[Legolas_1204]

I am doing a drag analysis of rocket nose cone and I want to use 2D geometry instead of 3D geometry to save time. I know I have to use a 2D geometry with axisymmetric but how can I find the reference area for the geometry. In 3D we can find the reference area using projected area but how to find for 2D axisymmetric body.

[LuckyTran]

In 2D, the pressures are integrated over the reference depth (default is 1 m), which is another needed input. Now you have a force that acts over an area.

Hint: you can imagine your 2D body being extruded 1m in the third direction and treat it as a 3D body. This analogue doesn't always work because the reference depth need not be 1m and the reference area need not be the projected area.

[Legolas_1204]

Originally Posted by LuckyTran

In 2D, the pressures are integrated over the reference depth (default is 1 m), which is another needed input. Now you have a force that acts over an area.

Hint: you can imagine your 2D body being extruded 1m in the third direction and treat it as a 3D body. This analogue doesn't always work because the reference depth need not be 1m and the reference area need not be the projected area.

I understand what you are saying but what you are telling is applicable for planar surface like an aerofoil. But nose cone is a revolving surface so can we take reference depth of 1m in 3rd direction?
Or do we need to revolve the surface 360 degrees and then find the projected area? (In this case, I know that reference area is projected area)

[LuckyTran]

I missed the part where you mentioned a cone.


In that case, it's integrated over 1 radian

[Legolas_1204]

Originally Posted by LuckyTran

I missed the part where you mentioned a cone.


In that case, it's integrated over 1 radian

Okay so from what you are saying if my cone radius is r meters, reference area should be,

(pi * r^2) / 2*pi ??
for 2D geometry?

[sina_sls]

Originally Posted by Legolas_1204

Okay so from what you are saying if my cone radius is r meters, reference area should be,

(pi * r^2) / 2*pi ??
for 2D geometry?

Hello Shashwat,

I want to compute area in 2D axisymmetric. Do you figure out how can we compute it? Could you please share it?

Best Regards,

[LuckyTran]

The reference area comes from the 3D representation of the object unless you have a degenerate body such infinite parallel plates where there is no meaning of area.

The pressure being integrated over 1 radian is related to how you go about computing reference area except to understand how the coefficients are calculated using a surface integrated force in 2D.


If you are doing a 2D axissymmetric simulation of a cone, draw the 3D version of the one and get the reference area.

[sina_sls]

Originally Posted by LuckyTran

The reference area comes from the 3D representation of the object unless you have a degenerate body such infinite parallel plates where there is no meaning of area.

The pressure being integrated over 1 radian is related to how you go about computing reference area except to understand how the coefficients are calculated using a surface integrated force in 2D.


If you are doing a 2D axissymmetric simulation of a cone, draw the 3D version of the one and get the reference area.

Thank you for your response, I need area of red part of the picture which i attached , i have to draw 3D form?
bottom of geometry is axis.