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Complete course in geometry system of definitions, assumptions, and angles has been made to a... Centre point is the longest diameter of an ellipse pattern draftsman it can prevent you from hitting the wall the... To download or read a book to Kindle of triangles always add up more! 180 0 assistance to the pattern draftsman of Circles and ellipses spherical geometry and hyperbolic geometry drafting or geometry! Lines and planes we considered in Chapter 7 and hyperbolic geometry and elliptic Integral Corrections Landau... Problems are chosen to be of the ellipse explained with diagrams, pictures and an examination of the thickness plots!


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# very low variance explained after applying pca

I applied PCA on MNIST data and found that the first 64 components are able to retain 86% of variance.

Is there any problem while applying pca to a big dataset like MNIST. Because in most of the papers I have read to take the components which can explain upto 99% of variance. But it would be pointless to apply pca if such variance comes at 120 pca components.

It is entirely correct to apply PCA to a dataset like MNIST. Intuitively, corner pixels should almost never contain any information as to what digit is contained in the center of the image. So we should disregard them. You should expect similar results as with other datasets. PCA lowers the dimensionality of your data, thus allowing for a less complex model, however, this is at the cost of some information that is rejected when retaining only $n$ components.

When data is limited a less complex model will result in a lower bias and thus better accuracy when applied to novel data. However, the MNIST dataset is plentiful, thus the benefit of removing some features and losing even minimal information may cause your performance to degrade.

• Yes as a few of the feature columns in mnist is zero – Boris Mar 23 '18 at 9:58
• Yeah exactly, but honesty I would just leave them since your model's weights will just zero out as a consequence. A few additional features shouldn't hurt your performance when you have so much data available to you. – JahKnows Mar 23 '18 at 9:59
• Yes but so after doing dimensionality reduction the algorithms like svm run fast – Boris Mar 23 '18 at 10:01
• Oh yes! If you are not using deep learning, by all means PCA on MNIST is highly recommended. Precisely because of the high impact the number of features plays on the complexity of these algorithms. – JahKnows Mar 23 '18 at 10:02

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# Difference between revisions of "2016 AMC 10A Problems/Problem 25"

## Problem

How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600$ and $\text{lcm}(y,z)=900$?

$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$

We prime factorize $72,600,$ and $900$. The prime factorizations are $2^3\times 3^2$, $2^3\times 3\times 5^2$ and $2^2\times 3^2\times 5^2$, respectively. Let $x=2^a\times 3^b\times 5^c$, $y=2^d\times 3^e\times 5^f$ and $z=2^g\times 3^h\times 5^i$. We know that $$\max(a,d)=3$$ $$\max(b,e)=2$$ $$\max(a,g)=3$$ $$\max(b,h)=1$$ $$\max(c,i)=2$$ $$\max(d,g)=2$$ $$\max(e,h)=2$$ and $c=f=0$ since $\text{lcm}(x,y)$ isn't a multiple of 5. Since $\max(d,g)=2$ we know that $a=3$. We also know that since $\max(b,h)=1$ that $e=2$. So now some equations have become useless to us...let's take them out. $$\max(b,h)=1$$ $$\max(d,g)=2$$ are the only two important ones left. We do casework on each now. If $\max(b,h)=1$ then $(b,h)=(1,0),(0,1)$ or $(1,1)$. Similarly if $\max(d,g)=2$ then $(d,g)=(2,0),(2,1),(2,2),(1,2),(0,2)$. Thus our answer is $5\times 3=15$.

 2016 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Problem 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions


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# [Webkit-unassigned] [Bug 107856] mathml fences often look bad

bugzilla-daemon at webkit.org bugzilla-daemon at webkit.org
Sat Jun 21 02:30:43 PDT 2014

https://bugs.webkit.org/show_bug.cgi?id=107856

--- Comment #6 from Frédéric Wang (:fredw) <fred.wang at free.fr>  2014-06-21 02:31:02 PST ---
>
> Actually, as I read the spec, the <mfenced> in this case isn't even really legal, since there's no closing fence.

