I've taken a few measurements of an LC circuit and I need to solve for both L and C based on that. How do I solve this?
2.675e6 = 1 / (2 * pi * sqrt(L * (C + 100e-9))
5.8e6 = 1 / (2 * pi * sqrt(L * C))
You need pencil and paper:
Equation #1
5.8e6 = 1 / (2 * pi * sqrt(L * C))
sqrt(L * C)= 1 / (2 * pi * 5.8e6 )
L*C = 1 / (2 * pi * 5.8e6 )^2
Equation #2
2.675e6 = 1 / (2 * pi * sqrt(L * (C + 100e-9))
sqrt(L * (C + 100e-9))= 1 / ( 2 *pi *2.675e6 )
L * (C + 100e-9) = 1 / ( 2 *pi *2.675e6 )^2
Subtract #1 from #2
L * (C + 100e-9) - L*C = 1 / ( 2 *pi *2.675e6)^2 - 1 / (2 * pi * 5.8e6 )^2
L * 100e-9 = 1 / ( 2 *pi *2.675e6)^2 - 1 / (2 * pi * 5.8e6 )^2
L = 1e7 * (1 / ( 2 *pi *2.675e6)^2 - 1 / (2 * pi * 5.8e6 )^2 )
and than from #1
C = ( 1 / (2 * pi * 5.8e6 )^2 ) / L
Related
I am a) new to stackoverflow and b) an advanced beginner to R ;-)
i saw some bird artworks of Yeganeh with the associated functions in the web Drawing Birds in Flight With Mathematics and wanted to reproduce them in R to experiment a bit with colouring and so on.
However, while this one yielded a quite good result:
k <- 1:9830
X <- function(k) {
sin(pi * k / 20000) ^ 12 *
(0.5 * cos(31 * pi * k / 10000) ^ 16 *
sin(6 * pi * k / 10000) + (1 / 6 * sin(31 * pi * k / 10000)) ^ 20) +
3 * k / 20000 + cos(31 * pi * k / 10000) ^ 6 *
sin((pi / 2) * ((k - 10000) / 10000) ^ 7 - pi / 5)
}
Y <- function(k) {
-9 / 4 * cos(31 * pi * k / 10000) ^ 6 *
cos(pi / 2 * ((k - 10000) / 10000) ^ 7 - pi / 5) *
(2 / 3 + (sin(pi * k / 20000) * sin(3 * pi * k / 20000)) ^ 6) +
3 / 4 * cos(3 * pi * ((k - 10000) / 100000)) ^ 10 *
cos(9 * pi * ((k - 10000) / 100000)) ^ 10 *
cos(36 * pi * ((k - 10000) / 100000)) ^ 14 +
7 / 10 * ((k - 10000) / 10000) ^ 2
}
R <- function(k) {
sin(pi * k / 20000) ^ 10 *
(1 / 4 * cos(31 * pi * k / 10000 + 25 * pi / 32) ^ 20 +
1 / 20 * cos(31 * pi * k / 10000) ^ 2) +
1 / 30 * (3 / 2 - cos(62 * pi * k / 10000) ^ 2)
}
bird <- data.frame(x = X(k), y = Y(k), r = R(k))
library(tidyverse)
library(ggforce)
q <- ggplot() +
geom_circle(aes(x0 = x, y0 = y, r = r),
data = bird,
n = 30) +
coord_fixed() +
theme_void()
the following code yielded some weird result which should basically be related to the difference in the function. (x-A(k))+(y-B(k))=(R(k)) for the parrot below, whlie the bird above "simply" consisted of the k-th circle (X(k), Y(k)) and the radius of the k-th circle R(k)
k <- -10000:10000
A <- function(k) {
(3*k/20000)+(cos(37*pi*k/10000))*sin((k/10000)*(3*pi/5))+(9/7)*(cos(37*pi*k/10000))*(cos(pi*k/20000))*sin(pi*k/10000)
}
B <- function(k) {
(-5/4)*(cos(37*pi*k/10000))*cos((k/10000)*(3*pi/5))*(1+3*(cos(pi*k/20000)*cos(3*pi*k/20000)))+(2/3)*(cos(3*pi*k/200000)*cos(9*pi*k/200000)*cos(9*pi*k/100000))
}
R <- function(k) {
(1/32)+(1/15)*(sin(37*pi*k/10000))*((sin(pi*k/10000))+(3/2)*(cos(pi*k/20000)))
}
parrot <- data.frame(a = A(k), b = B(k), r = R(k))
q <- ggplot() +
geom_circle(aes(x0 = a, y0 = b, r = r),
data = parrot,
n=30) +
coord_fixed() +
theme_void()
q
Any help would be very much appreciated. Cartesian coords already applied as [explained here] (https://www.wikiwand.com/en/Hamid_Naderi_Yeganeh). From the visual point of view, it seems like the function is plotted properly but the "view" on it needs to be changed...
