如何算三角形的cotangent

公式为:

cotC=a2+b2−c24A

\cot C = \frac{a^2 + b^2 - c^2}{4A}

证明
根据余弦定理

cosC=a2+b2−c22ab\cos C = \frac{a^2 + b^2 - c^2}{2ab}

根据面积公式
A=12absinCA = \frac{1}{2}ab \sin C
所以
sinC=2Aab\sin C = \frac{2A}{ab}

所以cotC=cosCsinC=a2+b2−c22ab⋅ab2A=a2+b2−c24A\cot C =\frac{\cos C}{\sin C} = \frac{a^2 + b^2 - c^2}{2ab} \cdot \frac{ab}{2A} = \frac{a^2 + b^2 - c^2}{4A}

alec jacobson 代码:

function C = cotangent(V,F,varargin)% COTANGENT compute the cotangents of each angle in mesh (V,F), more details% can be found in Section 1.1 of "Algorithms and Interfaces for Real-Time% Deformation of 2D and 3D shapes" [Jacobson 2013]% % C = cotangent(V,F)% C = cotangent(V,F,'ParameterName',parameter_value,...)%% Known bugs:%   This seems to return 0.5*C and for tets already multiplies by%   edge-lengths%% Inputs:%   V  #V by dim list of rest domain positions%   F  #F by {3|4} list of {triangle|tetrahedra} indices into V%   Optional (3-manifolds only):%     'SideLengths' followed by #F by 3 list of edge lengths corresponding%       to: 23 31 12. In this case V is ignored.%       or%       followed by #T by 6 list of tet edges lengths:%       41 42 43 23 31 12%     'FaceAreas' followed by #T by 4 list of tet face areas% Outputs:%   C  #F by {3|6} list of cotangents corresponding%     angles for triangles, columns correspond to edges 23,31,12%     dihedral angles *times opposite edge length* over 6 for tets, %       WRONG: columns correspond to edges 23,31,12,41,42,43 %       RIGHT: columns correspond to *faces* 23,31,12,41,42,43%% See also: cotmatrix%% Copyright 2013, Alec Jacobson (jacobson@inf.ethz.ch)%switch size(F,2)case 3% default valuesl = [];% Map of parameter names to variable namesparams_to_variables = containers.Map( ...{'SideLengths'}, {'l'});v = 1;while v <= numel(varargin)param_name = varargin{v};if isKey(params_to_variables,param_name)assert(v+1<=numel(varargin));v = v+1;% Trick: use feval on anonymous function to use assignin to this workspace feval(@()assignin('caller',params_to_variables(param_name),varargin{v}));elseerror('Unsupported parameter: %s',varargin{v});endv=v+1;endassert(numel(varargin) == 0);% trianglesif isempty(l)% edge lengths numbered same as opposite verticesl = [ ...sqrt(sum((V(F(:,2),:)-V(F(:,3),:)).^2,2)) ...sqrt(sum((V(F(:,3),:)-V(F(:,1),:)).^2,2)) ...sqrt(sum((V(F(:,1),:)-V(F(:,2),:)).^2,2)) ...];endi1 = F(:,1); i2 = F(:,2); i3 = F(:,3);l1 = l(:,1); l2 = l(:,2); l3 = l(:,3);% semiperimeterss = (l1 + l2 + l3)*0.5;% Heron's formula for areadblA = 2*sqrt( s.*(s-l1).*(s-l2).*(s-l3));% cotangents and diagonal entries for element matrices% correctly divided by 4 (alec 2010)C = [ ...(l2.^2 + l3.^2 -l1.^2)./dblA/4 ...(l1.^2 + l3.^2 -l2.^2)./dblA/4 ...(l1.^2 + l2.^2 -l3.^2)./dblA/4 ...];case 4% Dealing with tetsT = F;l = [];s = [];v = 1;while v <= numel(varargin)switch varargin{v}case 'SideLengths'assert((v+1)<=numel(varargin));v = v+1;l = varargin{v};case 'FaceAreas'assert((v+1)<=numel(varargin));v = v+1;s = varargin{v};otherwiseerror(['Unsupported parameter: ' varargin{v}]);endv = v+1;endif isempty(l)% lengths of edges opposite *face* pairs: 23 31 12 41 42 43l = [ ...sqrt(sum((V(T(:,4),:)-V(T(:,1),:)).^2,2)) ...sqrt(sum((V(T(:,4),:)-V(T(:,2),:)).^2,2)) ...sqrt(sum((V(T(:,4),:)-V(T(:,3),:)).^2,2)) ...sqrt(sum((V(T(:,2),:)-V(T(:,3),:)).^2,2)) ...sqrt(sum((V(T(:,3),:)-V(T(:,1),:)).^2,2)) ...sqrt(sum((V(T(:,1),:)-V(T(:,2),:)).^2,2)) ...];end% (unsigned) face Areas (opposite vertices: 1 2 3 4)if isempty(s)s = 0.5*[ ...doublearea_intrinsic(l(:,[2 3 4])) ...doublearea_intrinsic(l(:,[1 3 5])) ...doublearea_intrinsic(l(:,[1 2 6])) ...doublearea_intrinsic(l(:,[4 5 6]))];end[~,cos_theta] = dihedral_angles([],[],'SideLengths',l,'FaceAreas',s);% volumevol = volume_intrinsic(l);%% Law of cosines%% http://math.stackexchange.com/a/49340/35376%H_sqr = (1/16) * ...%  (4*l(:,[4 5 6 1 2 3]).^2.* l(:,[1 2 3 4 5 6]).^2 - ...%  ((l(:, [2 3 4 5 6 1]).^2 + l(:,[5 6 1 2 3 4]).^2) - ...%  (l(:, [3 4 5 6 1 2]).^2 + l(:,[6 1 2 3 4 5]).^2)).^2);%cos_theta= (H_sqr - s(:,[2 3 1 4 4 4]).^2 - ...%                    s(:,[3 1 2 1 2 3]).^2)./ ...%                (-2*s(:,[2 3 1 4 4 4]).* ...%                    s(:,[3 1 2 1 2 3]));%% To retrieve dihedral angles stop here...%theta = acos(cos_theta);%theta/pi*180% Law of sines% http://mathworld.wolfram.com/Tetrahedron.html%sin_theta = bsxfun(@rdivide,vol,(2./(3*l)) .* s(:,[1 2 3 2 3 1]) .* s(:,[4 4 4 3 1 2]));sin_theta = bsxfun(@rdivide,vol,(2./(3*l)) .* s(:,[2 3 1 4 4 4]) .* s(:,[3 1 2 1 2 3]));%% Using sin for dihedral angles gets into trouble with signs%theta = asin(sin_theta);%theta/pi*180% http://arxiv.org/pdf/1208.0354.pdf Page 18C = 1/6 * l .* cos_theta ./ sin_theta;%% One should probably never use acot to retrieve signed angles%theta = acot(C);%theta/pi*180otherwiseerror('Unsupported simplex type');endend

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