Motion in cartesian coordinates

[wonder1]

Hi all,

i am having a problem with question 3, as its not clear if i should use the Z value for the camera as 15 or 25 m or... Could you suggest me. Cheers

The goal of this project is to obtain some understanding of the camera’s motion in space.
Also, based on the camera’s motion, we will obtain an understanding of the type of engine
necessary to operate the camera’s cables.

Some general observations: The camera must be able to reach the ground for maintenance
(calibration during a game), but most of the time the camera will hover 15 to 25 metres above
the playing field. The position of the camera at time t will be denoted by C(t). In our model
we will assume that the cables are always stretched on a straight line (although in reality there
is a small amount of curving due to gravity).

Part 1: Motion in Cartesian coordinates
We introduce a coordinate system where the origin (0, 0, 0) coincides with the SW post. The
x-axis goes from the origin in the direction of the SE post. The y-axis goes from the origin to
the NW post. The z-axis goes straight up from the origin. The position of the camera will be
represented by

C(t) = (x(t), y(t), z(t)).

1. What are the coordinates of the four corners of the playing field?

2. What are the coordinates of the top of each post?

3. The position of the camera in space completely determines the amount of cable that each
post has released. Compute the length of each cable as a function of x, y, and z. (That
is, find the distance from the camera to the top of each post.)

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[MarkFL]

Hello and welcome to MHB, wonder! (Wave)

What you are being asked for question 3, is simply the distance from the camera to the top of each post. The distance formula for determining the distance $d$ between two points in 3Space, let's call them $P_1\left(x_1,y_1,z_1\right),\,P_2\left(x_2,y_2,z_2\right)$ is given by:

\(\displaystyle d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

So, assuming you have already correctly found the coordinates for the top of each post, and using the given position $C(t)$ of the camera at time $t$, can you now state the amount of cable that each post will have released at time $t$?

 

[wonder1]

Hi Mark,

Thank you for your help.

I have found the coordinates of the top of poles, however I am having a hard time understanding what it means with finding the initial position C(t) of the camera at time t.. because when i substituted the values i got 80 and i know its wrong.

If i knew the Z value of camera i could work it out, i feel like i have overlooked something in question details but i don't know what it is.

i just re-read it initially all the cables are straight: does that mean at C (t) is the camera in the center of all the cables?

 

[MarkFL]

We aren't given the initial position of the camera, all we know is that $C(t)$ represents the position of the camera at time $t$. However, this is all we need to answer question 3. We are simply being asked to give the amount of cable that has been released from each post, which is equal to the camera's distance from the top of each post.

 

[wonder1]

I am sorry mark but i am just stuck. I have found the length between the the SW ground corner to at the Pole top of the NE to be 228.03 thinking i could work out a perpendicular length but i don't think its right and i tried using dot-point product but its not making sense.

 

[MarkFL]

I suspect you are overthinking this...:D

Let's begin with the SW post...the top of this post has coordinates:

\(\displaystyle (0,0,80)\)

At time $t$, the camera has coordinates:

\(\displaystyle (x(t),y(t),z(t))\)

And so the amount of cable $d_{SW}$ released from the SW post at time $t$ is:

\(\displaystyle d_{SW}=\sqrt{\left(x(t)-0\right)^2+\left(y(t)-0\right)^2+\left(z(t)-80\right)^2}=\sqrt{x^2(t)+y^2(t)+\left(z(t)-80\right)^2}\)

Can you do the same thing for the other 3 posts?

 

[wonder1]

;) ;)

That's fantastic Thank you Mark. Yes, I can do to the other poles.