Society of Robots - Robot Forum

hi
i cant im afraid but i do know of live robot shows by ROAMING ROBOTS
(see below for dates)

I would be happy to have a page on my website of future dates of robot meets and competitons - so i will have to see if i can find out more information on this.
best wishes
whatisarobot.co.uk

June 16th 2013 - Guildford Spectrum - Guildford

July 13/14th 2013 - Newport Centre - Newport - ( 2013 Championships ) ( 2 Days )

October 6th 2013 - Maidstone Leisure Centre - Maidstone - ( Winter Tour Round 1 )

October 27th 2013 - Doncaster Dome - Doncaster - ( Winter Tour Round 2 )

November 10th 2013 - GL1 - Gloucester - ( Winter Tour Grand Final )

First, with the Markov update that you suggest, the update of a cell would be incremental, not forcing it to 1.
And, if that cell would become unblocked, then further sweeps across that cell would pass through that cell, and start marking the cell as unoccupied.
This is why I explicitly suggested you update *all cells* that a sweep goes through (in a line) with the first N-1 cells being updated as "unoccupied" and the last cell updated as "occupied."
"Update" probably means re-evaluate the Markov estimation, rather than just slamming the value to 1/0.

I am not sure if I miss understood jwatte previous post but what I got from him was

and don't mark the cell as fully blocked until you get N positive confirmations that it's blocked

. This is rather not what I wanted to do (although it can do the job) but this will be problematic when come to storing into array and take average to see if occupied or not... What I was thinking of is apply Bayes theorem directly, so sorry to jwatte if I missed explaining that.

@jkerns
 

You can either clear the whole stored array of "blocked" cells before you update

.... don't you think this would waste a lot of time?

can you please explain what you mean by this

you can build in a "clear any point not written this cycle" into your sub-function, or you can use a "forgetting factor" where you decrement the values of each cell over time as  JWatte suggests.

... I am not really understand that part.

For example, this is what i am having right now, but I believe these initial value could be chosen better.


P(occ) = 0.5 -> this is initial occupancy percentage for all cells
P(data|occ) = 0.1 -> this is the laser data does not tell there is an object (means empty)
P(notOcc) = 1 - 0.5 = 0.5 -> this is occupancy percentage of a cell not occupied
P(data|notOcc) = 0.5 -> this is probaility of a cell is not occupied at anypoint (now I still think this value could be better chosen because right now I have a very slow convergent rage)

Example.

If the laser scanner is telling that there is a object, so this means
P(data|Occ) = 0.9

therefore

P(Occ|data) = [P(Occ) * P(data|Occ)] / [(P(Occ)*P(data|Occ) + P(notOCC)*P(data|notOcc)]
                   = (0.5 * 0.9) / [(0.5 * 0.9) + (0.5 * 0.5)]
                   = 0.64

This 0.64 means there is a 64% chance that this cell is occupied. Now, if you repeat this step on the same cells, you can get it to converge. But again, the rate of converge is kinda slow!

Hannah,

I had great luck putting an ad on Craigslist. I got four dead wheelchairs donated to me, and was able to make one working wheelchair robot from them with a heap of spare parts left over. I put an ad in the "wanted" section. I asked for electric chairs working or not, and explained what I wanted to do with them. I had several generous people call me. If you have something like craigslist there, it couldn't hurt to try it.

You might also see if there are robotic clubs in your area. Many times they'll trade parts with each other. A bit further out there, but you could also write a letter to local hospitals or nursing homes asking for their no longer useable chairs instead of them getting scrapped.

Get creative. It worked for me, and I met some great people in the process. 

John