I think this testcase has been generated by a previous version of MathJax which incorrectly used <mfenced> to convert thinks like

\left{ ... \right.

into

<mfenced open="{" close=""> ... </mfenced>

which per http://www.w3.org/TR/MathML3/chapter3.html#presm.mfenced is equivalent to

<mrow>
<mo>{</mo>
<mrow>
...
</mrow>
<mo></mo>
</mrow>

I'm not sure it is very good semantically to have this empty <mo> as a closing fence. That's one more good reason not to use the <mfenced> element. Fortunately, MathJax stopped using that element in the most recent versions.

--
Configure bugmail: https://bugs.webkit.org/userprefs.cgi?tab=email
------- You are receiving this mail because: -------
You are the assignee for the bug.


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# 4.3. pyopus.optimizer.optfilter — Filter-based point acceptance¶

Filter support for constrained optimization (PyOPUS subsystem name: FILT)

class pyopus.optimizer.optfilter.Filter(hmax=0.0, debug=0)[source]

All points are stored in a dictionary with keys of the form (f,h).

The value stored alongside (f,h) can be anything.

A point (f,h) dominates (f0,h0) if

• h<h0 and f<=f0 or

• h<=h0 and f<f0

(f,h) is dominated by the filter if h>hmax.

No point in the filter dominates any other point in the filter.

hmax=0 results in extreme barrier behavior.

accept(f, h, misc)[source]

Checks a point against the filter.

If h>hmax the point does not dominate nor is dominated. Such a point is rejected.

If filter is empty, a point is accepted. Such a point is considered to dominate filter.

If point dominates any filter point it is accepted. Dominated filter points are deleted.

If a point is dominated by any filter point it is rejected. If a point does not dominate nor is dominated it is accepted.

Returns boolean tuple (dominates, dominated, accepted).

bestFeasible()[source]

Returns (f,h,misc) of the feasible point.

Returns (None, None, None) if no such point exists.

static dominates(f, h, f0, h0)[source]

Returns True if (f,h) dominates (f0,h0).

leastInfeasible()[source]

Returns (f,h,misc) of the infeasible point with lowest h.

Returns (None, None, None) if no such point exists.

mostInfeasible()[source]

Returns (f,h,misc) of the infeasible point with highest h.

Returns (None, None, None) if no such point exists.

orderedHlist()[source]

Returns h values in increasing order.

orderedPoints()[source]

Returns the f and h values ordered by increasing h.

position(f, h)[source]

Returns the position of h in the ordered list of h values.

0 … feasible point 1 … best infeasible point 2 … second infeasible point …

Returns None if the point is not in the filter.

reset(hmax=None)[source]

Resets the filter.

Points are stored in a dictionary with h as key. Dictionary values are tuples of the form (f, misc). There can be only one point for every h0 value.

updateHmax(hmax)[source]

Updates hmax and purges points with h>hmax.


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# Verifing trig identities

cos(x-(pie/2))=cos(x)tan(x)

I have to verify this identity and cant seem to figure it out. cos (x-y)=cos x * cos y + sin x * sin y
well since cos y = 0, it kind of eliminates that side of the equation and I end up with sinx * sin1

## Answers and Replies

Related Precalculus Mathematics Homework Help News on Phys.org
Cyosis
Homework Helper
It is sin(x)*1 not sin 1, because sin(pi/2)=1. So far you have cos(x-pi/2)=sin(x), which is correct. Now write out the right hand side of your original equation and see what you get.

Last edited:


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# Ngô Quốc Anh

## March 10, 2012

### An integral of 1/(1+|x|^2) over the whole Euclidean space

Filed under: Giải Tích 2, Giải Tích 3, Linh Tinh — Ngô Quốc Anh @ 2:46

Sometimes, we need a precise value for following

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}}, \quad \alpha>\frac{n}{2}.$

As such, I am going to calculate it and place the result here for future works.

