I'm trying to plot airfoil camber lines in excel, given the polynomial and some constants.
Here is the given data, the final to reach is on the right top, with the (x,y) coordinates being used to create a CAD model of a wing that will be used for a 3-bladed wind turbine rotor.
For the moment I get this:
Is there a fault in the given polynomial?
It looks like the authors reported the a parameters in reverse order. What was reported as a0 is in fact a3, reported a1 is a2, reported a2 is a1 and reported a3 is a0. To give a specific example, the parameters when z' = 10 mm should be
a0 = -0.8E-03
a1 = -8.44E-01
a2 = 2.99E-02
a3 = -3E-04
and the equation when z' = 10 mm should be
y' = yLE + [-0.8E-03]*(x'-xLE)^0 + [-8.44E-01]*(x'-xLE)^1 + [2.99E-02]*(x'-xLE)^2 + [-3E-04]*(x'-xLE)^3
Note that the parameters as reported are multiplied by scale factors (e.g. 10^4) and you need to divide the reported values by those scale factors to get the true values.
Here's my graph of the airfoil profiles
This could have happened if the authors used Excel's LINEST function to do data fitting, because LINEST reports curve fit parameters in high-to-low order (a[n], a[n-1], ... a[2], a1, a[0]). To me this is ordering counter-intuitive and can lead to confusion.
Hope that helps
Related
I want to calculate the Gross Value Added for different countries and industries using multi-regional input-output (MRIO) tables. However, I struggle to find a good explanation of how this is done based on the data available. The definition of the GVA (Gross Value Added) is the output of a country/industry less the intermediate consumption, and it is related to the GDP by:
GVA = GDP + subsidies - taxes
So far, I have used the "extensions" or "satellite accounts" that provide the Value Added (VA) disaggregated across different flows, i.e. example from Exiobase in the picture. The VA is the sum of all 12 to my understanding. However, based on the definition of the GVA, I have subtracted 1-3 since these are taxes (so GVA = sum of line 4-12). To me, this seems like the correct approach, but I have not succeeded in finding an explanation that could confirm/disprove. I also become uncertain due to the naming of the extension, i.e. "value added" sounding like "gross value added". Does anyone know the correct way of doing this?
Finally, in MRIO x is termed "gross output" being the total output to final demand + intermediate consumption:
x = Ax + y (Ax = intermediate, y = demand)
or
x = (I-A)^-1 * y = L*y (L = Leontif inverse/requirement matrix)
Does this mean that I can also derive the GVAs from x by subtracting the intermediate consumption? In my mind, this will just leave me with "y", but there might be a another smart way?
Thanks in advance!
From what I understand, yes you can !
You have to differentiate Z = Ax summed along its rows or along its columns
x - rowSum(Z) is the GVA.
x - colSum(Z) is the total final demand.
Regarding Exiobase, I don't have a real answer.
I found that summing all lines of the VA (keeping lines 1-3), I get "quasi" the same results as subtracting the row sum of Z to x.
Which is stange...
I’ve used Excel in the past but the calculations including the Poisson-Distribution took a while, that’s why I switched to SQL. Soon I’ve recognized that SQL might not be a proper solution to deal with statistical issues. Finally I’ve decided to switch to Matlab but I’m not used to it at all, my problem Is the following:
I’ve imported a .csv-table and have two columns with values, let’s say A and B (110 x 1 double)
These values both are the input values for my Poisson-calculations. Since I wanna calculate for at least the first 20 events, I’ve created a variable z=1:20.
When I now calculated let’s say
New = Poisspdf(z,A),
it says something like non-scalar arguments must match in size.
Z only has 20 records but A and l both have 110 records. So I’ve expanded Z= 1:110 and transposed it:
Znew = Z.
When I now try to execute the actual calculation:
Results = Poisspdf(Znew,A).*Poisspdf(Znew,B)
I always get only a 100x1 Vector but what I want is a matrix that is 20x20 for each record of A and B (based on my actual choice of z=1:20, I only changed to z=1:110 because Matlab told that they need to match in size).
So in this 20x20 Matrix there should always be in each cell the result of a slightly different calculation (Poisspdf(Znew,A).*Poisspdf(Znew,B)).
For example in the first cell (1,1) I want to have the result of
Poisspdf(0,value of A).*Poisspdf(0,value of B),
in cell(1,2): Poisspdf(0,value of A).*Poisspdf(1,value of B),
in cell(2,1): Poisspdf(1,value of A).*Poisspdf(0,value of B),
and so on...assuming that it’s in the Format cell(row, column)
Finally I want to sum up certain parts of each 20x20 matrix and show the result of the summed up parts in new columns.
Is there anybody able to help? Many thanks!
EDIT:
Poisson Matrix in Excel
In Excel there is Poisson-function: POISSON(x, μ, FALSE) = probability density function value f(x) at the value x for the Poisson distribution with mean μ.
