How do I calculate the acceleration using only 3D distances?

11 次查看（过去 30 天）

显示更早的评论

lil brain 2022-4-15

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1697685-how-do-i-calculate-the-acceleration-using-only-3d-distances

评论： Les Beckham 2022-4-18

采纳的回答： Les Beckham

acc_dist.mat

在 MATLAB Online 中打开

Hi,

I have a list of distances in each cell of acc_dist . The first three columns in each cell are xyz coordinates. So column one are all x, column two are all y, and column three are all z. The rows are the coordinate points measured at a speed of 72 Hz.

I am trying to compute the acceleration of these coordinate points by first calculating the 3D euclidean distance between each point and the adjacent point like so:

sqrt((x2-x1).^2+(y2-y1).^2+(z2-z1).^2)

And then I am trying to use the acceleration formula like so:

acceleration = dv/dt

dv = change in velocity

dt = change in time

How would I go about calculating the acceleration for the entire length of each list?

Thank you!

4 个评论
显示 2更早的评论隐藏 2更早的评论

Walter Roberson 2022-4-15

Use gradient() twice.

However... acceleration is a signed vector quantity, derived from velocity which is a signed vector quantity. This is a problem for you because sqrt((x2-x1).^2+(y2-y1).^2+(z2-z1).^2) is not signed and not a vector. You are calculating speed, not velocity.

lil brain 2022-4-15

编辑：lil brain 2022-4-15

@Walter Roberson I dont think I follow could you elaborate on that?

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Les Beckham 2022-4-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1697685-how-do-i-calculate-the-acceleration-using-only-3d-distances#answer_943810

编辑：Les Beckham 2022-4-15

在 MATLAB Online 中打开

acc_dist.mat

load acc_dist.mat

xyz = cell_of_double_pre_ballsCopy{1};

dxyz = diff(xyz); % difference between adjacent points in xyz coordinates

v = sqrt(dxyz(:,1).^2 + dxyz(:,2).^2 + dxyz(:,3).^2) * 72;

a = gradient(v, 1/72);

t = 0:1/72:(length(a)-1)/72;

plot(t,v)

plot(t,a)

13 个评论
显示 11更早的评论隐藏 11更早的评论

lil brain 2022-4-15

在 MATLAB Online 中打开

Sorry you are right this wasnt very clear of me.

Essentially I am trying to apply your code above to a larger file cell array with 19 cells (the file I uploaded only has 2 cells). I am trying to build for loops that perform all of your steps on each of the cells in my cell array. But I am having trouble making it work.

Here is what I have so far:

% differences
n = numel(cell_of_double_pre_ballsCopy);
dxyz = {};
for d = 1:n
    dxyz{d} = cellfun(@diff,cell_of_double_pre_ballsCopy,'UniformOutput',false); % unsure if this is the correct way of applying the code to every cell in the array
end
dxyz_doubles = cellfun(@cell2mat, dxyz, 'uni',false ); % need to convert the cell array into a cell array of doubles
% velocity
n = numel(dxyz_doubles);
vel = {}; %velocity
for v = 1:n
    vel{v} = sqrt(dxyz_doubles{v}(:,1).^2 + dxyz_doubles{v}(:,2).^2 + dxyz_doubles{v}(:,3).^2);
end
% acceleration
n = numel(vel);
acc = {}; %acceleration
for v = 1:n
    acc{a} = gradient(vel{a},1/72);
end

Can you help me with this?

Walter Roberson 2022-4-17

在 MATLAB Online 中打开

@Les Beckham

"Apply this function to every element of this array" is one of the fundamental array operations in theory.

Consider for example,

A = [1 3 5]
A = 1×3
     1     3     5
B = A.^2
B = 1×3
     1     9    25

From a theoretical perspective, this is not "squaring the vector": from a theoretical perspective, it is "Apply the function x->x^2 to each element of A, returning an array of the results.

This is frequenty called the "map" operation. And having a brief call that says that you are mapping a function over all of the elements of an array, is often significantly clearer than using a for loop.

Les Beckham 2022-4-18