Missing one equation to solve the linear system

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The discussion revolves around a geometric problem involving a quadrilateral formed by four rigid rods, where sides AD and BC can be rescaled while maintaining the length of side CD. The main challenge is to find the coordinates of points C and D such that the slopes of AD and BC change minimally, ideally remaining unchanged. Participants clarify that the slopes of AD and BC should not change during the rescaling process, leading to confusion about the implications of "minimally changed." The conversation highlights the complexity of maintaining the quadrilateral's shape while adjusting the lengths of the sides, with suggestions for analytical solutions and considerations of angles involved. Ultimately, the problem remains open-ended, with no definitive solution provided.
skrat
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Homework Statement


So imagine 4 rigid rods connected together to form a dashed quadrilateral as shown in the picture.
IMG_20180704_211046.jpg

Now AB is fixed, can not be changed in anyway, while all other sides (AD, BC and CD) are connected but can move freely. The initial conditions (dashed quadrilateral) are given.
Now we rescale the dashed BC and AD sides to given values. Note that the length of CD remains constant!

Find coordinates of points C and D of newely generated quadrilateral in such way that the slope of AD and BC is minimally changed.

Homework Equations

The Attempt at a Solution


Ok so the way I see the problem I have 4 values to determine
$$(x_C, y_C)\text{ and } (x_D, y_D).$$
First 3 (out of 4) equations are
$$ ||\vec{BC}|| = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2}$$
$$ ||\vec{AD}|| = \sqrt{(x_D - x_A)^2 + (y_D - y_A)^2}$$
$$ ||\vec{DC}|| = \sqrt{(x_C - x_D)^2 + (y_C - y_D)^2}$$

The forth one is the one I am struggling with. I know that the slope equals ##\tan \varphi = \frac{\Delta x}{\Delta y}## but I'm not really sure what to do with that. I'm quite sure I need to find the relation between ##\vartheta## and ##\varphi## first.

EDIT: This probably doesn't lead to a linear system of equations. I assume Lagrangian multiplier will be needed or some other method for finding multivariate extremes. I will kindly ask the moderators to change the topic title.
 

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skrat said:
Now we rescale the dashed BC and AD sides to given values.
It's not clear what you mean by that.

in such way that the slope of AD and BC is minimally changed.
By the "slope of AD" , do you mean ##\tan (\theta)##? It isn't clear why ##\theta## would need to change at all when you "rescale" AD and BC.
 
Stephen Tashi said:
It's not clear what you mean by that.
Lengthen. From ##||\vec {BC}||## to ##||\vec{BC}||'##.
Stephen Tashi said:
By the "slope of AD" , do you mean ##\tan (\theta)##? It isn't clear why ##\theta## would need to change at all when you "rescale" AD and BC.
Sorry, it is not clear. You are correct, ##\tan \vartheta## is not allowed to change when lengthening the sides. It's totally unnecessary to specifically point that out.
 
skrat said:
Find coordinates of points C and D of newely generated quadrilateral in such way that the slope of AD and BC is minimally changed.
So by "minimally changed" you meant "unchanged"?
 
vela said:
So by "minimally changed" you meant "unchanged"?
That would be amazing, but I don't see how that would be possible except in special cases.

Note that the length of CD is constant, while sides AD an BC are (except in special cases) NOT paralel. So from my understanding, the angles are unchanged if and only if AD and BC are parallel and if both of the sides are lengthen by the same amount.
 
Ah, OK. I was thinking parallelogram instead of quadrilateral.

When you said ##\tan\theta## is not allowed to change in post #3, that's the same as saying the slope of BC can't change, but in the original post, you say the slope of BC can change.

If both ##\theta## and ##\phi## can change, what precisely do you mean by minimally changed?

When you talk about rescaling BC and AD, do you mean by the same factor, or could they change length independent of each other?
 
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vela said:
When you said ##\tan\theta## is not allowed to change in post #3, that's the same as saying the slope of BC can't change, but in the original post, you say the slope of BC can change.
Yeah that's a result of my bad english and poor problem statement.
So
1. We start with the dashed quadrilateral from the original post.
2. Then we rescale the BC and AD - almost never for the same factor, except in special cases. The BC side is rescaled along vector ##\hat{BC}## and AD along ##\hat{AD}##. So "the slope of BC and AD can't change" applies to this step ONLY. I think this is more clear now.
3. Since the AD and BC sides were lengthened (and since they are in general not parallel to each other), the length of CD is now either to long or to short - the OP says the length of CD has to remain constant - I have to rotate points C and D to change the distance between them. And I want to rotate them in such a way that the slope of BC and AD is minimally changed.

vela said:
If both ##\theta## and ##\phi## can change, what precisely do you mean by minimally changed?
That's the tricky part, left up to me to decide. Since ##\theta## and ##\phi## are not independent it might be sufficient to minimize their sum, if not, maybe sum of the squares of the slope, ...
If possible I would like to find an analytical solution to the problem, so the challenge is to find a condition that:
- minimally changes the shape of the quadrilateral (it was my interpretation that this means minimal change of the BC and AD slopes, maybe you have a better idea)
- possibly brings me to an analytical solution (but don't bother with that too much)
 
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Anybody?
 
skrat said:
minimally changes the shape of the quadrilateral (it was my interpretation that this means minimal change of the BC and AD slopes, maybe you have a better idea)
even if you fix on minimal change to slope, that is still open to imterpretation. How about rotation of each through the smallest angle, ψ (same angle for each but opposite directions)? Of course, there might be no such solution. Indeed, if one rod is extended much more than the other there may be no solution at all.
 

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