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Puzzle Shows Freely Available Rigging Force Help
- Thread starter TheTreeSpyder
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TheTreeSpyder
Branched out member
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- Florida>>> USA
Re: Puzzle Shows Freely Available Rigging Force He
Actually, i think i read that speed squared X Maas/2; maybe it was velocity, and my undeducated terminology isn't geting it again, sorry. But, replacing the word speed with velocity, weight with Mass; it is kinda there; and i don't think there is too much differance betwixt the terms, especially here in this analasys? Also, i couldn't remember how to make the squared sign etc. But, i plead ya look at the point that the weight, matters less than the speed; so Zer0 impact with more line tension can be better overall.
In all humility, i think i can stand by the principals i've tried to share. i'm glad to be given the right vocabulary and remember the lessons, but i really don't think the words change much of the operation; but am glad to be empowered to share it more correctly. i truly hope i haven't done more harm than good for my efforts.
Here is another drawing i prepared on how i have made all this work fer me; by raising the tensin in part or whole; beyond what is needed by slanting and or bending the line, i think we can safely predict that Nature will want to relieve that extra tension.
i'll try to look at the puzzle there some more, this isn't on a stump, but pinned to wall??
Actually, i think i read that speed squared X Maas/2; maybe it was velocity, and my undeducated terminology isn't geting it again, sorry. But, replacing the word speed with velocity, weight with Mass; it is kinda there; and i don't think there is too much differance betwixt the terms, especially here in this analasys? Also, i couldn't remember how to make the squared sign etc. But, i plead ya look at the point that the weight, matters less than the speed; so Zer0 impact with more line tension can be better overall.
In all humility, i think i can stand by the principals i've tried to share. i'm glad to be given the right vocabulary and remember the lessons, but i really don't think the words change much of the operation; but am glad to be empowered to share it more correctly. i truly hope i haven't done more harm than good for my efforts.
Here is another drawing i prepared on how i have made all this work fer me; by raising the tensin in part or whole; beyond what is needed by slanting and or bending the line, i think we can safely predict that Nature will want to relieve that extra tension.
i'll try to look at the puzzle there some more, this isn't on a stump, but pinned to wall??
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- Santa Cruz, CA
Re: Puzzle Shows Freely Available Rigging Force He
Ken,
Think of my diagram as a leaner. The roots are at point 'O'.
The tree does not lean against anything, it's just anchored
at the roots.
force and stuff...
You can replace velocity with speed. That's fine.
The units are the same (distance)/(time).
Weight is (mass)x(gravity) and is a force.
Gravity is an acceleration.
force, momentum, energy, power, moment, work,
pressure are all different things.
For example:
Btw, (foo^2) means (foo)x(foo) and (a*b) means (a x b)
velocity (speed) = distance / time
acceleration = distance / (time^2)
force = mass*acceleration
pressure = force / area
kinetic energy = 1/2 * mass * velocity^2
potential energy = mass * height * gravity
linear momentum = mass * velocity
moment and work = force * distance
power = work / time
Ken,
Think of my diagram as a leaner. The roots are at point 'O'.
The tree does not lean against anything, it's just anchored
at the roots.
force and stuff...
You can replace velocity with speed. That's fine.
The units are the same (distance)/(time).
Weight is (mass)x(gravity) and is a force.
Gravity is an acceleration.
force, momentum, energy, power, moment, work,
pressure are all different things.
For example:
Btw, (foo^2) means (foo)x(foo) and (a*b) means (a x b)
velocity (speed) = distance / time
acceleration = distance / (time^2)
force = mass*acceleration
pressure = force / area
kinetic energy = 1/2 * mass * velocity^2
potential energy = mass * height * gravity
linear momentum = mass * velocity
moment and work = force * distance
power = work / time
TheTreeSpyder
Branched out member
- Location
- Florida>>> USA
Re: Puzzle Shows Freely Available Rigging Force He
That is about as clear as i've seen that greek stuff expressed, thanks a lot! i thought that velocity meant direction of speed as a value whatever?
i'm still not sure i understand your puzzle deal; but this is what came immediately to mind; and won't fade as some type of answer. So here it is, might even learn something in spite of meself!
Once again, i don't have the education of answering in exact numbers, only witnessed patterns; and really couldn't see what you were getting at, but this is what i see.
Thanks a lot,
KC
That is about as clear as i've seen that greek stuff expressed, thanks a lot! i thought that velocity meant direction of speed as a value whatever?
i'm still not sure i understand your puzzle deal; but this is what came immediately to mind; and won't fade as some type of answer. So here it is, might even learn something in spite of meself!
Once again, i don't have the education of answering in exact numbers, only witnessed patterns; and really couldn't see what you were getting at, but this is what i see.
Thanks a lot,
KC
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- Location
- Santa Cruz, CA
Re: Puzzle Shows Freely Available Rigging Force He
KC,
Yes, velocity implies speed with a direction.
I have no idea what you wrote but I think you understand
some of the principles. You need to write fewer words and
write more equations. All the problems you showed were
static systems. To solve them you need only 2
equations.
1. The sum of the forces is zero.
2. The sum of the moments is zero.
That's all.
Assume the y axis is vertical-- up is positive,
down is negative.
The x axis is horizontal-- right is positive,
left is negative.
1. Sum of the forces at point 'O'
vertical (y axis) -100 + Fy = 0 i.e. Fy = 100
So Fy is 100 but in the opposing direction.
horizontal (x axis) 0 + Fx = 0 i.e. Fx = 0
There are no forces in the x axis.
2. Sum of the moments at point '0'.
A bending moment (or just moment) is a force acting
over the distance perpendicular to the direction of
the force (ok that's not very clear, see the diagram).
