You can find the magnitude of gravity by looking up the table, assuming it is 9.81. We take out the direction of gravity and set the magnitude directly to 9.81.
The purpose of this adjustment is that the known quantities in the equations solved in the previous section also include the error of gravity, and the error needs to be adjusted in this way.
The pictures in the paper are as follows
Let’s adjust the direction of gravity
g
^
?
\vec{\hat{g}}
g^?
b
1
?
,
b
2
?
\vec{b_{1}},\vec{b_{2}}
b1?
w
1
,
w
2
w_{1},w_{2}
The size of w1?,w2? can adjust the gravity direction.
g
=
∣
∣
g
∣
∣
?
g
?
+
w
1
b
1
?
+
w
2
b
2
?
g=||g||·\vec{g} + w_{1}\vec{b_{1}} + w_{2}\vec{b_{2}}
g=∣∣g∣∣?g
g
g
g becomes
B
?
W
B·W
B?W instead,
W
∈
2
×
1
W∈2×1
W∈2×1 refers to the weight change,
B
∈
3
×
2
B∈3×2
B∈3×2 refers to two vertical vectors
You need to first find two vertical vectors and then repeat the operation in the previous section for multiplication. However, here is an iterative operation to solve. The operation in the previous section is actually equivalent to providing an initial value.
Matrix multiplication in the code is actually equivalent to dot multiplication
a
?
?
t
e
m
p
?
=
∣
∣
a
∣
∣
?
∣
∣
t
e
m
p
∣
∣
?
c
o
s
θ
\vec{a}·\vec{temp}=||a||·||temp||·cosθ
a
a
?
\vec{a}
a
c
o
s
θ
cosθ
cosθ
a.transport*temp=cosθ
Then the vector
a
?
\vec{a}
a
c
o
s
θ
cosθ
cosθ obtains the length of vector a, this
a
?
\vec{a}
a
g
?
\vec{g}
g
reuse
t
e
m
p
?
\vec{temp}
temp
a
?
\vec{a}
a
a
?
\vec{a}
a
a
?
\vec{a}
a
a
?
×
b
?
\vec{a}×\vec{b}
a
The coefficient in the code is divided by 100, and the result is also divided by 100 because if the coefficient is divided by 100, the result will be amplified 100 times to meet the requirements, so the result must also be divided by 100. This is to amplify the data for the stability of the numerical solution.