In order to evaluate the above integral, we need to use the so-called co-area formula. We first write

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \int_0^{ + \infty } {\left( {\int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} } \right)dr}.$

Note that

$\displaystyle \begin{gathered} \int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} = \frac{1}{{{{(1 + {r^2})}^\alpha }}}\int_{\partial {B_0}(r)} {dS} \hfill \\ \qquad\qquad\qquad= \frac{1}{{{{(1 + {r^2})}^\alpha }}}\text{Area}({B_0}(r)) = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}\end{gathered}.$

Therefore,

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\int_0^{ + \infty } {\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}} dr = {\pi ^{\frac{n}{2}}}\frac{{\Gamma \left( {\alpha - \frac{n}{2}} \right)}}{{\Gamma (\alpha )}}$

where we have used

$\displaystyle\int_0^{ + \infty } {\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}} dr = \frac{1}{2}\frac{{\Gamma \left( {\alpha - \frac{n}{2}} \right)\Gamma \left( {\frac{n}{2}} \right)}}{{\Gamma (\alpha )}}.$

In particular, by the duplication formula, we get that

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^n}}}} = {\pi ^{\frac{n}{2}}}\frac{{\Gamma \left( {\frac{n}{2}} \right)}}{{\Gamma (n)}} = \frac{1}{{{2^n}}}\frac{{2{\pi ^{\frac{{n + 1}}{2}}}}}{{\Gamma \left( {\frac{n}{2} + \frac{1}{2}} \right)}} = \frac{{{\sigma _n}}}{{{2^n}}}$

where $\sigma_n$ is the volume of $n$-sphere of radius 1 in $\mathbb R^{n+1}$.

Another particular formula occurs when $\alpha=\frac{n+2}{2}$. We then have

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^{\frac{n}{2} + 1}}}}} = {\pi ^{\frac{n}{2}}}\frac{{\Gamma \left( 1 \right)}}{{\Gamma \left( {\frac{n}{2} + 1} \right)}} = \frac{{{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2} + 1} \right)}} = {\omega _n}$

where $\omega_n$ is the volume of $n$-ball of radius 1 in $\mathbb R^{n+1}$.

Keep in mind that the $n$-sphere of radius 1 is nothing but the boundary the $n+1$-ball of radius 1. Therefore, $\sigma_n$ is nothing but the surface area of the $n+1$-ball of radius 1. Interestingly, we have the following well-known formula

$\displaystyle {\omega _n} = \frac{{{\sigma _{n - 1}}}}{n}.$

As such, the previous formula can be expressed as

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^{\frac{n}{2} + 1}}}}} = \frac{{{\sigma _{n - 1}}}}{n}.$

1. Hi Ngo, how can you do with the same integral above times log(1+|x|^2)? Give me an advice, please!
Txs

Comment by Fab — April 23, 2012 @ 21:21

• Unless there is some motivation, otherwise, it is not worth considering that integral. As you may know, the one I considered above comes from blow-up analysis for PDEs.

Comment by Ngô Quốc Anh — April 23, 2012 @ 21:47


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# Tag Info

## New answers tagged inertia

1

This happends because of the friction generated between the paper and the coin. Remember, friction, at low speed, is not affected by the speed. So when you pull the paper, at any speed, the friction will have always the same value. The distance covered by the coin is always $x=\frac{1}{2}\frac{F}{m}t^2$, where, $F$ is the force of the friction (constant), $... 2 The forces are the same (assuming the the coin is moving relative to the paper in both cases), is the friction force between the coin and the paper, and is in a first approximation at least, independent of the relative speed between paper and coin. The difference between the two cases is that is that the force acts during a shorter time in the fast motion ... 1 There is a limit on how much frictional force the penny can apply to the paper in order to remain attached to it. It is limited by its mass based on$F_f = \mu _s F_N = \mu _s * m_{penny}*g$. If you accelerate the paper, you are applying a force$F = ma \$ and the penny's frictional force will try to oppose to keep the penny on the paper (inertia). For a ...

