I already said my brain isn't smart enough, I'm sorry. If the explanation is only about cells then this question cannot be answered, but if the explanation is about initial position of blocks, question will be answered (maybe). And if we talk about blocks in one cell, it would be better if we give demo not only 2 blocks but 3 or more blocks in one cell including fixed blocks too. For example:
Great demo !!! I love. Different variants. It is interesting. Thanks! Regarding fixed blocks, we do not consider them because the star does not act on them. In your demonstration, you can see that two wedges with the ground - also do not give a passage. Therefore, we did not take them into the experiment. Thanks! Very good demo
For a fixed wedge why can't it be passed? even though the metal wedge underneath, should have been hit by a star effect that made it passable (invisible). So this so-called wedge effect dominates over the star effect? But if it was reversed or replaced with a movable wedge, the result would be different, since both were hit by the star effect. And my brain still can't understand it well.
@Denis Nazin I'm trying to understand. What is meant by physical is this? Spoiler https://mekoramaforum.com/threads/physical-dimensions-of-mekorama-blocks.635/ and stars are considered to have no physical? (air/wind/void) And if 2 blocks are in one cell, then there is only one "blok property" occupying that cell? As for the physical, can one block or a combination of 2 blocks or produce a new block?
@Labyrenight Thanks for the answer! Judging by your questions, you understood me very well - thank you! Yes - I came to this conclusion. I want to thank you for the link - it's a very good topic. And she, in my opinion, is not fully disclosed. You can continue to explore the properties of the blocks. Regarding your last question. It is very interesting! It seems to me that when the blocks are superimposed, their dimensions FOLD, or are summed up. But B only sees the properties (physics) of one block. Accordingly, it behaves as - how it should interact with this block. But he stumbles upon the total dimensions of the two blocks. This can be seen very clearly in my demo card. Look closely at how B walks up the ladder, which is stacked with cubic blocks. He only sees the stairs. BUT! B trips over cubic blocks or bounces when descending. B does not raise his legs and body - because this is not necessary on the stairs, and therefore he stumbles over a real cubic block. We can use this NEW behavior B. This behavior creates new movements B. Which are not characteristic of him when he is on double or triple blocks. It is very interesting
I just looked again at the demo where B goes up and down the stairs, which are superimposed on the cubic blocks. On the contrary - B rises perfectly on cubic blocks. B never climbs so well on autopilot !! And when he descends, he jumps.
[BDAY] Perfect Angel Day Sowon (December 07, 1995) Yuju (October 04, 1997) Spoiler Spoiler: GFRIEND Birthday Level Eunbis Day Ye-sis Day Spoiler: Video
I don't understand this trick at all. The first level below demonstrates the stars that are enabled/disabled due to their position; the second level demonstrates partial/full failure of the stars that take effect as the ball rolls.
@935 It is about the position of the Star against the movable Block. Take a look at the cube in this demo. If you put the robot (body) on the brick, then the star will be active in all positions. If you put the robot (body) in the water, then the star will be active if you put it on the grass and win. The order is: brick<stone<dirt<metal<draggable<water<grass<win Or try making lots of star shadows. After that remove the stars one by one, then the shadow will be a star, see the order.
I have to thank you @935 I just realized the sequence has to do with the trigger position for the sensor. I will research it further.
The order is starting from the brick, the direction is from left to right. after that dirt, then stone. if it reaches the top then return from the bottom again, grass, the direction is the same to the right. and continues onwards. I can't make a good picture. Your explanation with arrows will be easier to understand. Thank you.
Hello Friend! I would like to help you with great pleasure! But for now I can't, because I don't understand everything that you say and everything that you understood! When I understand, I can make a photo-demo. It's a pity that we speak different languages. If we knew the same language, it would be easier to explain to each other! To understand such subtleties that you are talking about, you need to clearly formulate the idea - I still do not understand the location and dependence on it well.
It's about positioning the star against the movable block, so the starpass can work. I'll give an example on the x and y axes. There are 7 layers. We count from bottom to top. Each layer contains 16 blocks, counting from left to right. For example, the first layer we give the number 1-16 then the second layer 17-32 and so on until the last layer. Here's the order. So if I put the robot in that position. The star that can work is in the position of the number after the robot. For example if I put a star on a brick, the starpass will not work because it is in the number before the robot. whereas if I put a star on the grass then the starpass will work.
And for the z-axis the direction is from back to front. After the numbering reaches the top, then the next number is in front of number 1. So if i put stars in all areas behind the robot then starpass will not work, and if i put stars in all areas in front of robot then starpass will work.
@Labyrenight Wow! Thanks a lot!!!! This is a good explanation! I understand! What about the slider??? So the slider has no effect? This effect is only affected by the location of B!? Does the rest not matter?