Thanks in advance!
I construct a user-function:
find.c <- function(q) { f <- function(c) {
(log(7.2 + 6 * c * q - 6) * q^6 * (1-q)^(6-6) * factorial(6) / factorial(6) / factorial(6-6)
+ log(7.2 + 6 * c * q - 5) * q^5 * (1-q)^(6-5) * factorial(6) / factorial(5) / factorial(6-5)
+ log(7.2 + 6 * c * q - 4) * q^4 * (1-q)^(6-4) * factorial(6) / factorial(4) / factorial(6-4)
+ log(7.2 + 6 * c * q - 3) * q^3 * (1-q)^(6-3) * factorial(6) / factorial(3) / factorial(6-3)
+ log(7.2 + 6 * c * q - 2) * q^2 * (1-q)^(6-2) * factorial(6) / factorial(2) / factorial(6-2)
+ log(7.2 + 6 * c * q - 1) * q^1 * (1-q)^(6-1) * factorial(6) / factorial(1) / factorial(6-1)
+ log(7.2 + 6 * c * q - 0) * q^0 * (1-q)^(6-0) * factorial(6) / factorial(0) / factorial(6-0)
- log(7.2)
)}
g <- uniroot(f, lower=0, upper=100, extendInt = "yes")[1]
g}
And I tried to plot:
plot(x = seq(0, 1 , 0.001), find.c(x))
then it gives me: "Error in uniroot(f, lower = 0, upper = 100, extendInt = "yes") :
f() values at end points not of opposite sign"
Also is ther any way I can use 'for' loop to simplify this function? I tried may way with for but function inside function makes very complicated.
I agree with #r2evans suggestion, but I think you can restructure this to be simpler without the nested function.
First, define f with both c= and q= as arguments:
f <- function(c,q) {
(log(7.2 + 6 * c * q - 6) * q^6 * (1-q)^(6-6) * factorial(6) / factorial(6) / factorial(6-6)
+ log(7.2 + 6 * c * q - 5) * q^5 * (1-q)^(6-5) * factorial(6) / factorial(5) / factorial(6-5)
+ log(7.2 + 6 * c * q - 4) * q^4 * (1-q)^(6-4) * factorial(6) / factorial(4) / factorial(6-4)
+ log(7.2 + 6 * c * q - 3) * q^3 * (1-q)^(6-3) * factorial(6) / factorial(3) / factorial(6-3)
+ log(7.2 + 6 * c * q - 2) * q^2 * (1-q)^(6-2) * factorial(6) / factorial(2) / factorial(6-2)
+ log(7.2 + 6 * c * q - 1) * q^1 * (1-q)^(6-1) * factorial(6) / factorial(1) / factorial(6-1)
+ log(7.2 + 6 * c * q - 0) * q^0 * (1-q)^(6-0) * factorial(6) / factorial(0) / factorial(6-0)
- log(7.2)
)}
Then loop over each of the q= values:
mapply(function(...) uniroot(...)[[1]],
list(f), q=seq(0,1,0.001), lower=0, upper=100, extendInt="yes")
# [1] 0.000000 1.076564 1.076460 1.076375 1.076291 1.076206 1.076122
# [8] 1.076037 1.075953 1.075868 1.075784 1.075699 1.075614 1.075530
# etc
I have an algorithm written as follows but I need to write that code into R. I have included the algorithm and the R code. I am not sure if that is represented well enough. To write the R code in the sequential order is not straight forward. I am sorry for not providing all the values of the variables here. I am not sure of the output yet which is the reason I am unable to show the required. It is more of a theoretical question.
