/sci/ - Science & Math

[Anonymous No.1838303]

Can /sci/ solve this?
I ran into this problem a few years ago.
"A tower made of steel needs the longest cross brace possible with material w wide"
Looks easier than it is.
I have been unable to solve it.
note* the angle from corner to corner of the large rectangle is different from the angle of the brace.

[Anonymous No.1838320]

it stumped a college math prof.

[Anonymous No.1838358]

meh. I can probably do this. Motivate me.

[Anonymous No.1838386]

>>1838358
I've wrote programs/formulas to fit pipes together at any type of junction imaginable. some of the formulas have been over 100 characters long(1-1/2 lines on a 80 wide screen).
This simple looking problem has bugged me for years.
If anyone can solve it, I will be very impressed.
I'll even post my email just in case:
derpp live.com

[Anonymous No.1838401]

x^2 + y^2 = L^2

u jelly?

[Anonymous No.1838419]

>>1838401
wrong
>note* the angle from corner to corner of the large rectangle is different from the angle of the brace.

length of the brace - L
did you look at it?

[Anonymous No.1838498]

This problem cannot be solved without additional information (assuming x and y are the lengths of the sides of the box as opposed to the lengths of the sides of the lower right-triangle). You would need to know the distance of the edge of the box to the rectangle's lowest corner.

[Anonymous No.1838526]

>>1838498
I've tried simultaneous trig equations but only worked myself in circles

[Anonymous No.1838540]

Easy calculus.

[Anonymous No.1838551]

the cross brace would be the area of x*y...

next.

[Anonymous No.1838586]

big and small triangles are proportional
small:
(X-A)^2+(Y-B)=W^2
large:
A^2+B^2=L^2

..... now what?

[Anonymous No.1838632]

>>1838586
I think you're right, but how can you be sure that the triangles are similar? They certainly LOOK to be, but is there some math assumption?

[Anonymous No.1838678]

>>1838632
yes

[Anonymous No.1838736]

>>1838678
Ah, cool!

[Anonymous No.1838752]

>>1838678

that's not proof - it's close, but you marked the wrong angles.

[Anonymous No.1838764]

>>1838752
how did I mark wrong angles?

[Anonymous No.1838815]

>>1838764

you have the right idea, you're using the wrong angles to prove it. too busy arguing with morons to go into it at length, just look at the "proof" a bit harder and tell me why you chose an angle that isn't ninety degrees to be the ninety degree angle...

[Anonymous No.1838885]

>>1838815
the 90 is on the small rectangle
small triangle:
small angle=o
large angle=180-90-o=90-o (all angles in a triangle add up to 180 and it's a right triangle)

where are the mathfags?

[Anonymous No.1838905]

>>1838885

i'm with you on the math portion, but the final proof will be when that short distance between the corner and first line of W is known so that you can solve for

A squared + B squared = the square root of C (geometry) - once that's proven and both 'ends' of the X - Y rectangle are shown to be equal, then the rest of the equation can be filled in to solve for whatever.

[Anonymous No.1838920]

Can someone re-phrase the question?
I don't understand what it's asking.

[Anonymous No.1838958]

>>1838920
on the pic:
given X,Y,W, solve for L

[Anonymous No.1839048]

>>1838958

Isn't it more... on the pic given X, Y, W solve for maximum L? Shouldn't there be some calculus here?

[Anonymous No.1839056]

>>1838905
any retard can us that

[Anonymous No.1839067]

>>1839048
i agree but how do you express that? there are still too many unknowns

[Anonymous No.1839076]

I'm calling the small sections a and b, so that
a^2 + b^2 = w^2
(x-a)^2 + (y-a)^2 = L^2
because of the complementary angles,
w/a = L/(y-b)
w/b = L/(x-a)
Also, because the areas of all the sections must add up to xy,
xy = ab + wL + (x-a)(y-b)
xy = ab + wL + xy - ay - bx + ab
0 = 2ab + wL - ay - bx

surely some combination of these equations will solve it...

[Anonymous No.1839094]

>"A tower made of steel needs the longest cross brace possible with material w wide"

This is BAD architectural practice. The most stable results come from the shortest brace distance plus tension created by other braces in the structure.

Even if it is a math question, it is a POOR one.

