Is it possible to generate a UIBezierPath Curve from an SVG elliptical Command A.
it would be great if we can give an example of UIbezier Curve (Cubic / Quadratic cureve) that corresponds to the following Command :
"M600,350 l 50,-25
a25,25 -30 0,1 50,-25 l 50,-25
a25,50 -30 0,1 50,-25 l 50,-25
a25,75 -30 0,1 50,-25 l 50,-25
a25,100 -30 0,1 50,-25 l 50,-25"
Thank you !
Related
I have a data set for which I am plotting a graph of time vs intensity at a particular frequency.
On the x axis is the time data set which is in a numpy array and on the y axis is the intensity array.
time = [ 0.3 1.3 2.3 3.3 4.3 5.3 6.3 7.3 8.3 9.3 10.3 11.3 12.3 13.3
14.3 15.3 16.3 17.3 18.3 19.3 20.3 21.3 22.3 23.3 24.3 25.3 26.3 27.3
28.3 29.3 30.3 31.3 32.3 33.3 34.3 35.3 36.3 37.3 38.3 39.3 40.3 41.3
42.3 43.3 44.3 45.3 46.3 47.3 48.3 49.3 50.3 51.3 52.3 53.3 54.3 55.3
56.3 57.3 58.3 59.3]
intensity = [1.03587, 1.03187, 1.03561, 1.02893, 1.04659, 1.03633, 1.0481 ,
1.04156, 1.02164, 1.02741, 1.02675, 1.03651, 1.03713, 1.0252 ,
1.02853, 1.0378 , 1.04374, 1.01427, 1.0387 , 1.03389, 1.03148,
1.04334, 1.042 , 1.04154, 1.0161 , 1.0469 , 1.03152, 1.22406,
5.4362 , 7.92132, 6.50259, 4.7227 , 3.32571, 2.46484, 1.74615,
1.51446, 1.2711 , 1.15098, 1.09623, 1.0697 , 1.06085, 1.05837,
1.04151, 1.0358 , 1.03574, 1.05095, 1.03382, 1.04629, 1.03636,
1.03219, 1.03555, 1.02886, 1.04652, 1.02617, 1.04363, 1.03591,
1.04199, 1.03726, 1.03246, 1.0408 ]
When I plot this using matplotlib, using;
plt.figure(figsize=(15,6))
plt.title('Single frequency graph at 636 kHz', fontsize=18)
plt.plot(time,intensity)
plt.xticks(time[::3], fontsize=12)
plt.yticks(fontsize=12)
plt.xlabel('Elapsed time (minutes:seconds)', fontsize=18)
plt.ylabel('Intensity at 1020 kHz', fontsize=18)
plt.savefig('WIND_Single_frequency_graph_1020_kHz')
plt.show()
The graph looks like -
I want to fit a gaussian curve for this data, and this is the code I used,
def Gauss(x, A, B):
y = A*np.exp(-1*B*x**2)
return y
parameters, covariance = curve_fit(Gauss, time, intensity_636)
fit_A = parameters[0]
fit_B = parameters[1]
fit_y = Gauss(time, fit_A, fit_B)
plt.figure(figsize=(15,6))
plt.plot(time, intensity, 'o', label = 'data')
plt.plot(time, fit_y, '-', label ='fit')
plt.legend()
And the best fit i obtain looks like this -
Where am I going wrong? How can I make the best fit curve fit the data better?
By inspection one obweve that the curve isn't symmetrical. This is even more visible in logarithmic y-scale zoomed in the range close to the peak.
This draw to think that the Gaussian model (which is symetrical) cannot be fitted correctly.
Also one observe that the part of the curve around the peak isn't far from to be linear. Thus a better model might be made of the combination of two exponential functions, for example :
I suppose that you can code this function in your nonlinear regression sofware. The above rough values of the parameters can be used as starting values for the iterative calculus.
You need to define a more flexible model (more parameters) and to define reasonable initial values for them:
def f(x, a, b, mu, sigma):
return a + b * np.exp(-(x - mu) ** 2 / (2 * sigma ** 2))
popt, pcov = curve_fit(f, time, intensity, p0=[1, 1, 30.3, 2])
plt.plot(time, intensity)
plt.plot(time, f(time, *popt))
plt.show()
I've a basic example from a PDF i'm editing.