-4

This is a wonderful question and filled with interesting albeit incomplete or inaccurate answers. Launching an "orbital body" concerns two issues and two issues ONLY namely mass and intertia. So before we "launch" anything we must first understand THE EARTH IS MASSIVE...which works AGAINST the "inertial reality" of a conical shaped item which albeit is ...

2

This is the third time you've asked this question. The moment of inertia of a system depends on the choice of axis. If one is talking about the moment of inertia of an object with no other qualification, it is assumed that the axis goes through the center of mass, but the orientation of the axes in the body needs to be specified. In the case of a sphere, ...

3

You are correct that it will move, but also remember that conservation of momentum still holds, so, if you start out at rest, and push a large massed object, you have: \begin{align}\Sigma p_{0} &= \Sigma p_{f}\\ 0 &= mv + MV \\ v &= - (M/m)V \end{align} So, if your mass is small relative to the house, you will recoil from the house at a ...

Top 50 recent answers are included


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# Constraint for angle between connected rods

## Recommended Posts

I would like to create a constraint for holding the angle between two connecting rods to a certain angle.  The constraint force will only act on one of the objects.  The constraint is intended to create a motor on one of the connected objects as well as a rigid connection between the two when the motor is not activated that can be violated if enough external force is applied.  The constraint does not need to keep the objects physically connected -- the objects are kept physically connected by a separate constraint connecting them at end of the rods that I already have working.  The angle constraint will create a motor by varying the target angle in the constraint equation.  I am using sequential impulses.

I am having trouble coming up with the correct constraint equation.  I thought about having C = dot(normalize(P_1_center - P_1_end), normalize(P_2_center - P_2_end)) - 1 where the P_2 terms would be treated as constants when the derivative is taken due to the constraint force only applying on the first object as the first object is the object with the motor.  However I am unsure if this will work, or where to go next.  I believe that the velocity bias result for this constraint will be what ultimately rotates the object, but I am unsure if it will work as a function of dot product, or if I will be able to cap the motor force and still achieve correct results.  As well, the issue with using the dot product to find the angle is that it does not discriminate on the axis of the angle -- I would like to restrict the axises with which the objects can rotate with respect to each other.

I store the rotation as a quaternion about each object's center of mass, so another option instead of dot product is to look at the quaternions directly and use C = R1-R2, but calculating the quaternion between two objects as the constraint function involves quaternion multiplication and quaternion inverse to give another quaternion, of which I am unsure how to use in the velocity bias term because my angular velocities are in euler angles.  I can't figure out how to convert the quaternion result from C to a change in impulse as a quaternion can't be applied directly to an angular velocity.

Does anyone have insight on how to solve this?

Edited by Vexal

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You can convert a quaternion to axis-angle representation by doing something like this:

axis.angle = 2 * acos(quat.w) * (3.14159 / 180);    // angle of rotation
axis.x = quat.x;
axis.y = quat.y;
axis.z = quat.z;

normalize(axis);

Maybe that will help.

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I need the angles for each degree of freedom separately.  I found a function to get the euler angles from a quaternion, but I'm not sure if it works for all quadrants.  As well, I'm not sure if this approach is the right way to go about this in the first place.  I was trying to follow this: http://www.dyn4j.org/2010/12/angle-constraint/ but it doesn't say anything about how to actually apply the bias.  My stem now appears to mostly kind of work, but it's very unstable and can take a very long time to converge to a correct state, even with 1000 iterations of sequential impulses.

Edited by Vexal

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This is what I have so far.  It is still incredibly unstable:

https://youtu.be/8nay2RbWenk

Using angle constraints and sequential impulses to achieve motors connecting rods.

The bottom two segments are connected by positional constraints only; the rest of the segments are connected as rigid motors with a set target angle using angle constraints, as well as connected by positional constraints.  The middle segment moves its motor by varying the target angle of the constraint equation over time for either the x or y angle when I press a key on the keyboard. The rest of the segments connected by angle constraints are set to stay at x rotation = .5 radians.  I also apply force to the bottom two segments in this video as well as apply force to the rest when shooting it with a gun.

