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I'm getting a quartic in ##c##, is there a way around that...perhaps something elegant I'm missing?
.Scott said:At 45 degrees, c is h/sqrt(2)
At 0 and 90 it's 0.
Seems sin(theta)h/sqrt(2) to me
##h## and ##\theta## are the parameters. All those other variables ##y,x,z## are just intermediate variables to define the triangles which make up the system of equations..Scott said:Perhaps I don't understand what you are trying to do.
It seems to me that given h and theta, everything else (including Z and C) can be computed.
What are you taking as the givens.
.Scott said:Okay. So if you take the center of that square as your X,Y origin as well as the center of rotation, then the top of C is a point that simply follows a circle and the length of c is the y coordinate of that point.
Except in your left square, I would have drawn line segment 'c' from the top left corner point of the square down to the x-axis. That way, as the square continues to rotate counter-clockwise, it's still easy to tract 'C'. Also in both frames, I would start with a line segment that crosses through the origin and has a slope of 1. Then you can mark the angle between that 45-degree segment and the diagonal that goes to the top of C as theta.erobz said:Like this:
View attachment 323821The one on the left ##\theta = 90° , c = \frac{h}{2}## , and the one on the right ##\theta = 60°, c= ?##
I'm confused. The angle I have labeled is ##\theta## in both diagrams. It's referenced from horizontal, it represents the angle the right side of the square makes w.r.t. horizontal..Scott said:Except in your left square, I would have drawn line segment 'c' from the top left corner point of the square down to the x-axis. That way, as the square continues to rotate counter-clockwise, it's still easy to tract 'C'. Also in the left frame, you left out the theta. It should go all the way to that diagonal line segment.
Well, they couldn't possibly work together because wouldn't that imply ##\cos \beta ## is algebraic?.Scott said:Also, I tried to merge my ##cos(pi/4-\theta)h/\sqrt 2## into your equation - but Wolfram Alpha gave out on me.
erobz said:View attachment 323817
I'm getting a quartic in ##c##, is there a way around that...perhaps something elegant I'm missing?
Can you (anyone) find which equation in the system in post #3 is not true? Surely, I've bungled one of them ( atleast)? And they seem so straight forward its driving me bonkers... I've updated the diagram to be more clear, and just reposted the system.pasmith said:Rotating a square with vertices at [itex](\pm \frac12 h, \pm \frac12h)[/itex] counterclockwise through an angle of [itex]\theta \in [\frac14 \pi, \frac34 \pi][/itex] about its centre [itex](0,0)[/itex] will move the vertex at [itex](\frac12h , \frac12 h)[/itex] to what you have as [itex](-|x|,c)[/itex]. This is a linear transformation, so we should have [itex]c = f(\theta)h[/itex] for some [itex]f(\theta)[/itex].
To calculate [itex](-|x|, c)[/itex], we can use a rotation matrix: [tex]
\begin{pmatrix} -|x| \\ c \end{pmatrix} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} \tfrac12 h \\ \frac12 h \end{pmatrix}.[/tex] Hence [tex]
c = \tfrac12 h (\sin \theta + \cos \theta) = \frac{h}{\sqrt{2}} \cos\left(\theta - \frac\pi{4}\right).[/tex]
I see how it works out, thanks for taking the time to explain that. I agree with what you are saying..Scott said:View attachment 323823
A bit messy - but I measuring theta from that x=y segment
Theta is 90 degrees on the left panel. It's about 60 on the right.
The point is that theta+pi/4 is the angle from x,y=1,0 to the top of C.
erobz said:Can you (anyone) find which equation in the system in post #3 is not true? Surely, I've bungled one of them ( atleast)? And they seem so straight forward its driving me bonkers... I've updated the diagram to be more clear, and just reposted the system.
What I don't understand is when is solve the system I'm getting a quartic in ##c##. If that were the case, it means that the solution of some quartic is ##\cos \beta##. Is that possible?pasmith said:I think these are correct. I just don't think they help you. Eliminating [itex]x = z\cos\theta[/itex] and [itex]y = z \sin \theta[/itex] leads you to a system in [itex]c[/itex] and [itex]z[/itex], but it's not an easy system to solve.
