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It is a problem about aerodynaic drag.

The following is the formula.

$d{\vec{f_{drag}}} = -0.5C_{D}{\rho}{V^2}{({\hat{n_{i}}}\boldsymbol{\cdot}{\hat{v})}{\hat{v}}dA_{i}}$

${\vec{T_{drag}}}=\int{\vec{r}}{\times}d{\vec{f_{drag}}}=-0.5C_{D}{\rho}{V^2}\int({\hat{n_{i}}}\boldsymbol{\cdot}{\hat{v})(\vec{r_{i}}\times{\hat{v}})dA_{i}}=0.5C_{D}{\rho}{V^2}\int({\hat{n_{i}}}\boldsymbol{\cdot}{\hat{v})(\hat{v}\times{\vec{r_{i}}})dA_{i}}$

The position vector here corresponds to the position of the center of pressure of surface with respect to the center of mass.

We want to determine the torque on the object.

Assuming that we have an object which is a Rectangular Prism.

The length is C, the width is A,and the height is B.

Assumption1: The center of mass is in the center of the Rectangular Prism.

Assumption2: The center of pressure is located at the geometric center of the plane.

Choose a reference frame where the center of mass is located at the origin.

There are six planes in a Rectangular Prism. We define a plane which is perpendicular to x-axis in the positive part is A+x.

In the following are the areas of each plane.

$A_{+x}{\quad}A_{-x}{\quad}A_{+y}{\quad}A_{-y}{\quad}A_{+z}{\quad}A_{-z}$

With their normal vector in each column in the matrix below.

$\hat{n}=\begin{bmatrix} 1 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 \end{bmatrix}$

The velocity of the center of mass is :

$\vec{V}=\left\| V \right\|{\hat{v}}=V\begin{bmatrix} \alpha\\ \beta \\ \gamma\\ \end{bmatrix}$

We assume all the elements of velocity vector are positive.

The position vector of each plane is in the column of the matrix below.

$\vec{r}=\begin{bmatrix} C/2 &-C/2 & 0 & 0 & 0 & 0 \\ 0 & 0 &A/2 &-A/2 & 0 & 0 \\ 0 & 0 & 0 & 0 & B/2 &-B/2\\ \end{bmatrix}$

To symplify the calculation, we would ignore some constants.

For every plane of the object, torque is produced only when :

${{\hat{n}}\boldsymbol{\cdot}{\hat{v}}>0}$

So there are three planes, at most, producing torque on the object.

$\int({\hat{n_{i}}}\boldsymbol{\cdot}{\hat{v})(\hat{v}\times{\vec{r_{i}}})dA_{i}}={\alpha}A_{+x}\begin{bmatrix} 0 \\ {\gamma}C/2\\ {-\beta}C/2\\ \end{bmatrix}+{\beta}A_{+y}\begin{bmatrix} {-\gamma}A/2\\ 0 \\ {\alpha}A/2 \\ \end{bmatrix}+{\gamma}A_{+z}\begin{bmatrix} {\beta}B/2 \\ {-\alpha}B/2 \\ 0 \\ \end{bmatrix}=\begin{bmatrix} 0 \\ {\alpha\gamma}ABC/2 \\ {-\alpha\beta}ABC/2 \\ \end{bmatrix}+\begin{bmatrix} {-\beta\gamma}ABC/2 \\ 0 \\ {\alpha\beta}ABC/2 \\ \end{bmatrix}+\begin{bmatrix} {\beta\gamma}ABC/2 \\ {-\alpha\gamma}ABC/2 \\ 0 \\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

The torque ended up to be zero.

I want to build a model to run simulation on aerodynaic drag.

Based on the assumption, the result is so weird, zero torque,as there is always torque in practice.

What is the problem? Are the hypotheses too perfect to be true?

How can i examine my model to make sure that is correct and precise?

Any answer is appreciated, thank you.

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  • $\begingroup$ Torque depends on angle of attack, and if you have symmetric flow around the prism the torque is zero. $\endgroup$ – ja72 Mar 6 '18 at 1:30
  • $\begingroup$ Torque also arises from vortex shedding and this model does not account for this. $\endgroup$ – ja72 Mar 7 '18 at 4:39
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Assumption2 ensures that the net force vector for each face of the prism aims directly at the centroid. Therefore, there can be no torque.

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