If the moment wants to rotate the object counter
clockwise around point O, then it is positive.
So... sum of the moments at point O:
Mo + (100 x d) = 0 i.e. Mo = -100 x d
The moment is zero when the tree is vertical and maximum
when the tree is horizontal.
You can use this to solve any static problem. Pulleys,
levers, you name it.
KC,
Yes, velocity implies speed with a direction.
I have no idea what you wrote but I think you understand
some of the principles. You need to write fewer words and
write more equations. All the problems you showed were
static systems. To solve them you need only 2
equations.
1. The sum of the forces is zero.
2. The sum of the moments is zero.
That's all.
Assume the y axis is vertical-- up is positive,
down is negative.
The x axis is horizontal-- right is positive,
left is negative.
1. Sum of the forces at point 'O'
vertical (y axis) -100 + Fy = 0 i.e. Fy = 100
So Fy is 100 but in the opposing direction.
horizontal (x axis) 0 + Fx = 0 i.e. Fx = 0
There are no forces in the x axis.
2. Sum of the moments at point '0'.
A bending moment (or just moment) is a force acting
over the distance perpendicular to the direction of
the force (ok that's not very clear, see the diagram).
If the moment wants to rotate the object counter
clockwise around point O, then it is positive.
So... sum of the moments at point O:
Mo + (100 x d) = 0 i.e. Mo = -100 x d
The moment is zero when the tree is vertical and maximum
when the tree is horizontal.
You can use this to solve any static problem. Pulleys,
levers, you name it.
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TheTreeSpyder
Branched out member
- Location
- Florida>>> USA
Re: Puzzle Shows Freely Available Rigging Force He
i too must reread you thoughts, but they are similar.
My main point was that in my view the forces on the stump (will be as anything else, collectively zero from the 'event') should be calculated per position, and center is a more neutral zone.
Position (A) under the lean is compressed region forming pivot of balanced pulls on either side. One side is the load side, an acitve force of weight, length and lean; a value of Z(force). As the pivot between CG pulling down and (C) fibers pulling down (each = Zforce when they are each multiplied by their leveraged factor from (A); so (A)=2Z; supports both loads of downpressure of load and support, as if (A) was the pulley between 2 legs of line, load and control legs, and inheritted both loads as a central pulley does.
On the other side of pivot point (A)(extreme) is (C); the support; passive/responsive force self adjusting to the load of the tree's Zforce even as it is pitching in wind etc. dynamically loading per split second.
Because Z must balance on the support side; Z also=(A) to (C)distance X holding power at (C). Or tree falls. For when it is out of balance, tree as anything else wanders discontent until it is balanced by support/equal and opposite reaction. Everything is in balance, because Nature demands that every act be in balance; or the act continues until balance is found. Or something like that....
Seeing as Nature will use the smallest amount of resources the easiest way, i think using leveraged length at (C) is used as multiplier to reduce force needed from fibers. My position (B) your = (O) as a position in the stump, and not stump as a whole; therefore more neutral to the event forces i think. If the tree where to be taken by saw, these stress points of ABC in stump would be replicated in pattern in the hinge, as it stands under same forces and positions as stump did. How the pattern is shown in hinge can be adjusted by how the hinge meets the pre-existing force pattern in the stump, but will be a smaller version of the stump forces from that angle that the hinge addresses the preexisting lean Zforce's direction, from the pre-exiting point of the stump. i stole the neutral fiber idea from Stumper, and formed it to this. Though i learned through hinging, find same values in standing tree.
So i looked at (O) as a point and not a stump i guess!
Thanx alot, especially for clear collection of simple formulaes, and another way to express balanced forces.
i too must reread you thoughts, but they are similar.
My main point was that in my view the forces on the stump (will be as anything else, collectively zero from the 'event') should be calculated per position, and center is a more neutral zone.
Position (A) under the lean is compressed region forming pivot of balanced pulls on either side. One side is the load side, an acitve force of weight, length and lean; a value of Z(force). As the pivot between CG pulling down and (C) fibers pulling down (each = Zforce when they are each multiplied by their leveraged factor from (A); so (A)=2Z; supports both loads of downpressure of load and support, as if (A) was the pulley between 2 legs of line, load and control legs, and inheritted both loads as a central pulley does.
On the other side of pivot point (A)(extreme) is (C); the support; passive/responsive force self adjusting to the load of the tree's Zforce even as it is pitching in wind etc. dynamically loading per split second.
Because Z must balance on the support side; Z also=(A) to (C)distance X holding power at (C). Or tree falls. For when it is out of balance, tree as anything else wanders discontent until it is balanced by support/equal and opposite reaction. Everything is in balance, because Nature demands that every act be in balance; or the act continues until balance is found. Or something like that....
Seeing as Nature will use the smallest amount of resources the easiest way, i think using leveraged length at (C) is used as multiplier to reduce force needed from fibers. My position (B) your = (O) as a position in the stump, and not stump as a whole; therefore more neutral to the event forces i think. If the tree where to be taken by saw, these stress points of ABC in stump would be replicated in pattern in the hinge, as it stands under same forces and positions as stump did. How the pattern is shown in hinge can be adjusted by how the hinge meets the pre-existing force pattern in the stump, but will be a smaller version of the stump forces from that angle that the hinge addresses the preexisting lean Zforce's direction, from the pre-exiting point of the stump. i stole the neutral fiber idea from Stumper, and formed it to this. Though i learned through hinging, find same values in standing tree.
So i looked at (O) as a point and not a stump i guess!
Thanx alot, especially for clear collection of simple formulaes, and another way to express balanced forces.