The code
/P <</MCID 0>> BDC q
0.000008871 0 595.32 841.92 re
W* n
BT
/F1 12 Tf
1 0 0 1 56.64 759.96 Tm
/GS7 gs
0 g
/GS8 gs
0 G
[(n)4(a)4(m)4(e)] TJ
ET
Q
q
0.000008871 0 595.32 841.92 re
W* n
BT
/F1 12 Tf
1 0 0 1 109.7 759.96 Tm
0 g
0 G
[( )] TJ
ET
Q
works prefectly, producing "name" without quotes when I open the pdf.
Sadly, if I change the n with a c, something happens:
Same thing happen if i write [(N)4(a)4(m)4(e)] TJ (capital N) or with [\(Name)] TJ
What am I doing wrong?
Perhaps your font is subset, and so does not have a glyph for c. Your PDF viewer may be substituting a glyph from another standard font, but obeying the given metadata width for the c glyph in the font dictionary for your subset font, which will be 0 for a missing glyph. Hence the overwriting.
Edit: this should have been a comment, not an answer. Sorry.
I know that Table Text Extraction not everyone's cup of tea. But while reading PDF Stream data about table, there are certain things that I don't understand.
The PDF Code Stream of Table is:
q % Graphic State Starting Point
0 292.5 595.3 442.8 re % Rectangle x y Width Height
W* % Clipping Even Odd Rule
n % End without Filling
0 0 0 rg % Color of Stroke
161 735 m % Move to New Path
160.8 734.7 l 89.3 734.7 l 89 735 l 88.8 735.3 l 161.3 735.3 l
161 735 l % straight line
h % Close the Current Path
f* % Fill Path with Even Odd Rule
Q
And Underline is:
q % Graphic State Starting Point
1 0 0 1 451.5 759.5 cm % Current matrix
0.5 w % Width of Stroke
0 0 0 RG % color
0 -0.8 m % Move to New Path
72 -0.8 l % Straight Line
S % Stroke Line
Q % End of Graphic State
In underline the m modifies the cm and sets that to 451.5 759.5 and straight line is drawn from current point i.e. 451.5 to 72 points i.e 493.5 which is less that 0.8 to cm, i.e 759.5
I don't understand now, how the table line would be drawn from which point to which point.
Consider the line drawing section, where m is the move to operator and l is line-to:
% command coordinates
% =====================
161 735 m % Move-to a(161, 735)
160.8 734.7 l % line-to b(161, 735 -.3)
89.3 734.7 l % line-to c(90, 735 -.3)
89 735 l % line-to d(90, 735)
88.8 735.3 l % line-to e(90, 735 +.3)
161.3 735.3 l % line to f(161, 735 +.3)
161 735 l % line to g(161, 735)
h % close-path
f % fill
(There's some strange minor variations in x around 160 and 89 which are too small to register visually - rendering quirks?).
[Under very high resolution line will have arrows at end
<============ ... =======>
]
Other than that it's drawing a very thin long box with corners (89, 734,7), (161, 734.7), (161, 735.3), and (89, 735.3). The affect of the +/- ).3 points on the y-axis is most likely to give the effect of a slightly thickened line rather than a rectangle.
I have solid object that is spinning with a torque W, and I want to calculate the force F applied on a certain point that's D units away from the center of the object. All these values are represented in Vector3 format (x, y, z)
I know until now that W = D x F, where x is the cross product, so by expanding this I get:
Wx = Dy*Fz - Dz*Fy
Wy = Dz*Fx - Dx*Fz
Wz = Dx*Fy - Dy*Fx
So I have this equation, and I need to find (Fx, Fy, Fz), and I'm thinking of using the Simplex method to solve it.
Since the F vector can also have negative values, I split each F variable into 2 (F = G-H), so the new equation looks like this:
Wx = Dy*Gz - Dy*Hz - Dz*Gy + Dz*Hy
Wy = Dz*Gx - Dz*Hx - Dx*Gz + Dx*Hz
Wz = Dx*Gy - Dx*Hy - Dy*Gx + Dy*Hx
Next, I define the simplex table (we need <= inequalities, so I duplicate each equation and multiply it by -1.
Also, I define the objective function as: minimize (Gx - Hx + Gy - Hy + Gz - Hz).
The table looks like this:
Gx Hx Gy Hy Gz Hz <= RHS
============================================================
0 0 -Dz Dz Dy -Dy <= Wx = Gx
0 0 Dz -Dz -Dy Dy <= -Wx = Hx
Dz -Dz 0 0 Dx -Dx <= Wy = Gy
-Dz Dz 0 0 -Dx Dx <= -Wy = Hy
-Dy Dy Dx -Dx 0 0 <= Wz = Gz
Dy -Dy -Dx Dx 0 0 <= -Wz = Hz
============================================================
1 -1 1 -1 1 -1 0 = Z
The problem is that when I run it through an online solver I get Unbounded solution.
