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我需要计算两个平面之间的角度。一架飞机是手机，另一架是前臂飞机。我计算了这些飞机的法线，并在MATLAB中使用了公式atan2(norm(cross(var.n1,var.n2)),dot(var.n1,var.n2));。我希望看到以正峰值和负峰值为特征的手腕的屈曲/伸展角度，但通过这个公式，我只能得到正峰值。Matlab atan2两个3D矢量之间的角度
%% Script to compute the angles of wrist flexion/extension and adduction/abduction based on Vicon data
% REF: Cheryl et al., March 2008
% Order of the markers: 1.WRR 2.WRU 3.FAU 4.FAR 5.CMC2 6.CMC5 7.MCP5 8.MCP2
clc
close all
clear all
%% Initialization
% dir_kinematic = input('Path of the folder containing the kinematic files: ','s');
dir_kinematic = 'C:\Users\Utente\Desktop\TESI\Vicon\test.mat';
cd(dir_kinematic);
fileList = getAllFiles(dir_kinematic,0); % Get names of all kinematic files
f=1;
%% Conversion to angles
for f = 1:length(fileList)
if ~isempty(strfind(fileList{f},'mat')) % Take only mat files
% 0. Loading
load(fileList{f});
% 1. Filtering
frameRate = kinematic.framerate;
n = 9;
Wn = 2/(frameRate/2);
ftype = 'low';
[b,a] = butter(n,Wn,ftype);
kinematic.x = filtfilt(b,a,kinematic.x);
kinematic.y = filtfilt(b,a,kinematic.y);
kinematic.z = filtfilt(b,a,kinematic.z);
% 2. Create vectors
var.n=length(kinematic.x);
% Forearm plane
var.FAU_WRU=[kinematic.x(:,2)-kinematic.x(:,3),kinematic.y(:,2)-kinematic.y(:,3),kinematic.z(:,2)-kinematic.z(:,3)];
var.WRR_WRU=[kinematic.x(:,2)-kinematic.x(:,1),kinematic.y(:,2)-kinematic.y(:,1),kinematic.z(:,2)-kinematic.z(:,1)];
% Hand plane
var.CMC5_MCP5=[kinematic.x(:,7)-kinematic.x(:,6),kinematic.y(:,7)-kinematic.y(:,6),kinematic.z(:,7)-kinematic.z(:,6)];
var.MCP2_MCP5=[kinematic.x(:,7)-kinematic.x(:,8),kinematic.y(:,7)-kinematic.y(:,8),kinematic.z(:,7)-kinematic.z(:,8)];
% Transpose
var.FAU_WRU = var.FAU_WRU';
var.WRR_WRU = var.WRR_WRU';
var.CMC5_MCP5 = var.CMC5_MCP5';
var.MCP2_MCP5 = var.MCP2_MCP5';
% 3. Calculate angle of wrist flexion/extension
% Cross vector function for all time => create normal vector plane
var.forearm_n=[];
var.hand_n=[];
var.theta_rad=[];
for i = 1:var.n % Loop through experiment
% vector x and y of the forearm plane
var.v1=var.FAU_WRU(:,i); % take x,y,z of the vector for every time
var.v2=var.WRR_WRU(:,i);
% vector x and y of the hand plane
var.v3=var.CMC5_MCP5(:,i);
var.v4=var.MCP2_MCP5(:,i);
var.forearm_n= [var.forearm_n, cross(var.v1,var.v2)];
var.hand_n=[var.hand_n, cross(var.v3,var.v4)];
end
% Calculate angle
for i = 1:var.n
var.n1=(var.forearm_n(:,i));
var.n2=var.hand_n(:,i);
var.scalar_product(i) = dot(var.n1,var.n2);
%Equation (2) of the paper
var.theta_rad=[var.theta_rad, atan2(norm(cross(var.n1,var.n2)),dot(var.n1,var.n2))]; % result in radian
angle.flex_deflex_wrist{f}=(var.theta_rad*180)/pi;
end
% 4. Calculate angle of wrist adduction/abduction
% Projection vector onto plane
var.MCP2_MCP5_forearmproj=[];
var.WRR_WRU_forearmproj=[];
var.rad_ul_angle_rad=[];
for i=1:var.n
%take x,y,z of the vector for each time
var.v1=var.MCP2_MCP5(:,i);
var.v2=var.WRR_WRU(:,i);
% vector x and y of the forearm plane
var.vfx=var.FAU_WRU(:,i); % take x,y,z of the vector for every time
var.vfy=var.WRR_WRU(:,i);
%projection of vector MCP2_MCP5 and WRR_WRU onto forearm plane
var.squNorm1=(norm(var.vfx)*norm(var.vfx));
var.squNorm2=(norm(var.vfy)*norm(var.vfy));
var.MCP2_MCP5_forearmproj=[var.MCP2_MCP5_forearmproj,((((var.v1')*var.vfx)*var.vfx/var.squNorm1)+(((var.v1')*var.vfy)*var.vfy/var.squNorm2))];
var.WRR_WRU_forearmproj=[var.WRR_WRU_forearmproj,((var.vfx*((var.v2')*var.vfx/var.squNorm1))+(var.vfy*((var.v2')*var.vfy/var.squNorm2)))];
end
% Calculate angle
for i=1:var.n
var.n1=var.MCP2_MCP5_forearmproj(:,i)';
var.n2=var.WRR_WRU_forearmproj(:,i);
var.product=var.n1*var.n2;
var.rad_ul_angle_rad=[var.rad_ul_angle_rad, atan2(norm(cross(var.n1,var.n2)),dot(var.n1,var.n2))];% en rad
angle.rad_ul_wrist{f}=(var.rad_ul_angle_rad*180)/pi;
end
end
end
我想知道为什么我的角度始终为正？我需要看到积极和消极的峰... 感谢您的帮助！