Real World Geometry Problem Kicking My Old Butt!

[jim_s]

Ok, I'll preface this by saying that I haven't had any geometry in about 40 years...

Given H, L and t, I'm trying to find the length or angle noted with '?'.

As background, I need to mount a small fan that is t thick and L long, at such an angle that the bottom surface of a plate that that the fan is mounted to meets a vertical surface at height H, and such that the lower corners of the fan rest against the vertical surface and the horizontal surface (this would be the minimum-sized configuration). The thickness and length of the fan are variable (I know the value of each in advance, but I need to be able to substitute in different fan sizes and get the correct unknown length or angle), and the H height is a constant.

To further clarify (or muddy) things, consider the intersection of the line representing the plate, and the vertical surface, to be hinged at point H. A longer, or thicker, fan, would cause the far/lower end of the plate/line to change position on the X axis (as the two lower points of the fan will always need to contact the vertical and horizontal surface), as well as cause the vertical position of the fan lower corner's intersection with the vertical face to change. As I'm worried about the bottom surface of the plate, I'm ignoring the plate's thickness - a line should adequately represent the under side.

My intuition is telling me that this is a system of equations with 2 (?) unknowns, based on some geometry that Pythagoras and the Law of Sines might have some bearing on, but my rusty mind is having trouble piecing it together coherently.

I'm as much interested in the approach to the problem (more so, probably...) than the specific answer!

 

[jim_s]

Hi, Denis - I'm still here, and anxious to try to work through this - I truly appreciate your help!!

I've re-drawn the diagram (at a scale of 2x, just in case you were to print it out, so that I have some space to write on it), with the vertices labeled as you suggested. I labeled some external angles, just so things didn't get too cramped. I also labeled as 'x' and 'y', the lengths of the smaller triangle, for my personal convenience (these are values I'm trying to solve for). Apologies if the chosen labels offend geometric sensibilities, as they're also the traditional axis labels... < So, 'x' = BF, and 'y' = EB.

I've done some more playing with this (ok, like, kind of obsessively... ;-) in the mean time, and finally gave in, and went to WolframAlpha, to plug in some equations and numbers, so that I could at least get moving with the project that this is for. In reality, I can operate off a set of pre-calculated values for a set of known fan sizes, but I'd really like to understand how to derive a generalized solution that could work with an arbitrary set of values. (That having been said, I'm beginning to despair that it may be way more complicated to do so than I'd ever have imagined or feared!)

In my real-world problem, I know the following:

- length DE: this is the thickness of a small box fan - it can vary, from case-to-case (I'm working with a value of 10, which is very common - could also be 20 or 25)
- length EF = DG: this is the length of the fan - it can also vary from case-to-case (I'm working with a value of 60 - these will almost always be either 40 or 60)
- length AB: this is a fixed height that the under surface of a fan mounting plate must end up at. (I'm working with a value of 45) Obviously, not *any* arbitrary choice of ED and EF will work with this fixed height, but I'm considering fans with thicknesses between 10 and 25mm, and lengths between 40 and 60mm, so these should all work in my case, I think.

As you note, there are at least 2 similar triangles - ABC and EBF. Because the sides of the rectangle/fan are parallel, the angle BEF = angle A.

So, the formulas that I'm passing into WolframAlpha are:

sin(A) = DE/(AB - y), cos(A) = y/EF, sin(A) = x/EF, tan(A) = x/y, x>10

I included the last bounding condition, so WolframAlpha wouldn't return values that didn't apply to my real-world case.

Using the known actual values (fan size of 10x60mm, fixed height AB of 45mm), these become:

sin(A) = 10/(45-y), cos(A) = y/60, sin(A) = x/60, tan(A) = x/y, x > 10

Wolfram then returns an x value of 50.0949.

It also returns:

A ~= 2*(pi*n + 0.493989)
y ~= 60*cos(2*(pi*n + .493989))

for n in the set of integers

I'd like to understand the 'n' thing, as I can't find an integer value of n that actually gybes the Wolfram-calculated output.

From this magically-determined value of x, I can then back-calculate the remaining needed lengths and angles.

