So cool. New climber but I've worked under my father, who's been a climber for 30+ years, for about 4 or 5 years total now. I've been roping all that time. He's a master arborist and has always stressed that physics is vital to a safe take down. Thanks for the video!
Hey dude, i loved the manifold topics, where did you go? Do you mind doing a video on stereographic projections of the S^n sphere and proving its an n-dimensional manifold? :D Getting stuck on the details here.
Hey Reety, for better or for worse it takes me a lot of time and effort to make sure my videos are actually accurate and well explained, and things have been rocky in my life again, so I've had to take another break from things that take so much time and effort. I'm planning to come back and continue talking about manifolds, but it's going to be a couple months. The plan is to talk about tangent spaces, in a way that provides intuition on the topic. Possibly also discussing Lie and covariant differentiation, then discuss integration on manifolds. Covariant differentiation and integration require additional structure on the manifold, that will need to be introduced
@@ReetyTV I don't have a discord. Honestly, I don't know if I'd be able to answer your questions. I'm not a working mathematician or anything. I'm an arborist, and I just think this stuff is cool. I pretty much learn the material as I make the videos, and making the videos helps me understand the things I'm learning.
I spent some time this week trying to derive the 20% rule for dismantling a spar. ie when an arborist cuts down a spar, each piece should be 20% of the total height of the spar for it to land flat after a 3/4 rotation. I got height of last cut off the ground (s) is 4.45X length of the log(l), which is pretty close to the ideal 4.0, but I wonder if I did it correctly. I used total energy available in the first 90 deg, turning that into the rotational energy of a log spinning from one end PLUS the kinetic energy of the center of mass moving downward. I got w=sqrt((12g)/7l). That gave me angular velocity (w in radians), and downward velocity was wl/2 when the log was at 90 degrees, where l is length of log. From there I calculated how high the arborist has to be to allow another pi radians of rotation, which worked out to s=l*((7pi^2/24)+pi/2). I'm wondering whether at liftoff I should have made note that the moment of inertia changes from (ml^2)/3 to (ml^2)/12, but when I tried that I got a completely wrong answer. Reason? When the log leaves the stump, it should instantly begin to rotate about its center axis, so angular velocity should increase to conserve momentum, like a skater who pulls her arms in. Why doesn't that work out?
I could not find any other source or reference that describes the maximum or net force of the impact in terms or Fi + Fg... rather all of them say the F = m*a*(h/d)... Can you provide another source that talks about this inertial frame of reference with respect to a falling object?
Yes! Physics for Arborists that's great! when using this formula for negative rigging - would "h" be the distance of free fall or the total distance until completely stopped? I've been thinking a bit about physics and negative rigging lately, since the distance is measured from the piece's center of gravity - a long piece would fall a longer distance before the rope can stop it, but I don't really have a good understanding of how to calculate forces in a tipping motion, have you ever thought of this?
One comment: the Eg = mgh more accurately is Eg=mg(h+d). And you are assuming objects are rigid. I think the force you feel on impact if Fi, not Fnet. Otherwise when you place an object on top of your head and this is in static state you wouldn't feel any force (Fnet is 0).
I have never heard of stopping distance = 1ft. This seems incredibly facile, at best. To me, the whole equation falls to pieces when you make generalizations like this. SO many factors are playing around with each other when a body (be it yours or your friend's or a length of wood being rigged) comes to a stop, I can't see working with numbers that make such recklessly static assumptions. Where's rope stretch? Diameter? Systemic friction? Materials present? This topic is interesting, but not at all useful pragmatically for a practicing arborist. Theory versus praxis.
![HIWWHEE | OHM [Official MV]](http://i.ytimg.com/vi/OJJLtTLqLag/mqdefault.jpg)
So cool. New climber but I've worked under my father, who's been a climber for 30+ years, for about 4 or 5 years total now. I've been roping all that time. He's a master arborist and has always stressed that physics is vital to a safe take down. Thanks for the video!
Hey dude, i loved the manifold topics, where did you go? Do you mind doing a video on stereographic projections of the S^n sphere and proving its an n-dimensional manifold? :D Getting stuck on the details here.
Hey Reety, for better or for worse it takes me a lot of time and effort to make sure my videos are actually accurate and well explained, and things have been rocky in my life again, so I've had to take another break from things that take so much time and effort.
I'm planning to come back and continue talking about manifolds, but it's going to be a couple months. The plan is to talk about tangent spaces, in a way that provides intuition on the topic. Possibly also discussing Lie and covariant differentiation, then discuss integration on manifolds.
Covariant differentiation and integration require additional structure on the manifold, that will need to be introduced
@@tayloresch1909 Hey no worries man, is it possible I could ask you a few questions when i get stuck from time to time? Do you have a discord?
@@ReetyTV I don't have a discord. Honestly, I don't know if I'd be able to answer your questions. I'm not a working mathematician or anything. I'm an arborist, and I just think this stuff is cool.
I pretty much learn the material as I make the videos, and making the videos helps me understand the things I'm learning.
I spent some time this week trying to derive the 20% rule for dismantling a spar. ie when an arborist cuts down a spar, each piece should be 20% of the total height of the spar for it to land flat after a 3/4 rotation. I got height of last cut off the ground (s) is 4.45X length of the log(l), which is pretty close to the ideal 4.0, but I wonder if I did it correctly. I used total energy available in the first 90 deg, turning that into the rotational energy of a log spinning from one end PLUS the kinetic energy of the center of mass moving downward. I got w=sqrt((12g)/7l). That gave me angular velocity (w in radians), and downward velocity was wl/2 when the log was at 90 degrees, where l is length of log. From there I calculated how high the arborist has to be to allow another pi radians of rotation, which worked out to s=l*((7pi^2/24)+pi/2). I'm wondering whether at liftoff I should have made note that the moment of inertia changes from (ml^2)/3 to (ml^2)/12, but when I tried that I got a completely wrong answer. Reason? When the log leaves the stump, it should instantly begin to rotate about its center axis, so angular velocity should increase to conserve momentum, like a skater who pulls her arms in. Why doesn't that work out?
Excellent
Physics is fun😊
I could not find any other source or reference that describes the maximum or net force of the impact in terms or Fi + Fg... rather all of them say the F = m*a*(h/d)... Can you provide another source that talks about this inertial frame of reference with respect to a falling object?
Yes! Physics for Arborists that's great! when using this formula for negative rigging - would "h" be the distance of free fall or the total distance until completely stopped?
I've been thinking a bit about physics and negative rigging lately, since the distance is measured from the piece's center of gravity - a long piece would fall a longer distance before the rope can stop it, but I don't really have a good understanding of how to calculate forces in a tipping motion, have you ever thought of this?
One comment: the Eg = mgh more accurately is Eg=mg(h+d). And you are assuming objects are rigid. I think the force you feel on impact if Fi, not Fnet. Otherwise when you place an object on top of your head and this is in static state you wouldn't feel any force (Fnet is 0).
I have never heard of stopping distance = 1ft. This seems incredibly facile, at best. To me, the whole equation falls to pieces when you make generalizations like this. SO many factors are playing around with each other when a body (be it yours or your friend's or a length of wood being rigged) comes to a stop, I can't see working with numbers that make such recklessly static assumptions. Where's rope stretch? Diameter? Systemic friction? Materials present? This topic is interesting, but not at all useful pragmatically for a practicing arborist. Theory versus praxis.