[Anonymous No.1839107]

>>1839076
never thought of using areas
all other combinations, I would do a bunch of crunching just to end up with 0=0 of some shit

[Anonymous No.1839202]

>>1839094
<like this

width exaggerated in problem.jpg for clarity

[Anonymous No.1839204]

>>1839107
but just thinking of it intuitively, if you have the W section and are trying it at different angles in the corners, and connecting the ends of the W sections with an L line, there is only one angle at which the L line makes a 90 degree angle with the W line, and that angle determines the length of the L line. Putting that in mathematical terms seems like the surest way to do it.

[Anonymous No.1839215]

>>1839094
Isn't an architect more artist than engineer?

[Anonymous No.1839233]

>>1839204
It's hard to even draw correctly if you do the big rectangle first(trial and error)

[Anonymous No.1839243]

>>1839215

yes. and half the shit they dream up is borderline impossible.

that's why good engineers are worth their weight in salt; to fix the goofy shit architects dream up...

[Anonymous No.1839249]

ps. pheta y is equal to y/l

[Anonymous No.1839252]

>>1839243
agreed
engineers have a sense of "practical"

[Anonymous No.1839271]

whats that?

[Anonymous No.1839274]

>>1839215
>>1839243
>>1839252

Yes, but I do both. LEDFag here - I design shed structures and LED lights to work with the systems installed in the sheds.

But it still holds true - braces are best done via the shortest possible route.

[Anonymous No.1839292]

>>1839271
theta, the symbol for "the angle formed by'

[Anonymous No.1839302]

theta is unknown

[Anonymous No.1839307]

>>1839271

the symbol for hamburger.

[Anonymous No.1839316]

>>1839274

>>But it still holds true - braces are best done via the shortest possible route.

yes, but if only one brace can be used (lack of materials...)

[Anonymous No.1839332]

Solving for the angle first seems like a plan, but wolfram alpha couldn't isolate theta for me.
theta = arctan((y-cos(theta)*w)/(x-sin(theta)*w))

[Anonymous No.1839338]

>>1838586
I think the only problem with that solution is that even when you are given a value for x,y&w you still won't be able to solve for L

[╩E No.1839343]

>>1839249
explain your modus operandi

[Anonymous No.1839352]

>>1839316

No, in all situations, especially those where the pythagorean theorem is often used (floor braces between stories, etc.,) and in situations where you need to hold huge width structures.

[Anonymous No.1839362]

<span class="math">x_1^2 + y_1^2 = w^2[/spoiler]
<span class="math">x_2^2 + y_2^2 = L^2[/spoiler]
<span class="math">\frac{y_2}{x_2} = \frac{x_1}{y_1}[/spoiler]
<span class="math">x_1 + x_2 = x[/spoiler]
<span class="math">y_1 + y_2 = y[/spoiler]

We need an equation for L, written only in terms of x, y, and w.

<span class="math">(x - x_2)^2 + (y - y_2)^2 = w^2[/spoiler]
<span class="math">x^2 - 2xx_2 + x_2^2 + y^2 - 2yy_2 + y_2^2 = w^2[/spoiler]
<span class="math">L^2 = w^2 - x^2 - y^2 + 2xx_2 + 2yy_2[/spoiler]
<span class="math">L^2 = w^2 - x^2 - y^2 + 2(xx_2 + yy_2)[/spoiler]

That's as far as I got. I don't know how to use the proportion, but that's all that's missing so far...

[Anonymous No.1839371]

>>1839362
explain your notations

[Anonymous No.1839385]

>>1839362

You're missing the diagonal brace algorithms and tension specs in that entire set.

[Anonymous No.1839406]

>>1839362
thats what i spent half an hour trying to do with the angles and their relations to the other angles. all just trying to find "theta Y" but i just went in circles.

[Anonymous No.1839411]

>>1839362
>>1839076
OP here. I haven't actually worked on this problem in a while. I see some new ideas here. I think I'll try to dig out my previous work.I got too frustrated and quit.

[Anonymous No.1839419]

>>1839411
>thanx for doing my homework, guise!

[Anonymous No.1839422]

>>1839371
<span class="math">x_2[/spoiler] and <span class="math">y_2[/spoiler] are the x and y lengths of the large triangles formed in the corners, and <span class="math">x_1[/spoiler] and <span class="math">y_1[/spoiler] for the small triangles.