Can anyone please point me to what I'm doing wrong ?
Thanks in advance.
edit: I'm sure I messed up some signs somewhere (for example the Z should be defined as a max), but I'm sure I'm wrong when defining something more important.
There exists no unique solution to the problem as posed. You can only solve for the tangential projection of the force. This comes from the properties of the vector (cross) product - it is zero for collinear vectors and in particular for the vector product of a vector by itself. Therefore, if F is a solution of W = r x F, then F' = F + kr is also a solution for any k:
r x F' = r x (F + kr) = r x F + k (r x r) = r x F
since the r x r term is zero by the definition of vector product. Therefore, there is not a single solution but rather a whole linear space of vectors that are solutions.
If you restrict the solution to forces that have zero projection in the direction of r, then you could simply take the vector product of W and r:
W x r = (r x F) x r = -[r x (r x F)] = -[(r . F)r - (r . r)F] = |r|2F
with the first term of the expansion being zero because the projection of F onto r is zero (the dot denotes scalar (inner) product). Therefore:
F = (W x r) / |r|2
If you are also given the magnitude of F, i.e. |F|, then you can compute the radial component (if any) but there are still two possible solutions with radial components in opposing directions.
Quick dirty derivation...
Given D and F, you get W perpendicular to them. That's what a cross product does.
But you have W and D and need to find F. This is a bad assumption, but let's assume F was perpendicular to D. Call it Fp, since it's not necessarily the same as F. Ignoring magnitudes, WxD should give you the direction of Fp.
This ignoring magnitudes, so fix that with a little arithmetic. Starting with W=DxF applied to Fp:
mag(W) = mag(D)*mag(Fp) (ignoring geometry; using Fp perp to D)
mag(Fp) = mag(W)/mag(D)
Combining the cross product bit for direction with this stuff for magnitude,
Fp = WxD / mag(WxD) * mag(Fp)
Fp = WxD /mag(W) /mag(D) *mag(W) /mag(D)
= WxD / mag(D)^2.
Note that given any solution Fp to W=DxF, you can add any vector proportional to D to Fp to obtain another solution F. That is a totally free parameter to choose as you like.
Note also that if the torque applies to some sort of axle or object constrained to rotate about some axis, and F is applied to some oddball lever sticking out at a funny angle, then vector D points in some funny direction. You want to replace D with just the part perpendicular to the axle/axis, otherwise the "/mag(D)" part will be wrong.
So from your comment is clear that all rotations are spinning around center of gravity
in that case
F=M/r
F force [N]
M torque [N/m]
r scalar distance between center of rotation [m]
this way you know the scalar size of your Force
now you need the direction
it is perpendicular to rotation axis
and it is the tangent of the rotation in that point
dir=r x axis
F = F * dir / |dir|
bolds are vectors rest is scalar
x is cross product
dir is force direction
axis is rotation axis direction
now just change the direction according to rotation direction (signum of actual omega)
also depending on your coordinate system setup
so ether negate F or not
but this is in 3D free rotation very unprobable scenario
the object had to by symmetrical from mass point of view
or initial driving forces was applied in manner to achieve this
also beware that after first hit with any interaction Force this will not be true !!!
so if you want just to compute Force it generate on certain point if collision occurs is this fine
but immediately after this your spinning will change
and for non symmetric objects the spinning will be most likely off the center of gravity !!!
if your object will be disintegrated then you do not need to worry
if not then you have to apply rotation and movement dynamics
Rotation Dynamics
M=alpha*I
M torque [N/m]
alpha angular acceleration
I quadratic mass inertia for actual rotation axis [kg.m^2]
epislon''=omega'=alpha
' means derivation by time
omega angular speed
epsilon angle
px and py are the x and y coordinates of a point on a circle's circumference.
Given:
the center of the circle as: cx, cy
the radius of the circle as: r
px
How to calculate the value of py? Thanks!
Given px there are at most two possible values for py.
Look at the pythagorean theorem: (px-cx)^2+(py-cy)^2=r^2.
Let d=r^2-(px-cx)^2
If d>0 then you have two solutions. This gives py=sqrt(d)+cy, where the square root is positive or negative.
If d=0 then you have one solution py=cy, the left or right of the circle, depending on px
If d<0 you have no real points.
Although this is not programming... you know this equation right?
(x - h)^2 + (y - k)^2 = r^2
You have h and k from cx and cy
You have r
you have x from px
then is easy to solve it!