Sides of a triangle based on height?

Inx

So I'm trying to build something, and I need the sides to be triangles of a certain height - something like 30 inches, i'm willing to be flexible to an extent based on materials available - but I have no idea how long to make the individual sides. Is this a pythagoras thing? I was never very good at the maths.

Rend

Do you mean sides as in sides but not base = 30? Or sides as in sides and base = 30?

What exactly do you mean by height? Height of the triangle, length of each side, total perimeter of each of the three sides, etc?

DevoutlyApathetic

Another relevant question, this is a right triangle, right?

A right triangle has two sides that form a perfect L shape between them.

This is quite likely a Pythagoras thing. If the sides are are multiples of 3, 4 and 5 it will be a right triangle. There are other simple sets of such numbers as well.

Thanatos

You need certain information to get the measurements of a triangle. You need the length of at least one side, and either: 1) the length of the other two sides 2) the angle of one of one of the corners and the length of another side or 3) the angle of two of the corners.

Are you looking to do, like, equilateral triangles? A drawing of what, exactly, you're trying to do would be helpful.

DevoutlyApathetic

Okay, go here and look at the first picture.

If 'A' is the height you want and that looks right then take A and divide it by 5. This gives you a number, your ratio.

Now side 'B' is going to be 4 multiplied by your ratio found right before now.

Side 'C' is going to be 3 multiplied by your ratio found two sentences up.

That is probably the minimum amount of math I can get this down to.

Djeet

be aware, Pythagoras' theorem is only really useful* if it's a right triangle, one where one angle is 90 degree (perpendicular). Is your triangle perpendicular? Maybe send us a quick paint image to visualize your triangle and what you're measuring as "height" (could be a length of one leg of triangle, or could be a line from the vertex of one angle that intersects the 3rd leg normal (perpendicularly)).

*There are caveats. you can split up any triangle into 2 right triangles and use P's theorem, but you'd need more info on lengths or angle values.

Inx

As far as what I'm making, I'm trying to do like a waist-high incline pushup/pullup bar in order to do some physical therapy exercises in the home. Imagine two triangles on each side with a pipe running parallel and linking the vertices. Does that make sense? I'm having trouble doing it up in paint for some reason.

I imagine I could pull it off with a right triangle if the right angle was the vertex?

L Ron Howard

Show us with a top, side, and front view diagram in paint.

Shutdown

If it's not a right-angle triangle (and it sounds like it isn't) you need to figure out the height from the point to the base - that gives you a right-angled triangle. With that height and the opposite angle you; trig can figure out the sides (*). Do the same for the other side of the triangle.

(*) been a while, but it's SINE/COSINE/TANGENT stuff. I can't remember how that part works but I can remember that having just an angle and one side you can figure out everything on a triangle.

Triangles really are awesome.

Thanatos

Inx wrote: »

As far as what I'm making, I'm trying to do like a waist-high incline pushup/pullup bar in order to do some physical therapy exercises in the home. Imagine two triangles on each side with a pipe running parallel and linking the vertices. Does that make sense? I'm having trouble doing it up in paint for some reason.

I imagine I could pull it off with a right triangle if the right angle was the vertex?

Do you mean one triangle on each side (i.e. two triangles total), or, like, two triangles on each side, one on the ground and one mounted to a wall or something?

Inx

Shutdown wrote: »

If it's not a right-angle triangle (and it sounds like it isn't) you need to figure out the height from the point to the base - that gives you a right-angled triangle. With that height and the opposite angle you; trig can figure out the sides (*). Do the same for the other side of the triangle.

(*) been a while, but it's SINE/COSINE/TANGENT stuff. I can't remember how that part works but I can remember that having just an angle and one side you can figure out everything on a triangle.

Triangles really are awesome.

The ONLY thing I have is the height from point to base.

Thanatos wrote: »

[img][/img]
Do you mean one triangle on each side (i.e. two triangles total), or, like, two triangles on each side, one on the ground and one mounted to a wall or something?

One triangle on each side.

Here's my shitty attempt at mspaint

Derrick

A^2+B^2=C^2 is the equation for a right triangle.

http://en.wikipedia.org/wiki/Pythagorean_theorem

C is the hypotenuse, or the diagonal side.

So I'm imagining in this scenario that you've got a short side against the wall, a longer side from the wall to the bar, right?

So, you've got to determine how far from the wall you want your bar. Say you want it to be 3 feet out so you can get behind it.

So your B side is 3 feet, or 36 inches. Your C side you said you want to be 30 inches (the side you mount to the wall).

Your C side would be 46.86 inches.

A^2+B^2=C^2

30*30 + 36*36 = C*C

1296 + 900 =C*C

2196 = C*C

46.86=C



This is what I'm thinking ^

Slo

So, if I was building something like this, I'd go about it slightly differently.

I'd take 6 lengths of pipe, or square tube steel, take two of them, make an X, then put another piece along the bottom. Use bolts, washers, and nuts to fasten them together.

Do this twice, and throw a piece of pipe between them, and bam, you have yourself an inclined pullup/pushup bar.

You could do the same with some 2x4s if you're uncomfortable (or don't have the tool to) working with metal.

As for the numbers required, the easiest, cleanest way I always use to cheat, is use multiples of 3,4, and 5.

3, for your bottom, 4 for your Height, and 5 for your slope.
works with any multiplication.

So if you want 30" high, I'd cheat and do this instead
(3,4,5)x 8 = (21, 28, 40) now take these two Right triangles, and put them back to back.

So you have a 42" base, two 40" slants that aim towards each other, and bam, you got yourself a triangular prism.

uh, hope that helped?

Thanatos

So, here's the image:



Correct me if I'm wrong, but it looks like you're planning on floor-mounting this, and not wall-mounting it, right?

Blake T

If it is floor base and its an equilateral triangle the sides lengths are

30/sin(60)

(Sorry I'm on my phone and can't do the numbers)

This will make each side an equal length.

Is this what you are after?

This is based off thantos' drawing though, if that isn't correct, you should really just make your own sketch as its hard to exactly understand what you are trying to say.

Thanatos

Blake T wrote: »

If it is floor base and its an equilateral triangle the sides lengths are

30/sin(60)

(Sorry I'm on my phone and can't do the numbers)

This will make each side an equal length.

Is this what you are after?

This is based off thantos' drawing though, if that isn't correct, you should really just make your own sketch as its hard to exactly understand what you are trying to say.

That is his drawing, I just rehosted it.

@Inx Maybe it would be better for you to tell us what you want the final product to be, and what you have to work with. I'm thinking if all you need is a waist-height weight-bearing bar that you can move around, there may be better ways to do it.

Inx

Thanatos wrote: »

a waist-height weight-bearing bar that you can move around, there may be better ways to do it.

That's essentially the end goal. And yeah, it won't be wall mounted. I've considered a few other ideas but on the shoestring budget, materials, and knowhow I have at my disposal this was the simplest I could come up with. I'm open to suggestions, though.