Algorithm
VBDMAX = (va - VG) * 0.79 * (dep / D) ^ -1.21
VBOWMAX = -0.7 * VBDMAX
VBOWX = 0
' SKIP BOW IF -10D<X<15D OR OUTSIDE EDGE OF BARGES
If Y > B / 2 Then GoTo 200
If X < -10 * D Then GoTo 200
If X >= 15 * D Then GoTo 200
VBOWX = X * VBOWMAX / (10 * D) + VBOWMAX
If X <= 0 Then GoTo 200
VBOWX = X * (VBDMAX - VBOWMAX) / (5 * D) + VBOWMAX
If X <= 5 * D Then GoTo 200
VBOWX = -X * VBDMAX / (10 * D) + 15 * VBDMAX / 10
200 ' end bow
This is the R code that I have written
VBDMAX = (va - VG) * 0.79 * (dep / D) ^ -1.21
VBOWMAX = -0.7 * VBDMAX
VBOWX = 0
# SKIP BOW IF -10D<X<15D OR OUTSIDE EDGE OF BARGES
VBOWX <- ifelse ((Y>B/2 | X < -10*D | X>=15*D), 0,X*VBOWMAX/(10*D)+VBOWMAX)
VBOWX <- ifelse (X<=0 , X * (VBDMAX - VBOWMAX) / (5 * D) + VBOWMAX,
ifelse(x <=5*D, -X * VBDMAX / (10 * D) + 15 * VBDMAX / 10))
You can use ifelse constructs but you will need to nest those:
VBDMAX = (va - VG) * 0.79 * (dep / D) ^ -1.21
VBOWMAX = -0.7 * VBDMAX
VBOWX =
ifelse(Y > B / 2 || X < -10 * D || X >= 15 * D,
0,
ifelse(X <= 0,
X * VBOWX / (10 * D) + VBOWMAX,
ifelse(X <= 5 * D,
X * (VBDMAX - VBOWMAX) / (5 * D) + VBOWMAX,
-X * VBDMAX / (10 * D) + 15 * VBDMAX / 10
)
)
)
Understanding your question as how to translate "goto" statements to R, there are the following posibilities (if really needed) besides or in adjunction to the (often more appropriate) if/ifelse constructions as you already did:
a) entire code (for severe errors or if problem is solved): if (condition) stop("explain why...") or stopifnot(condition)
b) from within loops: see next and break
c) from within function: if (condition) return(), stopping the function here
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Evaluate
(z x^-1 y)^5 y^5
~~~~~~~~~~~~~~~~~~~~~~~ OVER
x^-4 z^-4
How would I evaluate this if X = 10, y = -3 and z = 3? I would like a step-by-step solution to help me fully understand it.
Numerator evaluates as (z*y*x^-1)^5 * y^5
further rewriting ((z^5*y^5)*y^5)/x^5
Denominator ((1/x^4)*(1/z^4))
Final Answer would be ((y^10)*(z^9))/x
as per your values it (3^19)/10
Exponentials have higher priority in most computer languages, so adding parentheses like this should make it clearer. I'm assuming that you're dividing the first polynomial by the second. It's simple algebra.
(z x^-1 y)^5 y^5
---------------- =
x^-4 z^-4
(y^10)(z^9)/x
You substitute the numbers.
Start with:
((z * x^-1 * y)^5 * y^5)/(x^-4 * z^-4)
Commute the exponent to the z factor: (A * B)^N => A^N * B^N
(z^5 * (x^-1 * y)^5 * y^5)/(x^-4 * z^-4)
Commute the exponent to the x and y factors: (A * B)^N => A^N * B^N
(z^5 * (x^-1)^5 * y^5 * y^5)/(x^-4 * z^-4)
Simplify the exponenet on the x factor: (A^N)^M => A^(N*M)
(z^5 * x^-5 * y^5 * y^5)/(x^-4 * z^-4)
Combine the y factors: A^N * A^M => A^(N+M)
(z^5 * x^-5 * y^10)/(x^-4 * z^-4)
Remove the negative exponent on x: 1/A^-N => A^N
(z^5 * x^-5 * y^10 * x^4) / (z^-4)
Remove the negative exponenet on z: 1/A^-N => A^N
z^5 * x^-5 * y^10 * x^4 * z^4
Combine the z factors: A^N * A^M => A^(N+M)
z^9 * x^-5 * y^10 * x^4
Combine the x factors: A^N * A^M => A^(N+M)
z^9 * x^-1 * y^10
Remove the negative exponent on x: A^(-N) => 1/A^N
(z^9 * y^10)/(x^1)
Simplify the x factor: A^1 => A
(z^9 * y^10)/(x)
And that's the algebraic form of your answer.
Next, subsitute the values:
3^9 * (-3)^10 / 10
Factor the exponents:
(3^3)^3 * (-3)^10 / 10
(3^3)^3 * ((-3)^2)^5 / 10
Evalutate the innermost exponents:
(3 * 3 * 3)^3 * ((-3)^2)^5 / 10
(9 * 3)^3 * ((-3)^2)^5 / 10
27^3 * ((-3)^2)^5 / 10
27^3 * 9^5 / 10
Continue evaluation exponents, breaking them down for simplicity:
27 * 27 * 27 * 9^5 / 10
27 * 27 * 27 * 9^5 / 10
729 * 27 * 9^5 / 10
19683 * 9^5 / 10
19683 * 9^2 * 9^2 * 9 / 10
19683 * 81 * 81 * 9 / 10
Then multiply the factors:
19683 * 81 * 729 / 10
19683 * 59049 / 10
1162261467 / 10
116226146.7
And there's your final answer.
You could also take advantage of the fact that X^N = (-X)^N if N is even by replacing -3 with 3 since 10 is even.