[Anonymous No.1839434]

>>1839362
(xx_2 + yy_2)
could you explain what this bussiness means?

[Anonymous No.1839457]

It seems ridiculous if this doesn't have a closed solution.

[Anonymous No.1839463]

Why not just find L as a function of {W, X, Y, and the angle between L and X}, then use calculus to find the maximum? (find dL/d(angle) = 0)

[Anonymous No.1839490]

>>1839419
problem from structural steel fab shop - years ago
I was doing blueprint detailing for the shop and actually had to tell them to figure it out once the frame was made.
Hurt my pride.

[Anonymous No.1839509]

Here's another idea out of left field. The diagonals on the brace have to be equal, and are both sqrt(w^2+l^2), so, with a and b being the corner pieces, such that a^2 + b^2 = w^2,
w^2 + L^2 = (X-2a)^2 + Y^2
= X^2 + (Y-2b)^2

[Anonymous No.1839523]

>>1839463
That doesn't make sense. L is already a function of X, Y, and W. The problem is finding the function. Only one angle works. With the problem. The problem is finding the angle, which is the same as finding L.

[Anonymous No.1839537]

>>1839362
you have 5 unknowns (x1, x2, y1, y2, L) and 5 (non-linear) independant equations

that would have a unique solution (maybe not algebric but at least numeric)


but what i don't get is that it looks like a problem of optimisation (you can have different values for L)

[Anonymous No.1839567]

Stand back everyone, I am a math wizard.

For given W, X, and Y, L is determined completely by the distance p along the x-axis that the brace meets the frame.

The distance q along the y-axis is determined by p^2 + q^2 = w^2, so q = sqrt(w^2 - p^2).

The distance L is, by geometry, equal to (y-q)^2 + (x-p)^2

Substitute the formula for q

L = (y - sqrt(w^2 - p^2))^2 + (x-p)^2

Where y, w, and x are constants. This is a function of one variable, which you want to maximize.

The Calculus says to take the derivative and find where it equals zero in order to maximize L. After trivial calculus and algebra, the closed-form solution so d/dl = 0 is:

p = x * w / sqrt(x^2 + y^2)

For a physical interpretation, put the X location of the joint such that it is along the x-axis proportional to the ratio of the width of the base over the hypotenuse of the frame (eg, p/x = w/sqrt(x^2 + y^2)).

Sanity check is w = 0; you put p in the corner and get the longest line that fits inside a box.

Pic Unrelated

[Anonymous No.1839574]

>>1839537
samefag here, i'm starting to think that it's not a problem of optimisation and that we just need to solve the system of equations

someone just have to check IRL with two sticks and two strings that L is unique

[Anonymous No.1839576]

a few limits to keep in mind once we get something worth checking, in the form of L = L(x,y,w). this is assuming 0 < w < y; not sure if anything would change for y < w < x.

lim(w-->y) L(x,y,w) = x
lim(w-->0) L(x,y,w) = sqrt(x^2 + y^2)

[Anonymous No.1839597]

>>1839574
You can solve for L as a function of one variable - namely, the distance between one of the corners and the intersection between the brace and the frame.

[Anonymous No.1839607]

>>1839537
It's really only 2 unknwowns, and they aren't independent, but constrained by x1=sqrt(w^2-y1).
Everything else can be determined from x1 and y1.

Or, rather, there is only one unknown, theta, from which everything else can be determined.

[Anonymous No.1839618]

>>1839607
Its more useful to work with distances than angles on this problem. I should know, because I fucking solved it already.

[Anonymous No.1839632]

>>1839567
ok i agree with this but when i said we have 5 unknowns i meant he introduced 5 unknowns (they are dependant to each other, that's why we actually have only one parameter to optimize L)

[Anonymous No.1839639]

>>1839567
OP here - you lost me at Calculus

the magic formula?
p = x * w / sqrt(x^2 + y^2)
p being the short leg of the small triangle

[Anonymous No.1839646]

>>1839567
You're not trying to maximize anything. You're equations only work if the LxW object is already fit into the space. I don't think you can use calculus on this.

[Anonymous No.1839659]

>>1839632
My post wasn't a reply to anyone other than the OP. Its just the solution, don't interpret it in the context of anyone else's posts.

