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Kinda whacky! Chalet-Tear!

PostPosted: Sun Jun 20, 2004 10:15 am
by Shrug53
Here is a rendering Mike did:

Image

It is a tear with a chalet top. I have have already bounced some ideas off of him about it but before I share my building opinions, I would love to hear what everyone else has to say. How would you build this and actually make it work?
There are a couple of requirements:

1. the entire folding section has to fit on to the flat part of the roof of a 4x8 tear (something like the benroy)

2. It must be completely wheatherproof in both the open and closed positions

3. It should stick out as little as possible when closed so as to maintain the nice tear shape

Lets hear your ideas! Then I will post my observations and maybe we can make this a real workable plan. Then Mike just needs to pick out a name for the new hybrid.

PostPosted: Sun Jun 20, 2004 11:18 am
by Shrug53
OK you math whizzes, here is a sketch based on some really basic dimensions:

Image

If you actually designed a tear for this, you could give it more flat roof, say four to five feet. But the height should still be two feet. C'mon math gang, if the base were four feet, how long would those panels need to be? How about with a five foot base?

PostPosted: Tue Jun 22, 2004 1:22 pm
by jimqpublic
Shrug53 wrote:... But the height should still be two feet. C'mon math gang, if the base were four feet, how long would those panels need to be?


4' base, 2' high isoscoles triangle has a 90 degree angle at the peak and 45's at the hinges, just like the Chalet. The two roof panels would be 2.83 feet or 2 feet, 10 inches. (Side of the isoscoles triangele is equal to the square root of the sum base/2 squared plus the height squared)

How about with a five foot base?


If you keep the height at 2' it's a bit flatter triangle, but the formula is the same. Roof panels become 3.20' or about 3' 2-3/8"

PostPosted: Wed Jun 23, 2004 11:31 am
by tdhombre
Hum - I think the length of the sides of the 3 ft. base triangle indicated would be 2.5 ft. (2'6")

Put a perpendicular at the mid-point of the 3 ft. base and the resulting right triangle would be 1.5 x 2 on the legs. Patharogean Theorem is 1.5 squared (2.25) plus 2 squared (4) is 6.25. The square root of 6.25 is 2.5 - the length of the hypotenuse (the longest side).

If the base is 5 ft then 2.5 squared plus 2 squared is 10.25 whose square root is 3.2 (ft.) or 3 ft 2 1/2 in.


jimqpublic has provided the length if the base is 4 ft.


My practical side says I don't want to stand up in the middle of my bed, though. :roll:

PostPosted: Wed Jun 23, 2004 1:03 pm
by campadk
Dave Klinzman wrote:My practical side says I don't want to stand up in the middle of my bed, though. :roll:


Think back.. way back to when we was kids jumping on the bed.

Could also bring your Christmas tree come to think of it..

PostPosted: Mon Jun 28, 2004 3:29 pm
by Forrest Ebbs
What if you combined the aft panel with the galley hatch? The big lateral hinge across the middle would draw the galley hatch upward and it would lock into place. It could even be on tracks. This would leave your galley exposed, but a roll-down snap canvas shield could temporarily cover it.

Just a thought.