pantheraba
More biners!!!
I love some of your words!!! You gotta be awesome at Boggle.
Added Wedge Push or Rope Pull input force direction to forward fall with sideLean
- Thread starter theTreeSpyder
- Start date
- Replies 30
- Views 3K
murphy4trees
TreeHouser
have a 3/4" double braid that hasn't been out this year. Mostly 5/8", got it used. Not sure what it is, but its a double braid guessing over 15,000 lbs
Often will just double up 1/2" true blue on the skid steer or pick up.
and that "dumb tree cutter" was an inside joke for those that go back to the old days at arboristsite, something that Rocky (squirrel) said around 2003..
Often will just double up 1/2" true blue on the skid steer or pick up.
and that "dumb tree cutter" was an inside joke for those that go back to the old days at arboristsite, something that Rocky (squirrel) said around 2003..
theTreeSpyder
TreeHouser
- Joined
- Feb 12, 2016
- Messages
- 691
- Thread Starter Thread Starter
- #29
Where i was going with this is:
Forward direction into face direction by wedge:
>> At start forces thicker hinge with faux load added to tree force forward
>> As folds /lifts off of wedge , extra wedge force relieved so back to tree load only
>> BUT w/thicker hinge than get w/o wedge
Hinge is 'exercised' stronger at 'birthing' for ensuing tour of duty
.
Wedge direction portion as a force against sideLean force portion is different to me:
>> Babies hinge against side load, by reliving/not adding load so this portion does NOT exercise hinge stronger
>> By temp relieve some side load,
>> BUT as tree folds/lifts off wedge, wedge input relieved also
So as forward wedge force relieved to less load to hinge ratio
>> the side load carried by wedge , 'springs' back/ impacts back to original load unrelieved.
Relieving extra SUPPORT (against side load) projected as logical opposite of relieving extra LOAD (forward)
.
So to theory, forward extra temporary load exercises hinge stronger,
>> imbues hinge with more strength for tour of duty until tearoff
>> contrasted to wedge force expensed against side load at start only
.
Rope about same except consideration of perhaps longer input time
>> even to arm wrestling some smallers all the way to ground on hinge into open face
>> also if get SPEED of movement , velocity can help then carry thru against sideLean
cuz E=MCsquared shows VELOCITY being squared as biggest consideration
(M as leveraged load of Mass itself)
>> thus, even as common hammer head is in faster/weaker force position on handle/lever
>> hits w/more force/not less as the now faster moving hammer head, speed squared giving higher impact as not simple but compound multiplier
.
Favor Tapered Hinge against sideLean, forced stronger with forward exercising hinge stronger thru added force at birthing
BUT , simple rectangular/strip hinge forced thicker can help too
>> by virtue of offering fibers in leveraged positions against sideLean within now expanded hinge field
But reapportionment of Tapered Hinge offers these positions more extremely.
.
So find angle portions against side load better expensed to face direction
>> to empower / birth stronger hinge , AND IT'S LEVERAGES AS ANOTHER MULTIPLIER
that don't have with temporary babying hinge by taking side load portion off of it
>> to then possibly impact back..
Forward direction into face direction by wedge:
>> At start forces thicker hinge with faux load added to tree force forward
>> As folds /lifts off of wedge , extra wedge force relieved so back to tree load only
>> BUT w/thicker hinge than get w/o wedge
Hinge is 'exercised' stronger at 'birthing' for ensuing tour of duty
.
Wedge direction portion as a force against sideLean force portion is different to me:
>> Babies hinge against side load, by reliving/not adding load so this portion does NOT exercise hinge stronger
>> By temp relieve some side load,
>> BUT as tree folds/lifts off wedge, wedge input relieved also
So as forward wedge force relieved to less load to hinge ratio
>> the side load carried by wedge , 'springs' back/ impacts back to original load unrelieved.
Relieving extra SUPPORT (against side load) projected as logical opposite of relieving extra LOAD (forward)
.
So to theory, forward extra temporary load exercises hinge stronger,
>> imbues hinge with more strength for tour of duty until tearoff
>> contrasted to wedge force expensed against side load at start only
.
Rope about same except consideration of perhaps longer input time
>> even to arm wrestling some smallers all the way to ground on hinge into open face
>> also if get SPEED of movement , velocity can help then carry thru against sideLean
cuz E=MCsquared shows VELOCITY being squared as biggest consideration
(M as leveraged load of Mass itself)
>> thus, even as common hammer head is in faster/weaker force position on handle/lever
>> hits w/more force/not less as the now faster moving hammer head, speed squared giving higher impact as not simple but compound multiplier
.
Favor Tapered Hinge against sideLean, forced stronger with forward exercising hinge stronger thru added force at birthing
BUT , simple rectangular/strip hinge forced thicker can help too
>> by virtue of offering fibers in leveraged positions against sideLean within now expanded hinge field
But reapportionment of Tapered Hinge offers these positions more extremely.
.
So find angle portions against side load better expensed to face direction
>> to empower / birth stronger hinge , AND IT'S LEVERAGES AS ANOTHER MULTIPLIER
that don't have with temporary babying hinge by taking side load portion off of it
>> to then possibly impact back..
I was just waiting for the guy in Frankie's videos to calculate the side lean retention strength of the hinge, after he calculated breaking the hinge conventionally. I think he said 3000 psi wood tensile at break. Missed opportunity. Kyle, math us! What is the side torque to snap the hinge in the video? If I do it it will make my head hurt. So I'm passing the buck to you. 
oh, Spidey, you're invited too but the answer has to be a hard number of foot lbs. Or the teacher gives no marks even if you showed your work.
oh, Spidey, you're invited too but the answer has to be a hard number of foot lbs. Or the teacher gives no marks even if you showed your work.
