[R5P7Duster]

I'm trying to figure out how to figure out what dimension the triangles need to be for my project. The base of them is 5" they are 7" long and the are meeting up with a 3" OD tube. Thanks for any help

[R5P7Duster]

Also I'm not very concerned with the fitment of the 3" tube as it will be welded and small gaps can be filled.

[Kiwi]

Is this what you need?

[holbieone]

give this a try

[neilw20]

How do you guys measure those angle so accurately?:violin:
With all that accuracy, how are the two presentations differing by quite a bit in one of the dimensions.

[Kiwi]

... how are the two presentations differing by quite a bit in one of the dimensions.

The specs don't list the number of 'triangles'. Looking at the picture, I believe there are three around 90 deg. (eight around 360deg.)

[holbieone]

How do you guys measure those angle so accurately?:violin:
With all that accuracy, how are the two presentations differing by quite a bit in one of the dimensions.

i drew a triangle based on 5.00" on the base and 7.00" long plus the 1.5" radius of the tube

that gets you the middle section ,but more info. is needed for both end sections

[Kiwi]

....i drew a triangle based on 5.00" on the base and 7.00" long plus the 1.5" radius of the tube....

What does the 0.8824" relate to?

[R5P7Duster]

Thanks for the posts. I was hoping to get an equation. When I try to mate the edges in solidworks they won't mate cause the angles are off by 1-2*.

[Perfect Circle]

The inverses of the trigonometric ratios are denoted in the three different ways, which are shown below.

x = sin y

y = sin-1 x

y = arcsin x

Arcsin x is a number whose sine is x.

Some rotations do not have values for the inverse trig. functions. 100o is an example that does not have an arcsin. The ranges of the inverse functions are listed below.

y = Arcsin x [-((PI)/2), (PI)/2]

y = Arccos x [0, (PI)]

y = Arctan x [-((PI)/2), (PI)/2]

Example:


1. Problem: Find Arcsin ((SQRT(2))/2).

Solution: Using a unit circle like the
one pictured below, you
can see that there is only one
angle in the Arcsin's
range that has a sine of
(SQRT(2))/2. That
angle is (PI)/4.

[Perfect Circle]

Angles are also called rotations because they can be formed by rotating a ray around the origin on the coordinate plane. The initial side is the x-axis and the ray that has been rotated to form an angle is the terminal side.


Reference angles are useful when dealing with rotations that end in the second, third, or fourth quadrants. A reference angle for a rotation is the acute angle formed by the terminal side and the x-axis. Example:


1. Problem: Find the reference angle
for theta


Solution: To find the measure of the
acute angle formed by the
terminal side and the x-axis
subtract the measure of theta from
180o.

180 - 115 = 65

The reference angle is 65o.
Once you have found the reference angle, use it to determine the trig. function values. Consider, for example, an angle of 150o. The terminal side makes a 30o angle with the x-axis, since 180 - 150 = 30. As the figure below shows, triangle ONR is congruent to triangle ON'R'; therefore, the ratios of the sides of the two triangles are the same, although the ratios may have different signs. (You could determine the function values directly from triangle ONR, but that is not necessary if you remember that the sine is positive and the cosine and tangent are negative in quadrant II.)

http://www.mathwarehouse.com/transfo...ns-in-math.php
:cheers:

[Kiwi]

Perfect Circle's formula is over my head.

This is how I would work out the angles:
Deg. the 'triangle' section on the 3'' tube = 45deg.
Tan(45/2) * (3/2) = 0.62132 ________________(3 = tube dia.)
(5/2) - 0.62132 = 1.87868 ________________(5 = 'triangle' side)

ATan(1.87868/7) = 15.023 deg + 90 = 105.023deg ____________(7 = 'triangle' length)
180deg - 105.023 = 74.977 deg.

Answer = 105.023 and 74.977 deg. when the 'triangle' end length is 45deg of the 3" tube.

[handlewanker]

Hi, I'd do this the simple way, if'n I was going to actually make it.

Draw a horizontal line across the top of your paper.

Draw a centre line down, about 8" long.

Draw two lines, 7" or 8" down from the top horizontal line, and 1.5" either side of the centre line.

Put your compass point in the top horizontal line centre point and cut two arcs 2.5" apart, (5" wide total) either side of the centre line.

Put your compass point in the end of the 2.5" arcs and cut two more arcs, 7" long across the 1.5" line at the bottom.

Draw a horizontal line through the point where the arcs cross the 1.5" lines.

You now have exactly one "triangular" side of the 8 sided figure.

Now draw two horizontal parallel lines the width of the skirt and 40" long (5" X 8 = 40").

At 5" intervals draw your triange figure you first drew against this "skirt" and hanging down.

The next step is to scribe this on metal, bend the skirt around to form an 8 sided octagonal shape, with the "triangles" standing up free, weld the seam, and then bend the 8 "triangles" inwards to meet the 3" diam at the centre, tack weld the sides, and weld them along the seams to form the "funnel" shape.

If you're very crafty you would have divided the circumference of the 3" diametre by 8 and make this the dimension of the bottom end of the triangle, (instead of the 1.5" spacing as originally suggested), beating each end slightly curved to make a perfect round hole for welding the tube into when you bring the 8 triangular sides together.

BTW, tack weld all seams before final welding, and make a few angle templates out of sheet metal to check your progress.