The angle constraint is C = Rot_2 - Rot_1
The position constraint is C = P_2 - P_1
Edited by Vexal

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I need the angles for each degree of freedom separately. I found a function to get the euler angles from a quaternion, but I'm not sure if it works for all quadrants.

Euler angles are order dependent and won't work for physics simulation this way.

What worked for me is taking the dot product of the 'rotation vector' with the axis of the coordinate frame, where rotation vector means representation like we use vor angular velocities or torques.

E.g.:

quat relativeError = rotation from first constraint space orientation to the second

float angle; vec3 axis; relativeError.ToAxisAndAngle(&angle, vec3 &axis); // like deftware said

vec3 rotationVector = axis * angle;

float xAngle = vec3(1,0,0).Dot(rotationVector);

float yAngle = vec3(0,1,0).Dot(rotationVector);

float zAngle = vec3(0,0,1).Dot(rotationVector); // yes, you could use rotationVector.z instead but just to show this works for any direction using the dot product :)

You see the 3 angles are order independent and it makes sense, but i'm uncertain if there is some kind of error involved.

However, I made a working powered ragdoll joint out of this so it's worth a try.

Edit:

You could download Newton http://newtondynamics.com/forum/index.php to see how it works in practice. The test setup is very similar to your video.

Probably has commented out the demo for testing purposes.

Open DemosSandbox project, edit newton-dynamics-master\applications\demosSandbox\sdkDemos\demos\StandardJoints.cpp

Uncomment at the very bottom:

//    AddJoesPoweredRagDoll(scene, dVector(0.0f, 0.0f, -25.0f), 0.0f, 20);
//    AddJoesPoweredRagDoll(scene, dVector(0.0f, 0.0f, 5.0f), 1.5f, 4);
//    AddJoesPoweredRagDoll(scene, dVector(0.0f, 0.0f, 15.0f), 0.0f, 4);

My joint is in the same file. Using the engine i don't need to know how constraint equations are made or solved from my given rows, but i guess you do or can figure it out from there.

Edited by JoeJ

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I don't think the order dependency of euler angles is the immediate issue. In most of the video, it is only applying force and rotating about the x axis, which should make ordering not applicable.

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You can do a swing-twist composition. First you need to compute the orientations of each rod. This really depends how you what you are simulating. Then for each pair of rods your build the relative quaternion. The relative quaternion is the rotation of the child as seen by the parent. Let q1 be the parent and q2 be the child. Then the relative quaternion simply becomes:

qr = conjugate( q1 ) * q2

Now you can decompose this into swing and twist rotations. I spare you the math, but the formulas are very simple:

qr = qs * qt

qt = normalize( Quaternion( 0, 0, qr.z, qr.w );

qs = qr * conjugate( qr )

From there you can extract angles.

HTH,

-Dirk

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You can do a swing-twist composition. First you need to compute the orientations of each rod. This really depends how you what you are simulating. Then for each pair of rods your build the relative quaternion. The relative quaternion is the rotation of the child as seen by the parent. Let q1 be the parent and q2 be the child. Then the relative quaternion simply becomes: qr = conjugate( q1 ) * q2 Now you can decompose this into swing and twist rotations. I spare you the math, but the formulas are very simple: qr = qs * qt qt = normalize( Quaternion( 0, 0, qr.z, qr.w ); qs = qr * conjugate( qr ) From there you can extract angles. HTH, -Dirk

Interesting you say that, because this was my first approach but no luck. Twist / swing worked perfectly for joint limits, but not for motors (unstable and jitter).

I concluded the reason is: Twist swing is also order dependent, although it's only two angles.

I remember now for the 3 axis method i have posted it works best to align the constraint space with the error axis, so one row do all the work and the other two just keep it stable.

Of course those things may depend on the solver as well.

Have you used the twist swing approach successfully for a motorized joint?

Maybe i have made a mistake, will try again...

