/sci/ - Science & Math

[Anonymous No.5929782]

Looking for someone who knows a lot of math to help with this one. I am willing to learn, I just need to know where to start.

Thanks in advance to anyone smart enough to know this.

[Anonymous No.5929831]

The third one is the same as the second in method. In this case the y value is constant as it rotates and the x value is the scalar projection on the xy plane. Take the answer you got on step two and multiple by the cosine of what the angle is.

[Anonymous No.5929838]

>>5929782
I don't think I understand completely. What variables are given except the position of p2?

[Anonymous No.5929848]

>>5929838
p1 is given intially as is p2. The poster want's to know how to calculate transformations in p1 based on rotation about p2. p2's coordinates don't matter at all.

Answer to first transformation:
y' = y * cos(theta)
x' = x* sin(theta)

Answer to second transformation:
y' = y * cos(theta)
x' = x

Answer to third transformation
y' = y
x' = x * cos(theta)

x'/y' is the coordinate after the transformation, x/y the coordinate before

[Anonymous No.5929855]

What the hell is the question? (s)

For 3d spacials/planes, you are running x:x:x ratios, from the planes you are working from.

The first one has no change in 2 of the three planes, It basically tells you to grab a dammed compass and do some basic math. (That thing with a pointy end and a pencil, joined to make a V shape. Used to make arcs. With a ruler, you can then figure out the change in the delta parameters for p1.)

The second one has "2 changes". If you want the simplest function, convert circles to spheres and do some dammed basic math.

The third has 3 changes. In this case, you have the appearance of an illusion. If you continued with the sphere, you would claim your diameter shrunk, which would be incorrect. In this case, you would need to work out a sphere + a circle that is tangently aligned to said sphere.

That one is super hard to do math. It's dammed. Instead, go to wikipedia and read up on 3-dimenional co-ordinate systems. They'll tell you that our distance for X in example one is really X:0:0 value, then number two would be X:Y:0 and number 3 is x:y:z.

That way you can keep drawing these little straight lines. Afterwards, they'll likely throw delta values at you and well... just throw the book at them. ^_^

[Anonymous No.5929858]

>>5929855
see >>5929848

[Anonymous No.5929859]

>>5929848
But where do you get theta from? Theta wasn't given. Obviously you can't solve any of those problems without further information, which is why I'm asking.

Also, your answers are wrong. The first answer is incorrect, because you're rotating around the origin, not the pivot. The correct transform is
y' = (y1 - y2)*sin(theta) + y2
x' = (x1 - x2)*cos(theta) + x2
Assuming that theta is the angle of rotation, x1/y1 are the coordinates of p1 and x2/y2 are the coordinates of p2.

The answers to the second and third problems are completely wrong.

[Anonymous No.5929867]

>>5929848

Is the x from answer 2 the resulting x' from answer 1?

I am probably wrong, but I feel like if it is not, then x' cannot equal x in answer 2. It has inherited the transformations from problem 1, so it would have to be different.

Maybe you are also confused because I should have mentioned that the transformations occur on the grey planes local axis. Which means the rotation along the x axis after rotating along the z should effect p2's x location.... At least I am pretty sure.

Idk, I am trying to teach myself all of this stuff, and it's not easy because I have no way to test it and know if I am correct... I mean, obviously I can know if my math is correct, but not if my logic is. I have no idea if I calculate this stuff that my result would actually be where p2 would be in the case of particular rotations.

[Anonymous No.5929874]

>>5929782


OP here. The rotations do not matter, nor does the location of p2. They are variable.

So for the sake of the thread, assume p2 is the origin, Z rotation is 30 degrees, Y rotation is 45 degrees and Y rotation is 20 degrees.

Also, as I stated in my previous post, these transformations are additive and they are based on the planes local axis.

[Anonymous No.5929875]

>>5929874

Jesus Christ, X is 45 degrees, Y is 20 degrees. Sry, I have not slept much recently....

[Anonymous No.5929949]

Why has no one used vectors to solve this?

[Anonymous No.5930434]

>>5929949

How? How do you get the location of p1 if you only know the magnitude and angles or rotation?

[Anonymous No.5930647]

bump

[Talimar No.5930840]

It's still all a form of "trig", except that you occasionally need to rotate your arcs to derive your angles, just so that you know which triangle you are working with. It's just a claim that allows you to know where you put things, relative to your co-ordinate system that is "going to never ever change."

In order to really answer the relationship, you need the degree in rotation. So, if for number 2, you rotated it along the axis 45 degrees, you can then figure out the ratio of the distance.

Most of the time, when they introduce this, they use the base distance D from the original value. So for the setup, if p1:p2 has a distance of "10", then the rotation will still fall on specific arcs when the 3rd plane = 0.

When you get to changing 2 planes, you derive two sets of values, and calculate such that each plane can still provide a x:y:0 value. It's like the more advanced algebra where you have 2 unknown variables that you want to find the specific answer to.

Or you could simply be smart and claim that because you are moving the plane itself that the two dots reside on, the distance between p1 and p2 will always be the same vector distance. When they actually go about and teach it correctly, then you start to learn about propulsion mechanics in 3 dimensions and the like. (and you get 3 co-ordinate systems to work with.)

Ie. if you a cup with a 3 cm diameter opening, turning it sideways will not somehow make the cup have a smaller top.