Microlighters

Hi Guys,

What is this essential piece of equipment used for?

Something that I find indispensable.

Before Trikenut shouts "Photo's pleese", here's the photo.

Regards
John ZU-sEXY

John Young wrote:Before Trikenut shouts "Photo's pleese", here's the photo.

Ya, ya..
I just like to see photos as it shows the thing visually!!

I've got a problem with 3-D could you be so kind as to show it at different angles please sir!

All I know is that what ever you do with it, there must be a way better piece of equipment for the job. Nothing flying related I would allso guess I need more clues!

Bad Nav wrote:I've got a problem with 3-D could you be so kind as to show it at different angles please sir!

Hi,

Surely - just for you ...

Regards
John ZY-sEXY

extra300s wrote:All I know is that what ever you do with it, there must be a way better piece of equipment for the job. Nothing flying related I would allso guess I need more clues!

Hi,

Definitely very flying related.

Regards
John ZU-sEXY

It looks like a chock - the same thing I paid R40 for 3 at GAME for 3 yellow blocks - no flipping over, no rolling on the tarmac (have not flown on grass for a while!)

Seriously - WTF?

On second thought - them bars look way too thin to hold back a 912S, it is a "wife satisfaction flimsy device"... Look - I park my aerie in this and if it bends I am safe to fly, if it does not bend I stay on the ground!

Not all of us fly GT450's - it is your "lift up my babe's nose and walk her to the hangar safely" device

Hi,

Nah, not a chock, nor a wife satisfier, nor a lifting the baby's nose device ....

Quack-Mate if he comes and quacks around here will probably work it out.

Regards
John ZU-sEXY

Tent Frame for the overnighters boet - so it is

Is dit dalk n rytoom vir n gans, of dalk n stellasie om vrugte op droog te maak??

John Young wrote:Hi,

Nah, not a chock, nor a wife satisfier, nor a lifting the baby's nose device ....

Quack-Mate if he comes and quacks around here will probably work it out.

Regards
John ZU-sEXY

Thank you for placing so much trust in my ability to analytically dissect and compute this complex phenomenon.
Herewith the answer to your riddle:


S=\frac{1}{2}bh

where S is area, b is the length of the base of the triangle, and h is the height or altitude of the triangle. The term 'base' denotes any side, and 'height' denotes the length of a perpendicular from the point opposite the side onto the side itself.

Although simple, this formula is only useful if the height can be readily found. For example, the surveyor of a triangular field measures the length of each side, and can find the area from his results without having to construct a 'height'. Various methods may be used in practice, depending on what is known about the triangle. The following is a selection of frequently used formulae for the area of a triangle.[4]

Using vectors

The area of a parallelogram can be calculated using vectors. Let vectors AB and AC point respectively from A to B and from A to C. The area of parallelogram ABDC is then |{AB}\times{AC}|, which is the magnitude of the cross product of vectors AB and AC. |{AB}\times{AC}| is equal to |{h}\times{AC}|, where h represents the altitude h as a vector.

The area of triangle ABC is half of this, or S = \frac{1}{2}|{AB}\times{AC}|.

The area of triangle ABC can also be expressed in terms of dot products as follows:

\frac{1}{2} \sqrt{(\mathbf{AB} \cdot \mathbf{AB})(\mathbf{AC} \cdot \mathbf{AC}) -(\mathbf{AB} \cdot \mathbf{AC})^2} =\frac{1}{2} \sqrt{ |\mathbf{AB}|^2 |\mathbf{AC}|^2 -(\mathbf{AB} \cdot \mathbf{AC})^2} \, .

Applying trigonometry to find the altitude h.
Applying trigonometry to find the altitude h.

Using trigonometry

The height of a triangle can be found through an application of trigonometry. Using the labelling as in the image on the left, the altitude is h = a sin γ. Substituting this in the formula S = ½bh derived above, the area of the triangle can be expressed as:

S = \frac{1}{2}ab\sin \gamma = \frac{1}{2}bc\sin \alpha = \frac{1}{2}ca\sin \beta.

Furthermore, since sin α = sin (π - α) = sin (β + γ), and similarly for the other two angles:

S = \frac{1}{2}ab\sin (\alpha+\beta) = \frac{1}{2}bc\sin (\beta+\gamma) = \frac{1}{2}ca\sin (\gamma+\alpha).

[edit] Using coordinates

If vertex A is located at the origin (0, 0) of a Cartesian coordinate system and the coordinates of the other two vertices are given by B = (xB, yB) and C = (xC, yC), then the area S can be computed as ½ times the absolute value of the determinant

S=\frac{1}{2}\left|\det\begin{pmatrix}x_B & x_C \\ y_B & y_C \end{pmatrix}\right| = \frac{1}{2}|x_B y_C - x_C y_B|.

For three general vertices, the equation is:

S=\frac{1}{2} \left| \det\begin{pmatrix}x_A & x_B & x_C \\ y_A & y_B & y_C \\ 1 & 1 & 1\end{pmatrix} \right| = \frac{1}{2} \big| x_A y_C - x_A y_B + x_B y_A - x_B y_C + x_C y_B - x_C y_A \big|
S= \frac{1}{2} \big| (x_C - x_A) (y_B - y_A) - (x_B - x_A) (y_C - y_A) \big|.

In three dimensions, the area of a general triangle {A = (xA, yA, zA), B = (xB, yB, zB) and C = (xC, yC, zC)} is the Pythagorean sum of the areas of the respective projections on the three principal planes (i.e. x = 0, y = 0 and z = 0):

S=\frac{1}{2} \sqrt{ \left( \det\begin{pmatrix} x_A & x_B & x_C \\ y_A & y_B & y_C \\ 1 & 1 & 1 \end{pmatrix} \right)^2 + \left( \det\begin{pmatrix} y_A & y_B & y_C \\ z_A & z_B & z_C \\ 1 & 1 & 1 \end{pmatrix} \right)^2 + \left( \det\begin{pmatrix} z_A & z_B & z_C \\ x_A & x_B & x_C \\ 1 & 1 & 1 \end{pmatrix} \right)^2 }.

Using Heron's formula

The shape of the triangle is determined by the lengths of the sides alone. Therefore the area S also can be derived from the lengths of the sides. By Heron's formula:

S = \sqrt{s(s-a)(s-b)(s-c)}

where s = ½ (a + b + c) is the semiperimeter, or half of the triangle's perimeter.

Three equivalent ways of writing Heron's formula are

S = \frac{1}{4} \sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}

S = \frac{1}{4} \sqrt{2(a^2b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}

S = \frac{1}{4} \sqrt{(a+b-c) (a-b+c) (-a+b+c) (a+b+c)}.

Any further questions?

I am just gonna "where the fox that!"

Might see it in action saturday morning and then I will spill the caboodle... JY obviously adores that frame... Will report back after Saturday!

Hi Guys,

Thanks for the "laffs".

I bought a Mr. Funnel. The only problem is that all fuel filler necks on trikes are at an angle. One can't use the funnel unless you have a Buddie to hold it for you.

So sEXY Grandpa John made the frame with the front legs set back to accommodate the trike. The frame is very light and easy to handle.

Now I can pour 25 litres of clean filtered fuel into the trike per minute "solo". It works very well.

Thanks again for the "laffs" - photo's below.

Regards
John ZU-sEXY