## PSN-L Email List Message

From: tchannel1@............
Date: Thu, 13 Mar 2008 10:28:52 -0600

```Good question, and here's what I found:   I had this issue on the copper =
vertical, the bottom hinge is where I tried to use a roller on roller, =
then I tried a foil hinge.
Both failed big time.  Because of the forces and the different angles of =
pressure, the roller on roller keep sliding off. I resolved this using =
the eyebolts, where that force or forces were retain as the rotation =
change and the resulting angles.  Since the roller was retained inside a =
eyebolt its contact point just move around the eyebolt finding a new =
spot on which to pivot.

Not to suggest you should use eyebolts, but they worked for me, but =
doing this may show you where the roller on roller need to be located.

As to you question, I think all the factors would change the roller on =
roller locations and angles, A, B, C and all the angles, change one and =
the roller would need to be in a different location and or angle or it =
would slide off.   Thats why if the roller was mounted on a swivel, it =
could be rotated in all directions until the apposing roller would not =
slide off.
----- Original Message -----=20
From: Jerry Payton=20
To: PSN-L=20
Sent: Thursday, March 13, 2008 8:55 AM

Hopefully, I can frame this question understandably.

In our standard garden-gate configuration:   Consider a traditional =
right triangle with side A (vertical) & B (base) with hypotenuse C.  If =
a mass is attached to the BC end, I assume that there is an applied =
force to keep point AB against the pivot, whichever method is used =
there.

My question is, "Does it matter what the angle of the hypotenuse is?  =
Would, say a 30 degree angle work as well as a 45 degrees, or is there a =
cutoff angle to maintain the horizontal force against the pivot?"

This could directly affect the height of A when constructing a Lehman. =
Of course, I think I remember (its bee a long time!)that an equilateral =
triangle is more stable.  Therefore, I assume it might depend upon the =
length of the arm (BC).

Regards, and "thinking too much"

Good question, and here's what I =
found:  =20
I had this issue on the copper vertical, the bottom hinge is where I =
tried to=20
use a roller on roller, then I tried a foil hinge.
Both failed big time.  Because of =
the forces=20
and the different angles of pressure, the roller on roller keep sliding =
off. I=20
resolved this using the eyebolts, where that force or forces were retain =
as the=20
rotation change and the resulting angles.  Since the roller was =
retained=20
inside a eyebolt its contact point just move around the eyebolt finding =
a new=20
spot on which to pivot.

Not to suggest you should use eyebolts, =
but they=20
worked for me, but doing this may show you where the roller on roller =
need to be=20
located.

As to you question, I think all the =
factors would=20
change the roller on roller locations and angles, A, B, C and all the =
angles,=20
change one and the roller would need to be in a different location and =
or angle=20
or it would slide off.   Thats why if the roller was mounted =
on a=20
swivel, it could be rotated in all directions until the apposing roller =
would=20
not slide off.

----- Original Message -----
From:=20
Jerry =
Payton=20

To: PSN-L
Sent: Thursday, March 13, 2008 =
8:55=20
AM

Hopefully, I can frame this question understandably.

In our standard garden-gate configuration:   =
Consider a=20
traditional right triangle with side A (vertical) & B=20
(base) with hypotenuse C.  If a mass is attached to the BC =
end, I=20
assume that there is an applied force to keep point AB against =
the=20
pivot, whichever method is used there.

My question is, "Does it matter what the angle of the hypotenuse=20
is?  Would, say a 30 degree angle work as well as a 45 degrees, =
or is=20
there a cutoff angle to maintain the horizontal force against the=20
pivot?"

This could directly affect the height of A when constructing =
a=20
Lehman.  Of course, I think I remember (its bee a long time!)that =
an=20
equilateral triangle is more stable.  Therefore, I assume it =
might=20
depend upon the length of the arm (BC).

Regards, and "thinking too much"
Jerry
```