PSN-L Email List Message

Subject: Re: Question Please
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
  Subject: Question Please


  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
Subject: Question Please

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

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