Respective to PI each fixed scope of points on a figure has a projective truncation error. To prove the length of the diagonal line in a quad this form is build. (2*(sqrt((2*a)^2 + a^2)) * a) / sqrt((3a)^2 + a^2) = d but how is it possible to add limited curvature as a variable in maths? The code shows in general how to calculate diagonal lines by orthogonal triangles. So how to prove that each given curvature can be integrated getting the length by a sum of discret points in the figure?
https://ide.c9.io/neub/quad_refferer_bow
please check the links
www.ncbi.nlm.nih.gov › NCBI › Literature › PubMed Central (PMC)
math.stackexchange.com/.../how-can-i-limit-the-amount-of-curvature-of.
tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx
hi
Sounds interesting. Would you like to explain the problem a little more.
what is meant by
1. a and d?
2. "scope of points wrt PI
3. orthogonal triangles
4 quad (quadrilateral?)
can you show an example figure? as it stands we get d= sqrt(2a) ..the ref web sites are not accessible.
Cheers
Hi
Please explain the problem in more detail and I think that we'll help you to find the solution.
Recently we've developed the algorithm of Determination of the Line Geometry by X-Y-Z positions which is given with some errors. The main result of the developed procedure is a line curvature.
Kind regards
I try to refine the explanaition of the problem. How can the sum of several lines with the length as result be equal to the length of a curvature using a general length of the diagonal of a square or a rectangle. In general the form of 2.0*(((sqrt(func_pow2(2.0*A)+sqrt(func_pow2(A))))*A)/sqrt(func_pow2(3.0*A)+(sqrt(func_pow2(A))))) when given A = (a*b) of any given rectangle or quadrat. The question then whould be is an amount of line-lengths similar to the bow-length of a certain curvature. Is there a form where the value of the curvature can be multiplied with the diagonal gaining this result? Can we prove that there is such an point to line equvalence in finite calculations in maths?
If I have understood correctly .. the perimeter of a cyclic quadrilateral(polygon) in terms of circle radius and angle subtended seems a posible candidate..
Cheers
Please see if this is of help:
1. assume hypotenuse of a rt angled triangle (so li is a diameter of some circle) then l= sqrt( a1^2 +b1^2) = a1*sqrt (( 1+ ( b1/a1)^2 )) = a1sqrt(( 1+m1^2) ) [ m1 is slope of line and I think can be related to curvature).
3 Similarly l2= a2sqrt( (1+ mwo line segments for simplicity say l1 and l22 Consider l1 as h2^2) ) we can choose m2 so that m2= m1 =m
4. Then l1+l2 = (a1+a2)sqrt( (1+m^2))
5. if end points give line segment ld.. then ld=d*sqrt( 1+md^2) [ md is slope of ld)
6 again set md =m then (l1+l2)/ld = (a1+a2)/d
7 l1+l2 =ld*( a1+a2)/d ie sum of segments is related to slope m which is hopefully related to curvature thro radius of circum circle of the triangles.
Cheers
Wüssten Sie einen Bereich von zahlen für diese Theorie, die in die Gleichung passen? Dann würde ich die Gleichung programmieren und sie geben mit die Werte eines reellen Bereiches - bspw. der Physik
m hat bereich von 0 bis Unendlichkeit. Unten ein beispiel
l1= 10 l2= 20 sagen Sie mal ld=15
dann mit m=1.73 und s= sqrt(1+m^2)= 2
so ld = l1/s +l2/s = 10/2 +20/2 = 15 ..
hoeffentlich es hilft
Cheers
kann man in die Parameterdarstellung von Vektoren - Krümmungen einrechnen?
Parson said action is system! https://de.wikipedia.org/wiki/Talcott_Parsons
http://homepage.univie.ac.at/irmgard.gorgas/handouts/physik/skriptum2.pdf versteh ich eher noch - gute Darstellung - tolle Arbeit - bin gespannt ob ich das mal begreife