Area

[Fabrice]

Hi,

in a cylinder a throttle.
In close position the throttle has an angle of 3° versus the axis of the
cylinder (so that the throttle shape is not round).

When the throttle is closed, the projected area (in the axis of the
cylinder) is 0 mm^2 (easy).

How can I calculate the projected (in the axis of the cylinder) open area
when the throttle opens?

When the throttle rotates, its equation changes in function of the angle of
rotation, doesn't it?

Thanks
Fabrice

[Lynn Kurtz]

On Mon, 9 May 2005 20:21:47 +0200, "Fabrice" <fa...@village.uunet.be>
wrote:

I assume you know or can measure the area of the open throttle body
cylinder, which I will call A. Call your 3 degrees alpha, and call
beta the varible throttle angle which varies from 3 to 90 degrees.
Then the formula for the open area is:

Open Area = A ( 1 - sec( alpha ) cos( beta ).

Notice that when beta = alpha you get 1 - 1 = 0 and when beta = 90
degrees, cos( beta ) = 0 so you get the full throttle body area A.

--Lynn

[Fabrice]

Hi,

thx for the answer.
For my understanding...to arrive at the final answer:
you used the area defined by an ellipse and ?

thx
Fabrice

"Lynn Kurtz" <kurtzDEL...@asu.edu> a écrit dans le message de news:
YgCAQv8bycVlWy...@4ax.com...

[Lynn Kurtz]

On Wed, 11 May 2005 20:42:06 +0200, "Fabrice" <fa...@village.uunet.be>
wrote:

>Hi,
>

>thx for the answer.
>For my understanding...to arrive at the final answer:
>you used the area defined by an ellipse and ?
>
>thx
>Fabrice

It doesn't really matter that it is an ellipse. I just used the fact
that the projection of one plane area A on another plane is equal to
Acos(theta) where theta is the angle between the planes.

--Lynn