v*****r
发帖数: 2325 | 1 lower back rounding is also called "pelvic nutation"
我查了一下
就是蹲到水平或者past parallel 的时候, 屁股尖开始往膝盖方向收
a little nutation is okay, but excessive nutation is bad for lower back...
i can only squat to parallel without much pelvic nutation, seems that really
have to keep chest up.
my hip joint flexibility is average |
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r*****8
发帖数: 2560 | 2 能看懂论文吗？顺着上次的连接，找了一篇：
http://faculty.ifmo.ru/butikov/Applets/Gyroscope.pdf
Precession and nutation of a gyroscope. by Eugene Butikov,
专门讲逆动（nutation）和陀螺转速的关系。大致就是转速足够大，而干扰力比较小，
可以想成轴向不变。
文章里把下表面看成摩擦力无穷大，等你都看懂了，做一些小调整，就是你说的情况了。
要写一篇论文才能讲清楚的问题，哪有人会在这里给你详细讲解。 |
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s*****c
发帖数: 753 | 3 I guess you read the wiki, but please pay attention to the unit.
章动的主要项目来自于月球交点的退行，两者有相同的周期，都是6798天(18.6年)，
在黄道上的黄经章动分量是17.24"，垂直于黄道的斜章动是9.21"。另一个较明显的周
期是183天(0.5年)，章动分量分别是1.3"和0.6"。有趣的是，周期在5.5至6798天之间
，变动幅度大于0.0001"(目前测量能力能达到的数值)的项目，似乎都避开了34.8至91
天的范围。因此，这就成为章动的长周期项和短周期项的分界，长周期项会在年历中提
及并算出供参考，短周期项就只能从及时的短期预报中查表来做即时的修正。
Note the unit is ", not O (degree). 1 degree = 60' (minutes of arc) and 1'
= 60" (second of arc).
So the maximum Nutation of 17.24"=0.0048 degree. |
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g**G
发帖数: 767 | 4 你一开始声明了一个string hello, 然后声明了o，这个o是一个object的引用，这个引
用指向hello这个string
然后调用mutate，传进去的其实是o这个引用地址的一个copy
你无论在nutate里如何操作这个引用地址，函数外的o的引用还是不会变，始终指向
hello
你没法通过函数改这个
因为string是immutable的
如果o指向的不是string是一个你自定义的类
那么你在mutate里改o本身的内容，是可以反映出来的 |
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o**a
发帖数: 86 | 5 first you can get the 90 deg pulse width at your desired spin lock level.
(spink lock level in hz, the nutation freq)
pw90sl = 1/(4*tpwr_sl) Note 90 deg takes 1/4 of a full circle.
Then use the following equation to get spin lock power in dB
tpwr - 20.0*log10(pw90sl/pw90)
tpwr_sl: spin lock power in hz
tpwr: regular hard pulse power in dB
pw90: regular 90 deg pulse width
Make sure you understand what you are doing. Be very careful with spin lock.
It's one of the few things that can smoke the pro |
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o**a
发帖数: 86 | 6 【 以下文字转载自 Physics 讨论区 】
发信人: ouba (comfortably numb), 信区: Physics
标 题: a question on probability density
发信站: BBS 未名空间站 (Thu Feb 25 12:26:21 2010, 美东)
Folks, thanks for your time first!
function f = x.cos(a)**2 + y.cos(b)**2 + z.cos(c)**2
I know x > y > z. a, b, c are the angles between a vector and x, y, z axes
of a cartesian coordinate respectively.
If written in nutation and azimuthal angles,
f = x.sin(theta)**2.cos(phi)**2 + y.sin(theta)**2.sin(phi)**2 + z.cos(theta)
**2.
This vector is und |
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o**a
发帖数: 86 | 7 Folks, thanks for your time first!
function f = x.cos(a)**2 + y.cos(b)**2 + z.cos(c)**2
I know x > y > z. a, b, c are the angles between a vector and x, y, z axes
of a cartesian coordinate respectively.
If written in nutation and azimuthal angles,
f = x.sin(theta)**2.cos(phi)**2 + y.sin(theta)**2.sin(phi)**2 + z.cos(theta)
**2.
This vector is undergoing random re-orientation, i.e, the probability
density at (theta, phi) position is 1/(4pi)*sin(theta).
It is easy to see that z <= f <= x. Is th |
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