laser range finder with camera and laser pointer

This looks like the most accurate solution. Using some tig and a camera and a laser pointer, mounted at known distances from each other, you can make a laser range finder. I may have to do this. It would be cool for mapping a room along with a ping sensor and would give me something to put in Wall-'s other eye.

@yatakitombi,

Three solutions have been provided including the formulas that you would use and explanations of the formulas to some extent. The short answer is that there is nothing built into the camera or ARC at this time, but it is doable with some scripting or programming, depending on what you choose to do.

Let us know if you have any other questions about this. If not, please mark as answered and give credit to whoever helped you the most.

I will be working on these in my limited spare time. I do think this is a feature that should either be added to ARC or something worth someone's time to develop outside of ARC. If anyone else wants to tackle it, I'm good sitting back and watching. If not, with some specific information about the camera lens, I could start working on it. @yatakitombi, since you have a need for this and have expressed an interest in ez-sdk, and have stated that you have a mentor, maybe this is something you could develop with the information provided here. Then you could share your code and solution with the community. It's what we do here.

All this math reminds me of clue:

If you have the known sizes at two known distances, then that's enough. A simple equation can give you the factor between the two distances, and therefore used as a multiplier to find out the distance in the future.

To avoid needing camera angles:

1. Calculate two distances, one for the Height and one for Width.
2. Use the largest of the two as the distance

^ That will give you the most accurate distance with easiest calculation. Because the size of the object can only get smaller if displayed on an angle, not larger. So getting the largest of the two will tell you how far the object is away

The other two solutions were if you dont have a known size at a known distance.

If you know the size at a known distance, the equation that Rich provided works. Still very interested in the laser range finder...

Each of these solutions has its own downfall.

solution 1 - dont know the size at a distance solution 2 - unlevel or uneven ground and cant determine where the base of the object is solution 3 - light that affects the color of the laser reflecting back or dark objects.

Ping sensors also have their own issues. Some materials will not reflect back the signal accurately.

IR also has its own issues.

This is why multiple types of sensors are used to get data. Also, if a size at a distance were not known, but another sensor were used to get the distance, and you would be able to get the size from the camera, you could then switch to using solution 1. you would have to have one of the other solutions in place to allow you to use solution 1 if you didnt know the specifics.

Using these methods to train each other would give you the best results, but would require the most code.

Sounds like a project for an advanced student in my robotics class next year.

Or, to put it quite Simply:

dY = \sum_{n=0}^{N-1} x_n e^{-\frac{2\pi i}{N} nk } \qquad k = 0,\dots,N-1 ?Nn (n = 0, ..., N - 1):

\omega_N^n = e^{-\frac{2\pi i}{N} n } and define the polynomial x(z) xn:

x(z) = \sum_{n=0}^{N-1} x_n z^n.

X_k = x(\omega_N^k) = x(z) \mod (z - \omega_N^k) :<math>X_k = \sum_{n=0}^{N-1} x_n e^{-\frac{2\pi i}{N} nk } \qquad k = 0,\dots,N-1. </math> :<math>z^{2M}-1 = (z^M - 1) (z^M + 1) ,</math> :<math>z^{4M} + az^{2M} + 1 = (z^{2M} + \sqrt{2-a}z^M+1) (z^{2M} - \sqrt{2-a}z^M + 1)</math p_{s,0}(z)&= p(z) \mod \left(z^{2^{n-s}}-1\right)&\quad&\text{and}\ p_{s,m}(z) &= p(z)\mod \left(z^{2^{n-s}}-2\cos\left(\tfrac{m}{2^s}\pi\right)z^{2^{n-1-s}}+1\right)&m&=1,2,\dots,2^s-1 \end{align}</math>''m''=0, the covered indices are ''k''=''0'', 2^''k'', 2·2^''s'', 3·2^''s'',, (2^''n''-''s''-1)·2^''s'', for ''m''>''0'' the covered indices are ''k''=''m'', 2^''s''+1-''m'', 2^''s''+1+''m'', 2·2^''s''+1-''m'', 2·2^''s''+1+''m'', , 2^''n''-''m'' :<math>\phi_{N, \alpha}(z) = \left{ \begin{matrix} z^{2N} - 2 \cos (2 \pi \alpha) z^N + 1 & \mbox{if } 0 < \alpha < 1 \ \ z^{2N} - 1 & \mbox{if } \alpha = 0 \end{matrix} \right. </math>:<math>\phi_{rM, \alpha}(z) = \left{ \begin{array}{ll} \prod_{\ell=0}^{r-1} \phi_{M,(\alpha+\ell)/r} & \mbox{if } 0 < \alpha \leq 0.5 \ \ \prod_{\ell=0}^{r-1} \phi_{M,(1-\alpha+\ell)/r} & \mbox{if } 0.5 < \alpha < 1 \ \ \prod_{\ell=0}^{r-1} \phi_{M,\ell/(2r)} & \mbox{if } \alpha = 0 \end{array} \right. </math> =2.61949494946494946168989119

Done:

Roll

Thanks for that. I needed a good laugh this morning.

If you dont want to ruin anyones vision with a laser and dont have superpower math skills you could put a ultrasonic right above your camera