Well, it was bugging me so I limbered up. First I tried to touch my toes but my hamstrings said "You just can't get there from here." Then I pumped out some chin ups and my elbow said "I can see you've been doing DRT again, pain be upon you." So I gave up on increased blood flow to my brain. I tried tp purge decades of clutter from my head and root out a diagram or equation. There was a faint little echo "sigma equals M Y over I" and then it was "oh great now I've got figure out what that really means. Enter google and I found section modes from the video, also the more familiar moment of inertia and then the cryptic second moment of area. Finally it was cleared up by each guy saying this is what I use and the other guys are wrong, or, well, different. Live and let live. Great, solved nothing. Tried my M Y over I with the video numbers and it almost worked but the units didn't work out. Digging a little deeper I found two formulas for moment of inertia of a rectangular cross section beam, one b h squared over 6 and the other b h cubed over 12. But aha! the over 12 guy fit the units correctly for my recollected equation and to corroborate it they actually used my equation.
So sigma is stress in psi, the video guy chose 4,000 psi out of a 3,000 to 12,000 range for wood. The center of a beam is the neutral axis where it's neither compression nor tension, happens to be the center for the rectangular shape. The Y is the distance out to where you're calculating the stress which turns out to be 1/2 of the 1 1/2" hinge thickness (b = 11.5", h = 1.5" for bending the hinge in the video). So in the video he gets 16,876 in lb = 1406 ft lbs to break the hinge using the b h squared over 6. I ran those numbers and got 1437.5 ft lbs. Note he said he got his inertia from a chart, probably rounded and also note the half thickness from neutral axis to face neatly explains the 6 vs 12 and units. So now it makes sense.
On to the side loading, reverse b and h so I = 1.5" x (11.5" cubed)/12 = 190.11 inches to the fourth, then rearrange sigma = M Y over I to get M = (4000 psi) x (190.11 inches to the fourth) over (11.5" / 2 ) = 132,250 in lb = 11,021 ft lb which is a lot bigger than 1437 ft lbs. So that's why floor joists are used on edge ..... So if he had 2000 lb tree over center by 2 feet, 4000 ft lbs is around a third of the break strength, if a side lean.
On topic, crushing hinge fibers: 2000 lbs / ( 1.5" x 11.5") = only 116 psi. Not sure how realistic his 2000 lbs was.
Your numbers may and should be varied to get a feel for real trees.
Intuitively, wedging a side leaner is bad juju because you're just exacerbating and taking away from your margin of hinge safety by adding tensile stress.
The real challenge now is to normalize this stuff, using log weight charts, species, log/stem length and get basic limits for " hinge will snap if it's leaned this far and you try to cut a hinge". Doesn't take much over centre to make torque with a heavy tree, but the hinge is also proportionate.
When falling (above was static analysis)
The other factor that makes it interesting is how chewy the wood is, so you don't just lose all your hinge geometry in one brittle instant. I can see the margin of safety working well with the progressive loss of fibres reducing the hinge cross section as the trunk falls. I.e if you're at 1/3 break stress, your hinge might still be holding when you've snapped 2/3 of the hinge fibres on the way down.
Class homework: look up trapezoid beam cross section, figure out its neutral axis and see how much strength advantage it does/doesn't create. Directionality issues aside. Yup D_ that means you. If my brain hurts yours ought to too.
So sigma is stress in psi, the video guy chose 4,000 psi out of a 3,000 to 12,000 range for wood. The center of a beam is the neutral axis where it's neither compression nor tension, happens to be the center for the rectangular shape. The Y is the distance out to where you're calculating the stress which turns out to be 1/2 of the 1 1/2" hinge thickness (b = 11.5", h = 1.5" for bending the hinge in the video). So in the video he gets 16,876 in lb = 1406 ft lbs to break the hinge using the b h squared over 6. I ran those numbers and got 1437.5 ft lbs. Note he said he got his inertia from a chart, probably rounded and also note the half thickness from neutral axis to face neatly explains the 6 vs 12 and units. So now it makes sense.
On to the side loading, reverse b and h so I = 1.5" x (11.5" cubed)/12 = 190.11 inches to the fourth, then rearrange sigma = M Y over I to get M = (4000 psi) x (190.11 inches to the fourth) over (11.5" / 2 ) = 132,250 in lb = 11,021 ft lb which is a lot bigger than 1437 ft lbs. So that's why floor joists are used on edge ..... So if he had 2000 lb tree over center by 2 feet, 4000 ft lbs is around a third of the break strength, if a side lean.
On topic, crushing hinge fibers: 2000 lbs / ( 1.5" x 11.5") = only 116 psi. Not sure how realistic his 2000 lbs was.
Your numbers may and should be varied to get a feel for real trees.
Intuitively, wedging a side leaner is bad juju because you're just exacerbating and taking away from your margin of hinge safety by adding tensile stress.
The real challenge now is to normalize this stuff, using log weight charts, species, log/stem length and get basic limits for " hinge will snap if it's leaned this far and you try to cut a hinge". Doesn't take much over centre to make torque with a heavy tree, but the hinge is also proportionate.
When falling (above was static analysis)
The other factor that makes it interesting is how chewy the wood is, so you don't just lose all your hinge geometry in one brittle instant. I can see the margin of safety working well with the progressive loss of fibres reducing the hinge cross section as the trunk falls. I.e if you're at 1/3 break stress, your hinge might still be holding when you've snapped 2/3 of the hinge fibres on the way down.
Class homework: look up trapezoid beam cross section, figure out its neutral axis and see how much strength advantage it does/doesn't create. Directionality issues aside. Yup D_ that means you. If my brain hurts yours ought to too.
