First Set Of Cuboids
My first set of cuboids basically followed the same pattern that I explained in the background information section, although in this set I displayed the pattern in the increase of one cube per set. Through examining the sets of cubes I learnt that the pattern of hidden faces raise by three numbers each set, whilst the number of faces in view also raise by three numbers. The results in my table where confirmed with a model of the sets of cubes built by Lego, thus I can begin formulating a formula:
(Each cube added)+3(towards total of hidden faces)
Or
(Each Face in view)-(total number of faces)=(hidden faces) this formula is based on the original formula I used earlier, which still can be used in this case.
I can use these formulas to predict future sets then compare them to models built using Lego:
These set predictions are confirmed with what I found through observing the model cuboids, therefore the formula was a success. I will therefore proceed with using a similar formula as used here and the same formula that I used in the background knowledge section with future sets of Cuboids. The successfulness of this idea of using the same formula will be reviewed in the conclusion section of my investigation.
Second Set Of Cuboids
I have decided for my second set of cubes I will line up the cubes in a similar fashion as the first set but with an extra cube attached to the top of the first cube, adding a set of two cubes each time to the end. I believe that adding one cube to the top of the cubes will change the results of the table but hopefully the same formula can be used.
In this set I have used a formula that I explained in the ‘background information’ section to allow me to find the hidden faces which was (all faces)-(number of viewable faces)=(hidden faces) e.g. 6-5=1 therefore the number of hidden faces would be 1. I have decided to use this formula in other sets as well as it will allow me to fill in the information into the chart.
I have learnt through observing the results of the table that the number of faces in view increases by five faces each set of cube, whilst the number of hidden faces increase by seven each set of cubes. This brings up an interesting point as both this set of cuboids and the first set increase by a certain amount of cubes constantly; therefore I will be able to apply the first formula I used. This set of cuboids may rise by a certain amount of cubes but it does not rise by the same amount as in the first set, which was three hidden faces each set.
Here are my formulas for my second set of cuboids:
(Each cube added after first cube)+7(towards total of hidden faces)
Or
(Each Face in view)-(total number of faces)=(hidden faces) this formula is based on the original formula I used earlier, which still can be used in this case.
As you can see I have basically used the same formula that I used for the first set although I have replaced the +3 with a +7 to compensate for the amount of cubes there are. The original formula that I used also works, and I am confident that it should work for all the other cuboids I do.
Here are my future predictions on a chart:
Both formulas equal to the above results, as does the Lego Models based on the drawn set. My discoveries through studying the first and second formula I have used have shown that a simple formula could be used to work out similar formulas, however I wish to challenge this hypothesis by constructing many different sets of cuboids. This concept leads me to my next set of cuboids.
Third Set Of Cuboids
As in the second set of cuboids, each set increases by two cubes although in this set the cubes are placed side-by-side rather then on top. I believe that changing the shape of the cubes will affect the results on the chart, yet I am not sure if the same formula can be used, as I have not done any cubes side-by-side in any previous cuboids.
I once again used the=(all faces)-(number of viewable faces)=(hidden faces) formula to complete the chart which displayed a pattern in the set. I have learnt through observing the results of the table that the number of faces in view increases by four faces each set of cube, whilst the number of hidden faces increase by eight faces each set of cubes.
As in the first set and second, this set increase’s by a certain amount of cubes constantly; therefore I will be able to apply the first formula I used. This set of cuboids may rise by a certain amount of cubes but it does not rise by the same amount as in the first set, which was three hidden faces each set or as the second, which was by seven hidden faces.
As this set of cubes and the second set both have a congregant appearance they both have the same results in the ‘Total number of faces’ section in the chart, which was that the numbers increase by twelve faces each set of cubes. This is completely unlike the first set, which increased by six. I have come to a conclusion that even though the set of cuboids may have the same number of faces the number of hidden faces and faces in view will differ due to the shape or rotation of the cuboids.
This set of cuboids the faces in view increased by four whilst the number of hidden faces increased by eight faces. The previous set of faces increased by seven hidden faces and five faces in view. There is a one-face increase in the results for hidden faces and one face decreased in the results of faces in view, in this set compared to the second set.
An increase and decrease in faces is due to the position of the shape, due to faces that are facing the base (therefore hidden) and the other faces that are not hidden. In this case in the second set of cuboids because the cubes are facing upright, it starts with only one cube in contact with the base, whilst in this set two cubes are facing the base. Therefore because in this set more cubes are facing the base, then more cubes will be hidden then in the second set.
