seperate compartment and its geometrical relations.

Now you can just mess around with the equations as you see fit - substituting this equation for the x variable on the last one....

Everyone will have to admit this is really cool stuff. This didn't come to me over night - I've just never had to put this into words before because I just visualise these, so I'm sorry if my explanations aren't perfectly clear.

Edit a and b where labeled wrong....

[12 piece bucket]

Neil,
Some things are just too far out there to deal with- too dangerous! You were brave to post this on the forum- I've had thousands of personal PM's on this exact issue- give me time Neil, give me time! Can't rush these things.

I would just like to say to Bucket.....we'll sometimes a picture is worth a thousand words... I'm coming for that chicken head!

Booger Eater.

[12 piece bucket]

Hey, you're the guy who 'couldn't help yourself.' [Post #16 (of 91!)]

So...Leave me out of this!

I'm perfectly at home watching the 'braintrust' work from the sidelines. Plus, Mathew clued me in before he launched this little TGM soiree. So, I claim disqualification by reason of competitive advantage. Or insanity. Take your pick!

And so it goes...

From the arena floor:

"We who are about to die salute you."

-- Gladiator's Salute

And from the grandstand:

And all that time I thought choking the chicken was . . . uh nevermind.

[Bagger Lance]

The formulas you provided help clarify the roll aspect. Thanks again Mathew.

I can see it more clearly, but what would be great is if you could provide a couple of graphic examples.

On each graphic, pick any angle for #3 accumulator. Say 20 degrees at impact. Then show how wristcock and roll are calculated at the release point based on Jen's plane and the turned shoulder plane. Again, pick any reasonable angles.

Examples would help a bunch.


Bagger

[golf2much]

Key
a = This angle is the inclined plane going into the ground plane (this isn't exactly true but to elaborate would require more complication) at an angle. If you create a plane that vertically goes through both the ground and the inclined plane at 90 degrees of its intersection which is the plane line, you can find the angle on this plane Vertical to the ground, or vertical to the inclined plane? Can't be both can it? SO if a= the angle of the inclined plane relative to the ground plane, why do you need the extra plane to calculate this angle?

b = is the angle of the relationship between the where the plane I quoted above from when it directly goes into the ground plane (low point) vs where it actually is in degrees - relative to the angle this makes on the inclined plane.So b really isn't an angle with a specific geometric location, but the difference between two other angles, or the difference between the same geimetric angle in two different positions, right? This would be the "how open does the book get" question? If it is the difference between the angle of Jen's plane to the IP at the low point and the angle of Jen's plant at the top

c = The hinge action plane off from the horizontal eg - Horizontal hinge action = 0 degrees and Vertical Hinge Action = 90 degrees

The equation is - ill write this like your doing it on a calculator

a divided by 90 times b minus c

To simplify then,

a= the angle of the inclined plane relative to the ground plane
b= the difference between ? Jen's Plane at position 1 and Jen's plane at posiition 2?
c = The hinge action plane relative to horizontal or parallel


x=(a/90)*B-C

A labeled graphic would be very helpful

G2M

[Mathew]

To simplify then,

a= the angle of the inclined plane relative to the ground plane
b= the difference between ?
c = The hinge action plane relative to horizontal or parallel


x=(a/90)*B-C

A labeled graphic would be very helpful

G2M

I accidently labeled a to mean b....

A graphic would be useful for both equations and will them soon....

The one that your not too sure on is the angle formed by the left arm plane vertical to the inclined plane relative to its low point position on the inclined plane (hence directly towards the ground plane)....

[golf2much]

I accidently labeled a to mean b....

A graphic would be useful for both equations and will them soon....

responding to your edit...

To simplify then,

B= the angle of the inclined plane relative to the ground plane
a= the difference between ? Jen's Plane at position 1 and Jen's plane at posiition 2?
c = The hinge action plane relative to horizontal or parallel


x=(a/90)*B-C

[Mike O]

Mike; "Lines of Intersection" (with the wall). Of course these will be parallel. Any two lines, starting parallel to one another(the book edges), when extended to infinity, their lines of intersection with a common plane (the wall) will also be parallel.

In your example, you are thinking of the surface plane of the book cover, not the line or edge of the plane When the plane of book cover 1 and the plane of book cover 2 intersect a common plane, like the wall, each transcribes a line. These two intersections will be parallel to one another, even though the direction of the two book covers approaching the plane (wall). Try it for your self. Take a hard v=cover book and hold it against a solid surface that you can write on. Hold the book open to the amount of your choice and while hilding it in that position, trace the edge of the book cover where it touches the hard surface. Try it several times, at different amounts of open or closed. The tracings will always be parallel.

G2M

G2M,
I understand your post- thanks. So the edge of the front book cover and the back book cover edge, as they touch or run through the wall - whether at an angle or when both surfaces are parallel to each other- create lines on the vertical wall that are always parallel to each other- even when the plane of the covers are not- that is when the "up and down" angle of the planes are different for the two- they both still create parallel plane edges on the vertical wall. Now, if the plane angles are not the same "side to side" and they go through the vertical wall then those plane angle edges will not be parallel.

Now, could you make another post and just tell me what edge of the swing plane and what edge of the #3 accumulator plane are parallel? I'm afraid of looking back at all of the other posts-as I have a fear of fog! So don't worry if your answer is really simple- cause that's actually what I'm looking for.
Thanks,
Mike O.

[golf2much]

G2M,
I understand your post- thanks. So the edge of the front book cover and the back book cover edge, as they touch or run through the wall - whether at an angle or when both surfaces are parallel to each other- create lines on the vertical wall that are always parallel to each other- even when the plane of the covers are not- that is when the "up and down" angle of the planes are different for the two- they both still create parallel plane edges on the vertical wall. Now, if the plane angles are not the same "side to side" and they go through the vertical wall then those plane angle edges will not be parallel.

Now, could you make another post and just tell me what edge of the swing plane and what edge of the #3 accumulator plane are parallel? I'm afraid of looking back at all of the other posts-as I have a fear of fog! So don't worry if your answer is really simple- cause that's actually what I'm looking for.
Thanks,
Mike O.

The plane Mathew defined as Jen's plane starts out at the top as vertical/perpendicular to the swing plane. As the left arm moves down towards impact, Jen's plane rotates to a point near impact that is parallel to the SP. Picture the book again; if the spine is aligned with the left arm plane and one cover is aligned with the swing plane, place the other cover at 90* to the swing plane. Now mimic the movement of your left arm and wrist as you approach impact and allow the upward pointing book cover to rotate as though the book is closing. The closing cover will move to and through a position where the once upward point book cover is now parallel to the SP. I'm still working on the relationship with the #3acc/wrist cock. Still some fog there for me too. Maybe once the graphic is finished it will be clear to me. I have some thoughts on this but I'm still thinking.

G2M