hey Honus! I've never made a bucket before nor bought one and decided that, after confronting my budget, your cardboard bucket must be the way to go this halloween... so i started my own project =P
i've used only left over pizza boxes and tape that had been scattered about my room... and, so far, it doesn't look so bad...
my question for you is how did you calculate and design the template for the triangles that make up your dome? (sorry, geometry was never my forte)
and, what are you going to do about the visor? (i'm just a poor college student with only 23 cents to spend =P)
Last edited by Germ; Oct 17, 2006 at 10:43 AM.
OMG that is awesome! What next, blaster out of popsicle sticks and bailing twine?
looks awesome! one suggestion... don't make the visor out of cardboard. WAY too hard to see out of
Did you generate lower portion off a template I don't know about - or create that freestlye?
However you did it --- it looks PHENOMINAL!
Thanks for the kind words everyone! The templates I used were the ones posted by Alan here:
http://www.thedentedhelmet.com/showthread.php?t=10497
Alan is THE MAN! I never would have been able to do this without all his hard work.
I didn't make any exact templates for the dome- I just kind of winged it. I put up a kind of abbreviated tutorial here:
http://www.instructables.com/id/EB96DTBQ13ES9J6YYX/
I'm making the visor out of some flexible clear plastic. I'm going to try and see if I can scrounge some window tint for it.
Jerome
wow! that´s is the most inventive thing I have see in time....this, and the thrash can armour are the living proof that a BOBA can be made in a lot of artistic ways................YOU RULE, MAN!!
I have a litle surprise too...when I have finish, this forum will we in shock!!!
thanks!
Nice work. Shows what a little work can do!
Once you seal it up, I think it would stand up to being RTV molded.
Very impressive work!
I think you're well on your way to proving that.
Thanks for the compliment. You have done an outstanding job, I just provided the tools, great work.
I created a set of dome templates similar to yours but I did have the patients to assemble them as well as you have. Give me a couple of days and I'll clean them up and post them for everyone in a PDF file.
Alan
After an attempt of two at winging the dome construction like you said in your tutorial i decided to use a littl of my math skills to figure out the precise measurment for the triangles...
basically, since I wanted 10 triangles per quadrant, I had to solve for the perimeter of the base, calculate each arc, then the vector points for each angle of the ellipse... After which I then had to repeat the process vertically... something i had never learned in any trig class.. =(
after a few checks and rechecks i found that it worked phenomenonally! Each triangle matched up to form a perfect dome... I made the template in AutoCad LT and printed them off by quadrant. Unfortunately I cannot figure how to make the template into a .pdf file...
Thanks sooo much Honus! Cardboard helmets are perfect for any individual that wants to make a cheap, fast, and easy bucket for almost any occasion. With help from your tutorial i was able to build my bucket in a matter of 2 days (not painted of course =P).
I thought the dome was more of a hemisphere than an ellipse. Anyway, I'm interested to hear where you get the measurements for the circumference and for the dome radius (assuming you can treat it like a hemisphere.)
You used 10 triangles per quadrant???? I haven't tried this method, but I've gotta believe the vertex angle per triangle must be really small. You probably have the exact number, but how were you able to cut the material as precise as it seems this measurement would call for?
Did you lay out the boundaries of each quadrant like Honus did in post #8? If so, then wouldn't you need to use right triangles for the area right next to the quadrant boundaries? It seems if you don't you'll have a gap there. I'd like to see how your helmet turned out.
I would like to see where the quads join...I agree w/ jimmy buffet right angles seem most logical
ok, i'm a little annoyed because was logged out while typing this post
and lost about a page and half of work so here i go with a more condensed
version: -_-
...a nice short explanation =)
(right now i'm home for fall break and all the work I did to solve these
problems are quickly scribbled on many sheets of printing paper at my
dorm so, in turn, every thing posted here is from memory and may or may
not be completely accurate)
Things to know before I get into this are: I used wizardofflight's
templates, the base of the the dome is an ellipse with a major axis of 9
inches and a minor axis of 8.5 inches, the max height is something around
3.75 and the template for the dome height is not exactly an ellipse.
To find the small variation in distance that is caused by the shape of
the dome height template I first had to recognize that the template is,
in fact, made up of two ellipses connected tangently. To solve for how
much this effected the final distance of each triangle. This means, that
I had to solve for the total circumfrence of each ellipse.
The calculations i used to solve this involved finding at what point did
the tangent line connect. To solve this I used:
s = sqrt[4h2+c2/4]+[c2/(8h)]
ln[(2h+sqrt[4h2+c2/4])/(c/2)]
where h = height, c = chord length, and s = length...
Next, was the more complicated part... I had figure the length of each
vector that comprised each individual triangle unit. (This would have
been made easy had the dome base been a circle... because then every
vector would be just the radius rather than some horribly confusing
variable) To solve this part I used what I was given to create a formula
that would work for all the quadrants (At this point I'm looking at a
sketched out version of the dome base on some graph paper)...
I know that every triangle (since I want ten) will differ by only 9 degrees.
I know that the ellipse is made up of two axis whose radii are 4.5 and 4.25
With this in mind I created this:
(understand i went about finding this using only general quadratic and
parametric equations and that a = axis along the y, b = axis along the x,
and x and y are what they are normally in refrence to graphing, im going
to say that v = theata or your given angle)
tan^-1(b*tan(v)/a) = l // remaining variable (y = tan(v)*x)
then the point common to the ellipse is found by:
a*sin(v) = x // this only works for quad 1 due to 0 < v < 2pi
b*cos(v) = y
(I inputed this forumla into my TI-83 as a prog and ran it for every
angle in the quadrant I and then (cheated) mirrored it.)
Now, the dome base is done...
... im tired of typing for now (sorry) ... but i promise i'll post how i
found the remaining arc length of the dome and how that was implimented
into the overall production of the triangles..
below is a screenshot of the 6-triangle version of the product: (in AutoCad):
Well folks I have updated the helmet template thread to include the flat pattern for the dome. Here is the link;
http://www.thedentedhelmet.com/showt...431#post141431
The dome is not a true ellispe as stated above. So I created a 3D mesh of the dome and then flattened it out. It is divided in to 3 pages with a total of 20 panels. Have fun!!
Alan
Thanks for specifying where you got the initial values from. OK, so it's not a hemispehere, and the vector from the center to a given point on the dome is going to vary in magnitude. I'm pretty impressed that you came up with the formulas to find the circumerference of each ellipse.
I think I was under the impression that you were going to be using isosceles triangles. This is evidently not the case since the dome base is not circular. So by keeping the vertex angle at 9°, you need to know the length of each leg of the triangle (well, you need 2 lengths since you have an angle.)
So basically you need arc length as a function of φ where φ increments by 9° (quadrant 1 would be for 0° to 90°, 2 would be 90° to 180°, etc.) and φ is the angle on the XY-plane. It seems you have vector magnitude as a function of θ. By using the law of cosines, you can find arc length. You can either write a short computer program for summing or use a calculator to integrate (I personally would go with the computer program.
Anyway, very impressive work! I'm interested to see your approach for finding arc length.
wow, simply amazing. i might just have to try this myself and see if i can get it done in time for the 31st. Thanks for the inspiration.
when you start on the gauntlets let us know!
yeah do a tutorial honus some of cant afford gauntlets
Just a thought, but could the templates be printed out on heavy cardstock and then just cut it out and use those pieces for the actual build up? Been trying to think it through a little bit before I start. That would seem the easiest thing to do by saving a step.