In e.g. cell AD313 in the table above there is the following calculation:
=POISSON(0;first value of A;FALSE)*POISSON(0;first value of B;FALSE)
, in cell AD314
=POISSON(1;first value of A;FALSE)*POISSON(0;first value of B;FALSE)
, in cell AE313
=POISSON(0;first value of A;FALSE)*POISSON(1;first value of B;FALSE)
, and so on.
I am not sure if I completely understand your question. I wrote this code that might help you:
clear; clc
% These are the lambdas parameters for the Poisson distribution
lambdaA = 100;
lambdaB = 200;
% Generating Poisson data here
A = poissrnd(lambdaA,110,1);
B = poissrnd(lambdaB,110,1);
% Get the first 20 samples
zA = A(1:20);
zB = B(1:20);
% Perform the calculation
results = repmat(poisspdf(zA,lambdaA),1,20) .* repmat(poisspdf(zB,lambdaB)',20,1);
% Sum
sumFinal = sum(results,2);
Let me know if this is what you were trying to do.
This is really weird. I calculate R^2 values with Excel in two different ways and the results differ hugely. Why?
1) First I use Excel to do a linear regression via a graph, and use the "Add Trendline..." right mouse button functionality to specify Intercept = 0. The R square value shows -3.253. The regressed equation is Y = -0.1321 * X
2) Then I use Excel to do a linear regression via LINEST function. I highlight 5x2 rows and in the top left cell, I type "=LINEST ([Y vector]; [X vector], FALSE, TRUE). The False means the intercept is 0, and the True means Excel should print additional regression statistical information. Then I press CTRL + SHIFT + Enter. This will show me additional statistics, such as R^2 value in the third left cell. Which turns out to be 0.11166. The regressed equation is Y = -0.1321 * X
My question is; what am I doing wrong in calculating R^2 with the graph? Python and statsmodels.api confirms that R^2 is 0.11166, and the regressed equation is Y = -0.1321 * X.
Y =
0.0291970802919708
0.141801551718973
0.145668034655723
0.0691229530946433
0.0431577486597426
0.133618351873374
X =
-0.35551988
-0.20577599
0.10780785
-0.25028796
-0.42762184
0.02442197
Your calculation is correct. Scatter plot does not return correct R^2 when the intercept is 0. This is an formula fo R^2
where
If you use standard regression model, you use average value of y as y̅. But when you assume that the intercept equals 0, you need to set y̅ as zero. If you use the average value of y instead of zero, you get the R^2 = -3.252767.
You can see the calculation here. The SStot wrong column uses average value of y as y̅. Then the R^2 value equals to -3.252767. If you use 0 (as I did in SStot right column), then you get 0.111.
It is an old bug described by Microsoft here:https://support.microsoft.com/en-us/help/829249/you-will-receive-an-incorrect-r-squared-value-in-the-chart-tool-in-excel-2003
You need to use the LINEST function to get correct R^2 value.
Me and my fellow engineers just got tangled up in this. Based on this discussion and what we observed, the R^2 is wrong all of the time except when Excel calculates the best y-intercept. Any other y-intercept (either forced through Zero OR user-defined), is wrong.
I use datas in excel to produce a graphic.
Then I make a regression, and have an equation. I'd like to know what value would be obtained from the regression (for example, x = 7,6 is the value for which I wanna know an estimation of y).
It is an approximation with a 6 degree polynome.
One wimple method would be this : I have the equation, so I could use it
However, I wondered if there is a fast method to do it? Like I enter 7,6 somewhere to have the result quickly?
if you are looking at a linear regression line (straight line) you could try the forecast formula
=forecast(X, Known Ys, Known Xs)
you could also build your own equation automatically from
=linest(...)
I found the following on a site describing the capabilities of the linest function in excel:
In addition to using LOGEST to calculate statistics for other
regression types, you can use LINEST to calculate a range of other
regression types by entering functions of the x and y variables as the
x and y series for LINEST. For example, the following formula:
=LINEST(yvalues, xvalues^COLUMN($A:$C))
works when you have a single column of y-values and a single column of
x-values to calculate the cubic (polynomial of order 3) approximation
of the form:
y = m1*x + m2*x^2 + m3*x^3 + b
You can adjust this formula to calculate other types of regression,
but in some cases it requires the adjustment of the output values and
other statistics.
or look at:
=trend
suggest here are 3 points on,(x0,y0),(x1,y1),(x2,y2)
O = (x0,y0)
e1 = (x1-x0,y1-y0)
e2 = (x2-x0,y2-y0)
3 can make a new cordinate (O,e1,e2)
here's a point (x,y)
how to calculate the point location in (O,e1,e2)?please write down the formula ,thanks.
once i remember,but now i forget.
Let's call new coordinates a and b.
In the old coordinate system the point will be O+a*e1+b*e2. Since it should be the same point (x,y), we have two linear equations:
x=Ox+a*e1x+b*e2x
y=Oy+a*e1y+b*e2y
Everything except a and b is known, two unknowns, two equations - solution exists if e1 and e2 are not parallel.
The system can be solved either by inversion of matrix ( (e1x,e2x) , (e1y,e2y) ), or by expressing a in terms of b from the first equation and substituting it into the second one.