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Yes, all my joints use a quaternion based framework. Angular constraints, limits and motors. It is basically a derivation of Claude Larcousieres work from his PhD thesis. You need the correct corresponding Jacobian though which are not trivial to compute. If you drive the velocities into the wrong direction it will obviously not work. Here are some joint derivations, but I am not sure if they are correct. I use a different notation, so I didn't verify.

http://www.mft-spirit.nl/portfolio/publications.php

http://www.mft-spirit.nl/files/MTamis_Constraints.pdf

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Posted (edited)

I am ashamed to admit that I am not fluent enough in math to follow the notation in that person's papers.  Basically, if it's written in LaTeX, chances are, I won't understand it.

What is the jacobian you used?  I am essentially using a two-row jacobian for the angular constraint:  I combine both a positional and angular constraint, so I have

J_angle = [0 -1 0 1]

J_position = [-1 skew(leverArm_1) 1 skew(leverArm_2])]

These two constraints are run as separate constraints but I am under the belief that they should work the same as a 2-row Jacobian when running with sequential impulses.  The motor is achieved by varying the constant in C so that the bias factor pushes it around its motor.

I believe that the constraint is working better now, but becomes unstable past a certain limit:

https://youtu.be/tvKEm9TST4M

Edit:  I actually just tested again, and I realize that it only becomes unstable if I rotate on more than one axis (in the above video, it rotates on segment about the x axis, and the other segment about y) -- if I use just the x axis, it does not become unstable at all, as far as I can tell.  I am unsure why rotating on the y axis and x axis together is stable up to a point, but then eventually is unstable.

Here is one more video demonstrating how it is stable in the x, and stable in the y at first:

https://youtu.be/h7hzwfgceeY

@JoeJ, am I correct in understanding that you are using quaternions for calculating the impulse using the Jacobian (in that the derivative of the quaternion is simply the euler angular velocity, without anything special done to it), but for calculating the error to use for the bias correction factor, you use the dot product instead?

Edited by Vexal

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I'm a bit more ashamed than you not knowing how a physics solver actually works :)

Newton engine builds the jacobians, what i give as input for a motor is angular acceleration about an axis.

In theory we would need only need one axis and angle to get any rotation we want, but newton likes to do this for 3 orthonormal axis, so everything is defined and stable. I guess something similar applies to you.

I refer to the orientation build from this 3 axis as 'constraint space' (not sure it's the proper term), and the primary axis where the moter acts as 'error axis' (axis from Dirks 'qr' quat).

The problem is: Newton wants constraints order independent, which makes sense because we want them to be solved at the same time.

But it is impossible to decompose a rotation to multiple rotations so you can combine (multiply) them together in any order and get the initial result. Whatever order you take, you get a different result.

So my trick here limits the problem, because at least the measurement of angles is guaranteed to be order independent, and by aligning constraint space with error axis any error becomes negligible.

However, if your (or Dirks) solver solves the constraints in given order, then you don't have this problem.

The problem in your video looks like a jump over 180 degrees or similar gimbal lock issue coming from eulers. I guess not your jacobians are wrong but more likely the angle measurement.

But i also can see some unnatural twisting, also a typical euler issue. Euler angles == evil. Both Dirks or mine suggestion should fix those things.

Other than that it looks good - seems you're not far off.

I tried to work on twist / swing again, but Newton seems somehow broken at the moment due to heavy changes for new softbody feature. I need to clarify this...

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Posted (edited)

@JoeJ

I changed my angle computation to figure out the error using your suggested method.  It seemed to make it better -- the method in which it fails now appears to be consistent -- it will correctly rotate unless the target angle for any degree of freedom crosses the 180 degree mark.  When moving the target back inside the working quadrant, it becomes stable again.  Aside from that, it appears to be stable, even when it's shot with the gun.

https://youtu.be/RdfKQvJOzmg

Edited by Vexal

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Posted (edited)

I remember that when you do an Euler decomposition the middle angle can vary a lot. So you need to be careful with this. Personally I have never used Euler angles for the above reasons.

Edited by Dirk Gregorius

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I realized that I might be using axis angle, not euler angles for angular velocity. I didn't realize they're not the same thing.

