Triangulation Help Please.. Pic enclosed

[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?
With all that accuracy, how are the two presentations differing by quite a bit in one of the dimensions.

[Kiwi]

Originally Posted by neilw20 ... 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]

Originally Posted by neilw20 How do you guys measure those angle so accurately? 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]

Originally Posted by holbieone ....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

[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.