A = sin-1(50.0949/60) = 56.607
y = 60*cos(A) = 33.0227
AC = 45/cos(A) = 81.76
BC = 45*tan(A) = 68.26

And, indeed:

(AB)^2 + (BC)^2 = (AC)^2

SQRT(45^2 + 68.26^2) = 81.76

So, it all works, through the magic of WolframAlpha, but I would really like to be able to reproduce that on my own, w/o having to rely on Stephen Wolframs black magic.

And. shazaam, I have 3D-printed fan shroud that prints flat, folds at thinned seams (underneath), and fits like a charm!

I'm not sure where you're coming up with the '18-24-30' designation - is that some type of standard triangle ratio? (I do remember 1-1-sqrt2 from the old days!

So, sorry for the book-length reply, but figured I'd get it all out on the table. I would love and welcome any thoughts, suggestions, commentary, etc that you might have. It'd be super cool to figure out how to reproduce Wolfram's magic value!

Thanks Much!!

 

[jim_s]

I appreciate the link, but that appears to be a different and simpler problem that the one I'm trying to solve. That discussion assumes that you know all of the lengths of the sides of one of the triangles, and then just find the sides of the similar triangle. As you, and the discussion note, this is a pretty trivial problem.

While my problem does involve similar triangles, the big difference is that I don't know all of the lengths of either of the triangles - In fact, I only know one of the side lengths of *any* of the triangles.

I know the height of the outer triangle ABC (AB).

I know the hypotenuse of the inner triangle EBF (EF).

I don't know the lengths of any of the other sides of those two triangles.

I know the height of the small triangle EDA (ED).

I know the length of the base of the small triangle CGF (GF).

The only angles I know are the right angles.

The physical idea here is that for the given/fixed height that the fan mounting plate (AC) will meet the fixed vertical surface of an enclosure (AB - with A being the top of the enclosure), there is one angle A at which the fan will sit flush with a line drawn from A down to the x axis at point C. Different fan sizes (length and thickness) will result in different values for angle A (and resulting different lengths for AC, EB, BF and FC). If you were to take a ruler, and anchor it via a pivot point at point A, then slide it around at different angles, thinking about where a fan of different length and/or thickness (ie, rectangle DEFG) would result in contact at point E with AB, and contact with point F on the X axis, as well as where a plate of unknown length would meet the x axis at point C (while still meeting the y axis at point A) that might help visualize things better. (If that still doesn't make sense, let me know, and I can send a few pictures of the little paper model that I built to try to work through this.)

I'm sorry if I didn't adequately explain this previously. I can see how it would appear to be a trivial thing to solve, but I think it ends up being far more complicated, and I'd love to figure out how to solve it via calculation, vs having to rely on Wolfram Alpha for a limited set of configuration values.

Thanks Again!

 

[jim_s]

Trying to reattach the diagram (it seemed to have somehow ended up inline, but not as an attachment?

 

[jim_s]

Ok, I can work with whatever labels you like! I've updated the diagram, to match your labeling.

Your summary of the points and angles is correct.

Plugging the following equations into WolframAlpha: (I have to use a single variable name like 'x' and 'y', as Wolfram treats something like BC as two variables)
sin(B) = 72/(300 - y)
cos(B) = y/350
sin(B) = x/350
tan(B) = x/y
x>72

results in the magical value of x = 280

You, too, can experience the magic by visiting: http://www.wolframalpha.com/input/?...)+=+y/350,+sin(B)+=+x/350,+tan(B)+=+x/y,+x>72

So, armed with x = 280 (CG on our diagram, but to keep Wolfram happy, I use x), we can deduce that that FC = sqrt(350^2 - 280^2) = 210. Pythagoras approves of the inner triangle according to Wolfram!

Knowing 2 sides of the inner triangle, I get the angle B (angle GFC = B) as sin-1(280/350) = 53.13 degrees. (Sorry, I now you said this can be done w/o angles, but trig is the one thing I remember from geometry, and I'm a hammer looking for a nail!

Now knowing that angle B is 53.13 degrees, and knowing that the adjacent side (BC) is 300, reapplying the trig hammer leads to AC = 300*tan(B) = 400.

