[just-right]

Hey,

I have used to the Function Calculator to receive the resultant force. However I can't seem to find the origin of this vector. It is possible to calculate the origin with use of geometry. Is there a way to receive the origin coordinates of the x, y and z forces from the program itself?

Excuse me for my English, it is not my native language.

All the best,

T

[ghorrocks]

Hi,

The force vector gives you the x/y/z components of the vector. If you want to find the line through which the force acts you will need to calculate the torque of the body about some axis and derive the location from that.

Glenn Horrocks

[just-right]

Hi,

Thank you for your reply. I assume the generated torques arround the model origin. The torque consists of two sub-torques generated by the force component and their arm. Every arm is used in two of the equations, they have relations.

Tx = Fy * rz + Fz * ry
Ty = Fx * rz + Fz * rx
Tz = Fx * ry + Fy * rx

I converted the eqautions to a matrix:

T = F * r

Reorganised it and applied Gauss elimination, however the denominator became zero. I then tried to solve it using Matlab. I used the following expression:

S=solve('Tx=(Fy*rz)+(Fz*ry)','Ty=(Fx*rz)+(Fz*rx)', 'Tz=(Fx*ry)+(Fy*rx)');

The solutions are:

rx
ry
rz

in the structure S.

I came to the conclusion that the vector origin of the forces is:

rx = 6.3611 m
ry = 19.1505
rz = -0.4128

I highly doubt the result since the tunnel is only 5 meters wide, 5.277 meters high and has a length of 12 meters. The maximum positive x in the tunnel is 4 meters. Maximum negative z is 0.277 meters.

The study involves a vehicle and the centre of aerodynamic forces is very important. Could somebody explain a correct way of approaching this porblem?

[ghorrocks]

Hi,

I reckon if you draw your vector from the point you calculate you will find it is nowhere near the surface. Obviously your maths is wrong. I don't have time to check it but you don't seem to be using the correct definition of cross product.

Glenn Horrocks