X, Y, and W are constants that some variable should be in terms of. There's only room for one variable in determining the geometry of the girder. Its a lot more convenient to choose the position of the intersection than the angle, because then figuring out L is very direct.

[Anonymous No.1839682]

>>1839659
See picture. It doesn't actually matter whether you figure out L(p) or L(q), since you're going to get the same answer you'd get if you solved for one and then substituted.

[Anonymous No.1839693]

>>1838303

Ummm, it's just sqrt (x^2+y^2), isn't it? Or am I missing something.

[Anonymous No.1839694]

>>1839646
No, you're wrong. L(p) is the length that will fit a LxW box into the frame, given that the box has a corner on the frame at (p,0). For any point p that we choose and given W, X, and Y, L(p) will be the one and only length that fits a WxL box into an XxY frame.

[Anonymous No.1839698]

>>1839682
derp, forgot picture

[Anonymous No.1839701]

>>1839659
how did you guessed that it was better to introduce the distance p along the x-axis that the brace meets the frame ?

[Anonymous No.1839717]

>>1839632
The absense of evidence is not the evidence of absense

there are known knowns and known unknowns, but there are also unknown unknowns; things that we dont know that we dont know

[Anonymous No.1839718]

>>1839693

Ohhh, is x and y the length of the rectangle sides? If so, my bad, I misinterpeted the problem.

[Anonymous No.1839720]

do we even have enough equations to ensure a unique solution?

six variables: x, y, a, b, w, L

i have four equations (the last set of 3 equalities is only two independent equations). that leaves two degrees of freedom. we'd need five equations if we wanted to have exactly one possible value for one variable once given all the others.

am i missing another equation?

[Anonymous No.1839726]

>>1839701
DERP

i meant how did you guessed that For given W, X, and Y, L is determined completely by the distance p along the x-axis that the brace meets the frame ?

[Anonymous No.1839754]

>>1839720
There are two independent variables. The trick is to maximize one (L) with respect to the other (a or b).

I mean, of COURSE there isn't only one variable that is independent, since you can always manage to stick a shorter brace inside the frame.

[Anonymous No.1839760]

>>1839694
Your solution doesn't work. It fails the area test: 2pq + WL = py + qx.

[Anonymous No.1839761]

>>1839720
i think your missing equations are just L=L and W=W which are known

but where do your fractions come from ?

[Anonymous No.1839767]

>>1839726
L and p are both determined by W, X, Y.

[Anonymous No.1839779]

>>1839726
I don't know how to explain it other than it was obvious to me.

Say you pick a point P on the X axis. Since you are given W, there is only one valid point on the Y axis for the near side of the cross brace (since different points on the Y axis have different distances to the point you picked).

Now that you have the point on the X axis and the point on the Y axis, you can draw a line between them, and then perpendicular lines from that line at the intersections with the frames. These lines hit the frames at exactly one other point. There's only one distance between two points, so L is determined only by which P you picked.

[Anonymous No.1839786]

>this thread

i'm really really ashamed of 75 posts for this shit

[Inox !I4rwj6vpbw No.1839789]

Oh, god you're a retard

[Anonymous No.1839804]

>>1839567
>L = (y - sqrt(w^2 - p^2))^2 + (x-p)^2
>Where y, w, and x are constants. This is a function of one variable, which you want to maximize.
There is only one p and one L that fits such that w and L are at right angles. What do you think you're maximizing? L can only be one length. You don't want to maximize p. You need to find the p such that the hypotenuse formed by W is perpendicular to that formed by L.

[Anonymous No.1839819]

>>1839779
L is determined by X Y and W. So is p. There are not multiple p's that can work.

[Anonymous No.1839821]

>>1839804
you can move W in the space, L can take differet vales

check it

[Anonymous No.1839824]

>>1839786
let's see your solution, smart ass.

[Anonymous No.1839830]

>>1839821
If you move W, then L doesn't adjoin up to the opposite W.

[Anonymous No.1839850]

>>1839830
that's because L is not constant, when you move W, L is changing

[Anonymous No.1839855]

>longest cross brace
>lay it flat, L = x
shittiest wording of a question I've ever seen
especially the picture, which asks a totally different question than the text

[Anonymous No.1839862]

>>1839760
Whoops, that's because L(p) is L^2, not L. Sorry about that minor fail / handwave over the fact that you can maximize (L(p))^2 and find the p such that L(p) is maximized.