The triangle you care about is the one with sides [itex]c[/itex], [itex]x[/itex] and [itex]h/\sqrt{2}[/itex]. This is a right-angled triangle, and you can work out the angle [itex]\alpha[/itex] of this triangle at the centre of the square: clearly [itex]\alpha + \theta + \frac \pi 4 = \pi[/itex]. Hence [tex]
c = \frac{h}{\sqrt{2}} \sin \left(\pi - \theta - \frac \pi 4\right).[/tex] This simplifies to the expression [itex]\frac12h(\sin \theta + \cos \theta)[/itex] I gave earlier.
It looks right to me. I observed that the line N.A. is ##c## above the bottom corner from symmetry, and hence by inspection of the vertical extent of two sides of the square ##2c=h(\sin\theta+\cos\theta)##. You can plug this, with ##\theta=60°##, into your quartic and show that it's a solution. Factoring out that solution gives you a cubic, the solutions to which are ##c=\pm h/\sqrt 2## and ##c=(\sqrt{3}-1)h/4##.erobz said:Can you (anyone) find which equation in the system in post #3 is not true?
I like you diagrams @erobz! What software to do you use to create you diagrams?erobz said:View attachment 323817
I'm getting a quartic in ##c##, is there a way around that...perhaps something elegant I'm missing?
Thank you for your reply @erobz!erobz said:Thanks! Thats Power Point.
It takes practice. It's not designed specifically for mathematical diagrams, but you can get by with it. If there is anything that is geometrically heavy I just draft it in an actual CAD software first, then bring it over as a picture and mark it up.Callumnc1 said:Thank you for your reply @erobz!
Wow, I didn't know that it could be used to produce such great diagrams! I might try using it then.
Many thanks!
Thank you for letting me know @erobz!erobz said:It takes practice. It's not designed specifically for mathematical diagrams, but you can get by with it. If there is anything that is geometrically heavy I just draft it in an actual CAD software first, then bring it over as a picture and mark it up.
erobz said:$$ 2c^2 + 2\left( 2c \tan \theta -h \sec \theta \right)\sqrt{\frac{h^2}{2} - c^2} -2\left( \frac{h^2}{2} - c^2 \right) -h^2 = 0 $$
Then Expand, collect like terms and isolate the root and square both sides:
$$ \left[ 2 \left( 2c \tan \theta -h \sec \theta\right) \sqrt{\frac{h^2}{2} - c^2} \right]^2 = \left( 2h^2 -4c^2 \right)^2 $$
If you expand all that I get a quartic.
Thanks for sharing! I wouldn't have thought of that...just not good enough.pasmith said:But it factorises as two quadratics.
Leave [itex]2h^2 -4c^2[/itex] as [itex]4\left(\frac{h^2}2 - c^2\right)[/itex]. A factor of 2 can then be cancelled, and, after squaring, both sides will have a factor [itex]\left(\frac{h^2}2 - c^2\right)[/itex]. Assuming this factor is non-zero, we can cancel it to get a quadratic [tex]
4\left( \frac{h^2}2 - c^2 \right) = (2c\tan\theta - h\sec\theta)^2.[/tex] Expanding this and simplifying some trig functions we get [tex]
4c^2\sec^2\theta - 4ch\sec\theta \tan\theta + h^2 \sec^2 \theta - 2h^2 = 0[/tex] and multiplying by [itex]\cos^2\theta[/itex] we get [tex]
\begin{split}
0 &= 4c^2 - 4ch\sin \theta + h^2 - 2h^2 \cos^2 \theta \\
&= 4c^2 - 4ch\sin \theta + h^2 \sin^2 \theta - h^2 \cos^2 \theta \\
&= (2c - h\sin\theta)^2 - h^2 \cos^2 \theta.
\end{split}[/tex] Thus we have the four roots [tex]
\frac{h}{\sqrt 2}, \quad -\frac{h}{\sqrt 2},\quad \frac{h(\sin \theta + \cos \theta)}{2},\quad \frac{h(\sin \theta - \cos \theta)}{2}.[/tex]
The formula for solving for c in a rotated square is c = √(a² + b²), where a and b are the side lengths of the square.
To determine the value of c in a rotated square, you need to know the side lengths of the square and use the formula c = √(a² + b²).
No, you cannot solve for c in a rotated square without knowing the side lengths. The formula c = √(a² + b²) requires the values of a and b to be known.
No, there is only one formula for solving for c in a rotated square, which is c = √(a² + b²).
The value of c represents the length of the diagonal of the rotated square, which is also the hypotenuse of the right triangle formed by the two sides of the square.