Here are the formulas for my third set of cubes:
(Each cube added after first cube)+8(towards total of hidden faces)
Or
(Each Face in view)-(total number of faces)=(hidden faces) this formula is based on the original formula I used earlier, which still can be used in this case.
Also
(Number of hidden faces increase in set 2)+1=(hidden faces) this formula is based on the comparison of set 2 and set 3, which could possibly be used in later formulas that are similar in shape.
Here are my future predictions on a chart:
Third Set:
Compared to Second Set:
Here are my future predictions on a chart:
Both formulas equal to the above results, as does the Lego Models based on the drawn set. My discoveries through studying the second and third formula I have used have shown that two shapes that are similar, yet different in rotation will affect the number of hidden faces. I have however found that the total number of faces was exactly the same in both cuboids and the increase of the number of hidden faces was increased by one in the third set, but as an affect of this the faces in view decreased by one to balance out the results, thus when added together the hidden faces and faces in view for both sets should equal the same number.
I will next try to expand on this theory of rotating the shape does effect the hidden faces, but this time by actually enlarging the current shape which leads me to the next set.
Forth Set of Cuboids
Through comparing the third set of cuboids with the second set, I learnt that the rotation of a shape could affect the number of hidden faces. In this set I wish to broaden on that discovery, by still keeping the cubes side-by-side but this time combined with the second step of having the cubes on top as well. As this set combines the formation of third and second sets, then four cubes at the end of each set will increase the cuboids. The increasing of the cubes in this set is unlike set two or three, which increased by two cubes each time. This will allow me to evaluate the results of the two formations combined.
It is possible that with the two formations combined from sets two and three, that the results will simply be the two results of sets two and three combined. This perception is what I wish to investigate in this set, along with the use of my past formulas.
I once again used the=(all faces)-(number of viewable faces)=(hidden faces) formula to complete the chart which displayed a pattern in the set. I have learnt through observing the results of the table that the number of faces in view increases by six faces each set of cube, whilst the number of hidden faces increase by eighteen faces each set of cubes.
As in the previous sets, despite the difference in size this set increase’s by a certain amount of cubes constantly; therefore I will be able to apply the first formula I used. This set of cuboids may rise by a certain amount of cubes but it does not rise by the same amount as in the first set, which was three hidden faces each set, the second, which was by seven hidden faces or the third set, which was by eight hidden faces.
This set seems to increase ten faces more then the third set of hidden faces. A huge enlargement in hidden faces could be due to the number of cubes on each set of cuboids, as there are more here then in set three or any other. Therefore as there are more cubes combined to form cuboids, then there will be more hidden faces.
In the number of faces in view, there may be a pattern forming as in set two the number of faces in view are five whilst in the third set four. As the formation of this set is based on a combination of set two and three, therefore more cubes. Thus the number of faces in view increases more then either sets, as does the number of hidden faces. As in set three the more cubes that are facing the base equals more hidden faces.
The idea of the hidden faces results matching that of set two and three combined was shown here to be false, which therefore shows that two sets of cuboids combined will not have the same number of hidden faces as to that when they are added together.
Here are my formulas for my forth set of cuboids:
(Each cube added after first cube)+18(towards total of hidden faces)
Or
(Each Face in view)-(total number of faces)=(hidden faces) this formula is based on the original formula I used earlier, which still can be used in this case.
I learnt that a formula that was used in step three would not of worked for this step, as the hidden faces results did not match that of step two and three combined.
Here are my future predictions on a chart:
Forth Set:
Compared to Second set and third set combined:
Here are my future predictions on a chart:
Through observing this chart, I can see that there is a match in the total number of faces after 168, although the hidden faces and faces in view are completely different. The difference in to charts would be because the second and third set is not actually combined to form one shape, they are just two different results added together. The forth set is actually a shape, this explains why there is a match with the total number of hidden faces. This set has once again shown hat the shape itself affects the number of hidden faces and the way it is positioned, not the number of total faces.
My next set of cuboids will be aimed at proving something I have covered in the previous sets, that the more faces facing the base equals more hidden faces. I also to find a formula that doesn’t rely on adding a certain amount of cubes to the end of it for the hidden face formula.
Fifth set of cuboids
15,25,35,45
27,89,211,321
62,122,110