Put a square root on L and come back. Nice catch.

[Anonymous No.1839864]

>>1839824
>implying there's a solution for this

[Anonymous No.1839866]

>>1839767
Nope. The relationship between L and P is determined by W, X, and Y.

[Anonymous No.1839869]

>>1839864
There is a solution. It got posted in this fucking thread.

[Anonymous No.1839870]

>>1839850
For any p except the correct p, the lines of the box between W and L are not perpendicular. Therefore maximizing L is a meaningless exercise. The trick is to find the only p such that W and L are perpendicular.

[Anonymous No.1839875]

>>1839869
No working solution has been posted in this thread.

[Anonymous No.1839876]

>>1839830
If you move the width end, then you need to go a different distance perpendicularly from it in order to hit the other end.

For example, see this picture.

[Anonymous No.1839879]

>>1839866
P and L are both determined by X, Y, and W, which is why there is a solution to the problem.

[Anonymous No.1839887]

>>1839567
OP here
I did 3 different drafts and

p = x * w / sqrt(x^2 + y^2)

doesn't check

[Anonymous No.1839891]

>>1839567
>>1839870
you seem to both have right

i fail to see at least two different configurations for the same W

someone draw two L for the same W

[Anonymous No.1839895]

>>1839876
The equations you used don't work in that scenario.

[Anonymous No.1839904]

>>1839891
obviously not the example when you take the complementary

[Anonymous No.1839905]

>>1839891
The only way to have more than one L for the same W is if you do this:
>>1839876
But there is only one diagonal L. Finding the diagonal L is the task. It's not a continuous function. At least for the equations used, the equations assume the diagonal solution, not any partial diagonals. The equation for L is wrong if partial diagonals are allowed.

[Anonymous No.1839917]

>>1839875
Oh, didn't answer the question properly.

Math wizard, back to the rescue!

continued from:

>>1839567

fixing my earlier statements, L^2 = (y-q)^2 + (x-p)^2

p = wx / sqrt(x^2 + y^2) maximizes L^2

and by p^2 + q^2 = w^2,

q = wy / sqrt(x^2 + y^2) maximizes L^2

(you can check this by substituting. w^2x^2 / x^2 + y^2 + w^2y^2 / x^2+y^2, is equal to p^2 + q^2 and w^2(x^2 + y^2) / (x^2 + y^2). Trust me, it works)

Now substitute into L:

L^2 = (y - wy/sqrt(x^2 + y^2))^2 + (x - wx/sqrt(x^2 + y^2))^2

simplify

L^2 = y^2 - 2wy^2/sqrt(x^2 + y^2) + w^2y^2 / (x^2 + y^2) + x^2 - 2wx^2/sqrt(x^2 + y^2) + w^2x^2/(x^2 + y^2)

L^2 = x^2 + y^2 - 2w(x^2 + y^2) / sqrt(x^2 + y^2) + w^2 (x^2 + y^2)/(x^2 + y^2)

finally,

L^2 = x^2 + y^2 - 2*w*sqrt(x^2 + y^2) + w^2

[Anonymous No.1839928]

>>1839870
No. No. No. No. No. You're just wrong.

You can always find some length G such that given two points on a rectangle, you draw lines of length G perpendicular from the line between those two point that end on the rectangle.

[­ No.1839935]

shut up everyone stating there are infinite solutions. we have to find the largest L for given X, Y and W
Let's called the marked angle φ.

L² = (X - W sin φ)² + (Y - W cos φ)
L² = X² - 2*X*W* sin φ + W² sin² φ + Y² - 2*Y*W* cos φ + W² cos² φ
[sin² φ + cos² φ = 1 ]
L² = X² + Y² - 2W (X sin φ + Y cos φ) + W²
to find the maximum of L² we derivate:
0= [L²] ' (φ)= 0 + 0 - 2W (X cos φ - Y sin φ) + 0
Y sin φ = X cos φ
φ = arctan (X / Y)

L = sqrt( X² + Y² + W² - 2W (X sin (arctan(X/Y)) + Y cos (arctan(X/Y)) ) )

any logical fallacies?

[Anonymous No.1839940]

>>1839895
Q actually starts taking complex at that point, but thats another topic altogether.