3^9 * (-3)^10 / 10
3^9 * 3^10 / 10
3^19 / 10
3 * 3^18 / 10
3 * (3^9)^2 / 10
3 * (3 * 3^8)^2 / 10
3 * (3 * (3^2)^4)^2 / 10
3 * (3 * ((3^2)^2)^2)^2 / 10
3 * (3 * (9^2)^2)^2 / 10
3 * (3 * 81^2)^2 / 10
3 * (3 * 6561)^2 / 10
3 * (19683)^2 / 10
3 * 387420489 / 10
1162261467 / 10
116226146.7
Ok,
I have a projectile that has its position defined such that:
a.x = initialX + initialDX * time;
a.y = initialY + initialDY * time + 0.5 * gravtiy * time^2;
I want to be able to predict which obstacles in my environment this projectile will collide with. I plan on checking the distance from A the closest point on the curve to the point P.
I figure that at the point A the tangent to the curve will be perpendicular to the vector AP, and that the tangent to the curve at A will simply be the velocity V of the projectile at that point.
AP dot V = 0
ap.x = initialX + initialDX * time - p.x;
ap.y = initialY + initialDY * time + gravity * time^2 - p.y;
v.x = initialDX;
v.y = initialDY + gravity * time;
=>
AP dot V =
( 0.5 * gravity^2 ) * t^3 +
( 1.5 * gravity * initialDY ) * t^2 +
( initialDX^2 + initialDY^2 + gravity * ( initialY - p.y ) ) * t +
( initialDX * ( initialX - p.x ) + initialDY * ( initialY - p.y ) )
From here I can see that this is a cubic function. I have spent some time researching online and found that there is a general equation that seems to work for certain values for finding the roots.
This is the process I have attempted to implement.
http://www.sosmath.com/algebra/factor/fac11/fac11.html
a = 0.5 * gravity^2;
b = 1.5 * gravity * initialDY;
c = initialDX^2 + initialDY^2 + gravity * ( initialY - p.y );
d = initialDX * ( initialX - p.x ) + initialDY * ( initialY - p.y );
A = ( c - ( b * b ) / ( 3 * a ) ) / a;
B = -( d + ( 2 * b * b * b ) / ( 27 * a * a ) - ( b * c ) / ( 3 * a ) ) / a;
workingC = -Math.pow( A, 3 ) / 27;
u = ( -B + Math.sqrt( B * B - 4 * workingC ) ) / 2; // Quadratic formula
s = Math.pow( u + B, 1 / 3 );
t = Math.pow( u, 1 / 3 );
y = s - t;
x = y - b / ( 3 * a );
When I plug x back into my original equations for the curve as the time, this should give me A. This seems to work well for certain values, however when p.y is above a certain value, I don't have a positive to take a square root of in the quadratic equation.
I don't have a full enough understanding of the math to understand why this is happening, or what I can do to resolve the issue.
Any help on this would be much appreciated.
UPDATE:
I have adjusted my algorithm to deal with complex roots, however I am still having trouble.
This is what I do now if the discriminant is negative:
a = 0.5 * gravity^2;
b = 1.5 * gravity * initialDY;
c = initialDX^2 + initialDY^2 + gravity * ( initialY - p.y );
d = initialDX * ( initialX - p.x ) + initialDY * ( initialY - p.y );
A = ( c - ( b * b ) / ( 3 * a ) ) / a;
B = -( d + ( 2 * b * b * b ) / ( 27 * a * a ) - ( b * c ) / ( 3 * a ) ) / a;
workingC = -Math.pow( A, 3 ) / 27;
discriminant = B * B - 4 * workingC;
then if discriminant < 0;
uc = new ComplexNumber( -B / 2, Math.sqrt( -discriminant ) / 2 );
tc = uc.cubeRoot( );
uc.a += B;
sc = uc.cubeRoot( );
yc = sc - tc;
yc.a -= b / ( 3 * a );
x = -d / ( yc.a * yc.a + yc.b * yc.b );
For some reason, this is still not giving me the results I expect. Is there anything that stands out as being wrong here?
Real polynomials can have complex number roots and if the roots are not real, they occur in conjugate pairs.
This implies cubics always have at least one real root.
Now if you get a complex root using your method, you can try to get the conjugate, mutiply and divide the constant of the cubic, take reciprocal to get the real root.
So if you had to take the square root of a -ve number, then it is same as multiplying the square root of its modulus by the imaginary number 'i'.
So if you represent your root as (m,n) denoting the complex number m + in. Then the other root is m - in = (m, -n) and m and n are real numbers.
The cubic can then be written as P(x) = (x^2 - 2m + (m^2 + n^2))(x-r).
So if P(x) = x^3 - a_1 *x^2 + a_2*x - a_3, then we have that r = a_3/(m^2 + n^2) (a_3 is the product of the roots, which is r(m^2+n^2))
A simpler way to get r would be to use the formula r = a_1 - 2m (a_1 is the sum of the roots, which is r+2m).
Check out: http://en.wikipedia.org/wiki/Complex_number