Trying to work Two Triangles together to get a desired number

irsmun

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I have been out of trig for...hmm more years than I'd want to say. 20+

I am trying to build a dump trailer and I need to figure placement of a cylinder for a desired dump angle. There are a lot of unknowns and I was hoping to use my good friend trigonometry to fill in these unknowns. Apparently we are not as friendly as I had hoped.

Here is what I know. The hydraulic cylinder when closed is 46 1/4 inches. It must be at a 15 degree angle in order to start. Since this made a simple right triangle, I was able to find the measurement of how many inches down the fixed point must be to be at the 15 degree starting angle. I needed to solve for side A. My answer is 12 inches.

angle-1.jpg
But notice that this puts the cylinder fixed point (which makes up part of the triangle I cannot solve for) down 12 inches below. Now here is where I start to lose my brain. I need the dump angle to be a desired angle between 50 and 60 degrees. I want the most angle I can get from this cylinder. Here is what I know. I know the cylinder is at a fixed point 12 inches below the bottom leg of the top triangle. That cylinder fully extended will be 82 1/4 inches. My desired angle for the sake of the example is 55 degrees.

Answers I need: What is horizontal distance from the fixed point to the pivot point (LINE B in the pic) of the new top triangle created by the dump bed at 55 degrees. What is the distance from the pivot point to the connection point of the cylinder? (LINE A in the pic) Really I'd like to also know the maximum angle I can achieve given the placement of the cylinder and its length and the distance from fulcrum point to achieve it.

I think a picture is in order if I am to have any hope of anyone knowing what the heck I just said. :shock:

trailer-angle.jpg

Do I have enough information to solve? or am I going to have to cut a stick and manually figure this out? :D

...and yes I know, its not a scale drawing. What can I say, I'm short on time.
 
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Yes, you are correct. It was drawn showing the line to look like the hydraulic cylinder. I was concerned no one was going to understand what I was talking about if I just drew two triangles...so I gave some artistic perspective. :D

82.25 inches is the length of the full line extended (UP, which is the red box + the red line) and 46.25 is the length retracted (DOWN, just the red box because the red line is now retracted inside the red box). The red boxes show the cylinder in the two positions. And yes, it is 12 inches (if my first math problem was correct) below the horizontal. That problem had to be figured first in order to find the fixed point so that it could start at a 15 degree angle. In the picture I show the 15 degree angle behind the cylinder (red box) in a reverse triangle for clarity. The cylinder is also at 15 degrees in the DOWN position.
 
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Yes, you are correct in all assumptions. I do need to calculate CD. When that's done BC will be able to be calculated easily. So are you saying I don't have enough information and am just going to have to plug in guesses for the lengths? Forgive me if you gave me the answer and I didn't see it. :confused: Honestly I have trouble sleeping at night thinking about this. I am good at process of elimination stuff, but this would be easier solved with an equation. Even though angle BCE will vary it is the max angle and can therefore be considered permanent in the up position.
 
That would be great. I know there has to be an equation out there. I sure need it or I will be using the stick method mentioned earlier.:(

That angle cannot be 90 because it needs to be 55 in order to get the length right.

Now it might be possible to do something like this:


................E




A .............B........................ C.............. D
..................(12) .....................................(12)
................F.......................................... G

Add more points and make a right triangle out of EFG with F also being 12 inches lower than line AD. Now we know that EFG is 90 degrees, line EG is 82.25....ahhh nevermind. We will still end up with only two pieces of information. We need an equation that works for a triangle that has a graduated angle increase so we can know for each unit of extension the angle changes X degrees...or something like that. Surely this is not impossible.

I don't know how you made your triangle out of letters. I had to use the periods to keep the spacing.:?
 
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So what do I do? I have never met a problem like this that math could not help with. I need to start welding but without knowing where to put it I cannot start. :(

To answer your question, E is the pivot point and A would be where the cylinder connects to the dump box. But if I knew were to put my cylinder 12 inches below the plane, where it would connect to the box is easy to find. It would be 46.25 inches from point that is 12 inches down in the down position.

I know how I can do this but its manual. I can cut a length of lumber 82.25 inches. Then attach another length of lumber at the rear pivot point set at 55 degrees. All I would need to do then is find the point where my 82.25 inch piece is 12 inches below the plane on one end AND is touching the 55 degree piece on the other end--mark and measure. It can only be in one position where both of those things will be true. Because their is only one position that this can actually all be true to form the correct triangle....why can't math solve this? Maybe trig is the wrong tool.
 
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Ok. I am getting a little confused on which point we are talking about since we have a couple of different diagrams. I drew a fresh one.

angle-3.jpg

Point A is the pivot point...permanent. Point C is the connected end of the cylinder to the frame...permanent. Points D and B are the same point of connection to the other end of the cylinder. One (point D) is just showing it in its retracted position of 46.25 inches and the other (point B) shows it in its fully extended position of 82.25 inches. Point D becomes point B when the cylinder is extended to its maximum length. Point D and B are the same points fixed to the extending end of the cylinder. As such that means lines CD and CB are the same lines in two different positions.

Maybe it would have been better to say that D and B are actually just the same point (B), in two different positions.

I guess it would be an isosceles. Since AB and AD will always be the same, but we still don't know their lengths, which is the question. The fact that we are dealing with a second triangle below the first and that triangle plays into the measurements of the second is what is throwing me. The goal is to get a desired degree angle out of BAE to calculate the placement of C given that CB is a known length. I think that would mean all other angles and lines would need to accomodate that end. I can't wait to get to the end of this. I will make an excel spreadsheet for it. :D

I need to know how far back from point A to weld point C so that when its fully extended to point B angle BAE is 55 degrees (or whatever I want to go for) Here is a picture of one partially raised for a visual. FYI It is turned the opposite direction from our diagram.

trailer-angle2.jpg

There should be only one way all these are true:

When BAE is 55 degrees CB is 82.25
given that A is a fixed point
C is a fixed point 12 inches down from E
CD (when down) is 46.25
AB and AD will be parallel when down and D and B will be the same point
EDC will be 15 degrees (down position) when D and B are the same point

No matter what I keep trying or how I flip it around, I always seem to be one piece of information short to solve.:(
 
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Nobody??? Come on. Somebody's brain can do this...I wish mine could. :D
 
Wow, ok. Now I have my hand raised. I actually tried this same route in my scribbling. Step 2 I am not comprehending. I see what you are doing, I just don't understand why you are doing it. You doubled u added w and 180 divided by 2 and got 132.5. I used a different route to arrive at this and got the same number.

I used the isosceles property of triangle ABD (that you generously pointed out) and figured angleABD and angleADB to be 62.5 degrees.
Since that is sitting on a straight line AE, I simple subtracted 62.5 from 180 to arrive at 117.5 degrees for angleBDC.
With that known I added 15 to it to account for the triangle underneath CDE and arrived at 132.5.

Can you explain in detail your method so I can see it? (the key was the isosceles all the time. I forgot all about that.)



STEP2: angles of triangle BCD
angleBDC = v = (2u + w + 180) / 2 ; 132.5
angleCBD = x = ASIN[eSIN(v) / a] ; 24.493
angleBCD = y = 180 - v - x ; 23.007
 

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