[Anonymous No.1839949]

>>1839935
>(X sin (arctan(X/Y)) + Y cos (arctan(X/Y))

That equals X^2 + Y^2. Check it.

Your answer agrees with >>1839917

[­ No.1839954]

>>1839935
forgot the pic. fucking mistyped captcha..

[­ No.1839961]

>>1839949
it took me a while to type this, you know...

[­ No.1839980]

>>1839949

no, it doesn't.

[Anonymous No.1839995]

>>1839928
You're totally missing the point. The line cannot end anywhere on the rectangle. The line must end on the correct one of the other pair of points in the opposite corner. If you try to just maximize the length of a line perpendicular to two points, then you get this:

[Anonymous No.1839997]

>>1839980

it equals sqrt(X^2+Y^2)

their answers agree exactly from what i can see.

[Anonymous No.1839998]

>>1839935
φ already equals to arctan (X / Y)

why did you needed to derivate to find this ?

[Anonymous No.1840021]

I think the phraseology of the "longest" is throwing people off. The longest is the one that has all 4 of its corners touching the rectangle. But there is only one solution, one length, and one angle, such that all 4 corners are touching the rectangle.

[­ No.1840052]

>>1839998
it's my way of determining the highest value of L.


>>1839997
woops, didn't see the sqrt there. you are right, there's an easier way to write it. I didn't know my term could be simplified any more.

[Anonymous No.1840058]

>>1839935
The logical fallacy I see is that those equations only hold true for a single theta, the that such that all 4 corners of the inner rectangle are touching the outer rectangle. So maximizing dL/dtheta doesn't make sense to me, as any theta but the correct one is giving something meaningless. Or rather it's giving a situation where your assumptions are no longer valid. That's why this approach doesn't make sense to me.

[­ No.1840070]

>>1840058
I thought it was given that all corners touch the outer rectangle.

[Anonymous No.1840094]

>>1840070
Right, given that all corners touch the outer rectangle, and given the width of the inner rectangle, there can only be one length.

[Anonymous No.1840121]

>>1840052
>it's my way of determining the highest value of L.

phi, as you defined it in the first place is already equal to arctan(Y/X)

i don't understand why you needed to demonstrate this if it was already stated

[Anonymous No.1840150]

>>1839917
doesn't check

I'm lost here now but here's some working numbers from drafting if that helps

x=5.12
y=4.3
w=2.98
L~= 4.36

[Anonymous No.1840241]

grrrr...

[Anonymous No.1840843]

If the brace is a perfect rectangle like in the drawing, it can only fit within the other rectangle in such a way that makes all triangles formed 30-60-90. So:

L = sqr((x-wsin30/sin90)^2+(y-wsin60/sin90)^2)

Further proof: run the dimensions in the picture through formula.

[Anonymous No.1840861]

Oh, forgot to add. If y>x than the equation becomes:

L = sqr((y-wsin30/sin90)^2+(x-wsin60/sin90)^2

[μαθμαν !!vnLoxcfyRGs No.1840916]

holy shit.
I am going to end this thread, now.
First, only one solution for L. If calculus is involved, it won't be for optimizing L.
Second, lots of work, mostly trig
x by y box
width w
assuming both box and strut are rectangles (90 degrees in every corner)
x' is the distance along the bottom to the point where the brace touches the bottom.
likewise, y' is distance along the left to the point where the brace touches the left
x-x'/y-y'=y'/x'
because of similar triangles.
also, w/x'=l/(y-y')
so x'/(y-y')=l/w=y'/x-x'
x'^2+y'^2=w
-y'^2+yy'-xx'+x'^2=0
x'=x+-sqrt(x^2-4(-y'^2+yy'))/2
x' is by def smaller than x, so we can drop one of these "solutions" and plug into another equation.
(x-sqrt(x^2+4y'^2-4yy'))^2+4y'^2=4w
now solve above for y', plug back into
2x'=x-sqrt(x^2+4y'^2-4yy')
to get x' and finally plug x',y',w,y into
l=wx'/(y-y')
voila, your one solution for l
The only thing that remains is for someone to use wolfram alpha to find the solution to y' and plug everything in.

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[Anonymous No.1841622]

>>1840843
how do you figure 30,60,90 only?
pic related

>>1840916
still no solution of L in terms of X,Y and W