Note, you can't make the angled sections and skirt out of one single piece of sheet and bend it round as they do with a round funnel, unless you make the bottom section in one piece first and the skirt as a second piece and join them togeter at final assembly.

You'll also find it's a very difficult job to bend a piece of metal with 8 side accurately in one piece and join by a single seam.

If you intend using thin sheet metal, (galvanised or zinc coated sheet metal), and are going to silver solder or braze the seams take care the zinc fumes are toxic.

Plain steel sheet or alluminium is best Tig welded or gas welded and Mig needs expertise.
Ian.

[Kiwi]

......Draw a horizontal line across the top of your paper.
Draw a centre line down, about 8" long.
Draw two lines, 7" or 8" down from the top horizontal line, and 1.5" either side of the centre line.
Put your compass point in the top horizontal line centre point and cut two arcs 2.5" apart, (5" wide total) either side of the centre line.
Put your compass point in the end of the 2.5" arcs and cut two more arcs, 7" long across the 1.5" line at the bottom.
Draw a horizontal line through the point where the arcs cross the 1.5" lines.
You now have exactly one "triangular" side of the 8 sided figure.......

With this "triangular side" you will have a gap around the 3" tube of over 1".

[Pro-Fab]

I'm trying to figure out how to figure out what dimension the triangles need to be for my project. The base of them is 5" they are 7" long and the are meeting up with a 3" OD tube. Thanks for any help

How many degrees are you wrapping around the tube? The dimensions that you give here will not work. If you are trying to make 90 degrees using three pieces, then they must be 30 degrees each. Is the 7" the length of the seam, or from the tube to the center of the triangle base? Either way, lay out your 3" tube, and measure from the center point. With 30 degree segments, the gaps where the seams meet the tube will be less than1/16". The drawings that have been offered will work but you will have to trim back your last piece so that you can finnish at the desired angle, whatever that may be.
Decide on the segment angle and the most important length only and you will be able to keep everything concentric.
A CAD program will obviously give you the most accurate measurements than laying out by hand, but you won't be able to duplicate them by cutting manually anyway.

[Dougt]

you might want to try egineering power tools a free download lots of info

[handlewanker]

OOPS Kiwi, you're right, the dimension of the two 1.5" lines apart should have been the circumference of the 3" tube divided by 8, which is 3.142 X 3 = 9.426 div by 8 = 1.17824" or 1.2" rounding up, which is the dimension of one of the triangle bottom sides, but this dimension is for before you weld the seams and after you beat each bottom part to the profile of the 3" tube diameter, so that you don't get the gaps at the corners that have to be filled in.

If the bottom dimension isn't required to be slightly curved to mate with the tube, then you'll have to draw a another circle, 3" diam, and draw an octagon round the periphery to obtain empiracally the dimension of the bottom edge of the triangle which will be slightly longer than the 1.2".

Then the two 7" arcs cut the lower horizontal at the two points that are 1.2" apart, not the 1.5" I stated previously.....didn't think about the circumference div by 8 at the time.

Unless you want to do it for just the drawing sake, the laying out method is when you want to actually go and cut metal.

In actual workshop conditions, the easiest way to do the job is to cut the skirt seperately in one length of metal and the 8 triangles as 8 seperate pieces of metal which is just straight guillotine work, as opposed to cutting out a series of triangular slots from one single piece of metal, which can only be done with a bench shear, or if in thin metal a hand shears (and cut your hands to ribbons) and a good eye for shearing along a line.

Then the lot is assembled and tacked progressively checking for symetry as you go.

I am curious to know why the design calls for an 8 sided figure.
Ian.

[Kiwi]

......draw a another circle, 3" diam, and draw an octagon round the periphery to obtain empiracally the dimension of the bottom edge of the triangle which will be slightly longer than the 1.2"......

Did you see post #6

......the two points that are 1.2" apart, not the 1.5" I stated previously.....

I'm confused, I interpreted from your instruction that the distant between the two points of the bottom edge as 3"???

......I am curious to know why the design calls for an 8 sided figure.

R5P7Duster has never stated how many sides or what angles. Looking at the picture it looks like 3 sides over 135deg.

[handlewanker]

Hi Kiwi, drawing #6 is right....I assumed it was a funnel shaped object with just a couple of sides shown hence the assumption of 8 sides, as it "looked" that way.

The bottom end of the triangle, just as you show in post #6, is derived from one side of the octagon, I must have been half asleep...zzzzzzzz!

If you use the dimension of the circumference of the 3" tube div by 8 to produce the length of the bottom side of the triangle, (which is 1.2"), you can curve the bottom end to fit round the tube, otherwise the length actual of the ocatgon will leave a small gap at the corner as stated.

Looking at the first post with the question of ...how, there is no indication of what the figure shape actually is, but by appearance is part of an 8 sided figure, with only 3 sides required.

The information given is that the top side is 5" across, and one side of the triangle is 7" long, this is the length along the actual side, not the height of the figure, so all we needed was the length of one side of an octagon (assumption 100%) round the 3" tube, which your post #6 gave it, and the triangle could just be marked out, no maths or angles required.

It would be interesting to know what the part was for.
Ian.

[handlewanker]

Dougt, thanks for the link to Engineering Power Tools, very usefull.
Ian.