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Posted (edited)

This is what I am doing to compute the velocity bias:

                const XMVECTOR targetAngles = XMVectorSet(xRot, yRot, 0, 0);
const XMVECTOR targetQuaternion = XMQuaternionRotationRollPitchYawFromVector(targetAngles);
const XMVECTOR quaternionBetweenObjects = XMQuaternionMultiply(this->object2->qOrientation, XMQuaternionInverse(this->object1->qOrientation));
const XMVECTOR quaternionError = XMQuaternionMultiply(quaternionBetweenObjects, XMQuaternionInverse(targetQuaternion));

XMVECTOR axisBetweenObjects;
float angleBetweenObjects;
XMQuaternionToAxisAngle(&axisBetweenObjects, &angleBetweenObjects, quaternionError);

const XMVECTOR targetAngularVelocity = XMVectorMultiply(XMVector3Normalize(axisBetweenObjects), XMVectorReplicate(angleBetweenObjects));
this->velocityBias = XMVectorMultiply(targetAngularVelocity, XMVectorReplicate(-FractalTreeResources::angleBB));

Whenever yRot in targetAngles is between the radian equivalent of -90 degrees and +90 degrees, it works correctly with any value for xRot.  If xRot is 0, any value for yRot works.  If xRot is non-zero, then as soon as yRot is no longer set to be in the range of -90 and +90, it explodes.  Moving yRot back into the working range un-explodes it and it becomes stable again, even after being unstable for a long time.  After this works, the values of targetAngles will come from my inverse kinematics solver, so it needs 3 degrees of freedom because the robot arm segments can have the ability to move on any axis the player sets.

Edited by Vexal

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Posted (edited)
Dirk at one point helped me go through derivation of some joints. Maybe you could open a new topic to figure out the calculus for deriving the joint (or just reuse this topic). It's not too complicated and could be a great learning experience. I would be happy to try and help with the derivation. I understand the mathematics can seem very farfetched, but the problem here is truly a math problem.

Edit: Oh and I couldn't help with quaternion based derivations, but I could help with dot product based ones. Dot product does not give a full range of motion, but can be good enough to work with. To get started, you need to remove 2 DOF (two axes) and leave the third axis unrestricted. This adds two rows to jacobian. Then we add in limit on last axis for another row (when it is active), and finally the motor as a non-holonomic constraint (constraints velocity, no position level equation). When all constraints are active (including the 3 linear constraints) in total the joint would have 3 + 2 + 2 constraints. Once a good design is found to remove these freedoms, the rest is calculus. Edited by Randy Gaul

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# Automorphism of a compact Riemann surface with finitely many points removed maps punctured disc to punctured disc

I wonder if the following is true: If we have a compact Riemann surface $X$ where finitely many points are removed, so $X'=X - \{x_1,...x_k\}$ and an automorphism $f$ of $X'$, is it true that $f$ maps a punctured disc around a point $p\in \{x_1,...,x_k\}$ to a punctured disc around some point $q\in \{x_1,...,x_k\}$? By a punctured disc around $p$ I mean an open set $U\subset X'$ such that $U\cup \{p\}$ is biholomorphic to a disc in $\mathbb{C}$ via a chart $(U\cup \{p\}, \phi)$ of $X$.

I know that this is not the case if $X$ is a non compact Riemann surface. For example consider $X=\mathbb{C}$ and $X'=\mathbb{C}-\{0\}$. Then the automorphism $f(z)=\frac{1}{z}$ maps the punctured disc $B_1(0)-\{0\}$ to $\mathbb{C}-\overline{B_1(0)}$ which is not a punctured disc.

• So essentially the question is: Can $f$ be extended to $X\to X$? – Hagen von Eitzen Jun 10 '17 at 16:09
• @HagenvonEitzen: You're right. The background to this question is exactly this statement. I have an idea of a proof for which I need to know if the above question is true (and if so how to prove it). – user454042 Jun 10 '17 at 16:21
• math.stackexchange.com/questions/2554046/… – Moishe Kohan Dec 6 '17 at 22:25