And, as much as I appreciate your help, I now hate you... ;-) (But I swear I'll play real nice at least until you explain to me how you pulled that off!

So, I think its safe to say that you're on the right track! (And its only taken you a matter of minutes to resolve what I've spent hours on, then only being able to resolve by resorting to Wolfram's Black Magic!)

I'm now carefully re-reading the thread that you previously pointed me to (which I dismissed as trivial, and which has a diagram that looks suspiciously like the one you presented here - luckily, I'm well-accustomed to the taste of crow... ;o) I'm trying to determine how you so confidently (and correctly, it seems!) determined that these are all similar angles, thus allowing the simple act of just scaling up the known values.

This is very cool - THANK YOU!!

 

[jim_s]

Ok, its amazing what a whack over the head with a hammer can do to bring clarity to things...

I now see how you determined that triangles BEF, FCB and ADG are similar:

- we know for for triangle BEF, one angle is 90 and the other angle is B. I'll call the remaining angle V
- we know that for triangle FCG, angle GFC = B (corresponding angle), and its a right triangle, so the remaining angle, FGC, can only also be V
- Then, looking at triangle ADG, we have corresponding angles (angle A of triangle ADG and angle CGF). So, this makes the corners of angle FCG B, 90 and A. We'd previously called angle FGC V, but V = A, so we also now know that triangle BEF consists of angles A, 90 and B, too. And, since triangle ADG consists of angles A and 90, the remaining mystery angle can only be B. They're all similar!

Next stop, applying the similarity rules...

 

[jim_s]

Sorry, I haven't lost interest in this - got pulled away last night due to a sick kid, and none-too-shockingly, I'm now sick myself tonight. <

Geometry will have to wait another day, I'm afraid - brain isn't working too well tonight...

So, did you come up with your 280 figure on paper, originally, or did you rely on supernatural means, like I did?

I got excited, having found a 'side splitter' formula (basically, an application of similar triangles, but applied to partial sides in a way that I hadn't thought about), but haven't had a chance to try it out yet.

 

[hoosie]

Fan problem

 

[hoosie]

Fan problem - working steps

 

[hoosie]

Fan problem - examples

 

[hoosie]

Fan problem - additional graphs

 

[jim_s]

Wow, Hoosie - this is awesome - its going to take me some time this evening to puzzle my way through it, but I'll definitely do so. Thanks a bunch for your detailed work and explanations!!

 

[jim_s]

Denis - no, this is all good - your work and hoosie's. My crutch has been, and continues to be, that I need to use WolframAlpha to find x. Then once armed with that, I was able to work backward via the angles, to get the other needed values, but I'm dead in the water without someone/thing spoon-feeding me that initial x value. It looks like both you and hoosie have found a way to solve the problem w/o having to resort to outside resources (ie, a solver like WolframAlpha). I still need to work through both of your processes, so that I can learn how it is that you've each done it. (Having the numerical result is my most immediate need, but understanding how to reach that result is my deeper interest.

Ultimately, I need an algorithm that I can write a small program for, that will calculate the needed values from a set of inputs, but again, understanding the process is as important to me as having the 'answer'. (This is all just fun/hobby stuff, but it provides me lots of learning opportunities, like this one!

I think I was close to following your path a night or two ago, but family illness (my son and I have both contracted a case of The Crud) has temporarily stalled things. I'm hoping there will be some further ganglial activity possible tonight (last night, there was very little), and that I can continue down the path you were laying out ahead of me.

You guys are awesome to take the time to help me through this. I really, really appreciate the help!!

 

[jim_s]

Once you have x, you can get b directly (no angles):
I see no alternative to using Wolfram...or a similar "calculator".
There is simply no easy way to solve a quartic.


Thanks for the problem...even if it gave me a headache
You owe me $1.79 for paper plus pencil/eraser usage

Oh, man - a little piece of my soul just died - a bit like learning that maybe Santa isn't for real, after all...


I'd thought that you'd managed to find a purely-human means to solve the problem. (So your original proposal of AC = 400 on 12/12 used some type of solver, or did you work through that set of values specifically?)

Sounds like I owe you for some aspirin, too!