Ken Bones wrote:Thanks Andy.
Now could you supply an idiots guide plz!
Ken.
But that was the easier guide. I have put below this message the full differential calc for centripetal acceleration, although it is 25 years since I last tried to use that
The easiest way is just is just to estimate g from your flying style. If you are going to to crazy 3D then anything from 10 to 20g likely.Sedate flying 1 or 2 g. Tight, fast loops, maybe 4 or 5g.
Then use this to give you your wing loading by multiplying your model weight by the g figure.
So your 10kg model at 5g will put 50kg load on the wings.
Regards, Andy
CENTRIPETAL ACCELERATION
In two dimensions the position vector \textbf{r} which has magnitude (length) r and directed at an angle \theta above the x-axis can be expressed in Cartesian coordinates using the unit vectors \hat{x} and y-hat:[10]
\textbf{r} = r \cos(\theta) \hat{x} + r \sin(\theta) \hat{y}.
Assume uniform circular motion, which requires three things.
The object moves only on a circle.
The radius of the circle r does not change in time.
The object moves with constant angular velocity \omega around the circle. Therefore \theta = \omega t where t is time.
Now find the velocity \textbf{v} and acceleration \textbf{a} of the motion by taking derivatives of position with respect to time.
\textbf{r} = r \cos(\omega t) \hat{x} + r \sin(\omega t) \hat{y}
\dot{\textbf{r}} = \textbf{v} = - r \omega \sin(\omega t) \hat{x} + r \omega \cos(\omega t) \hat{y}
\ddot{\textbf{r}} = \textbf{a} = - r \omega^2 \cos(\omega t) \hat{x} - r \omega^2 \sin(\omega t) \hat{y}
\textbf{a} = - \omega^2 (r \cos(\omega t) \hat{x} + r \sin(\omega t) \hat{y})
Notice that the term in parenthesis is the original expression of \textbf{r} in Cartesian coordinates. Consequently,
\textbf{a} = - \omega^2 \textbf{r}. ie a =r.w**2
I can't remember how you calculate this one.