I have started new work, in which I have to use some math. It is my firs job in engineer environment so I have to get used to it and remind myself some math from school.
I have an linear object, lenght 7700 cm, lets name it pipe, I have smaler objects of lenght 54 cm.
I need to figure out how many of these objects I have to put on line and what distance of themselves for this to be simetrical. Surface of smal vertical object have to start that way, that the lenght from it to the end of the pipe will be 100 cm. I can adjust the end surface of the pipe a little, for example like the item would start 110 cm between the end of it, not 100. The lenght between smaller objects have to be the same and about 1000 cm (more or less).
So the example I have been given: Lenght of linear object : 5200 cm Lenght of the smaller object : 54 cm Distance to the end of pipe from the start of smaller object : 100,5 cm In this example the answer is: distances between surfaces of smaller objects will be 935 cm and there will be 6 of them in total (5 more of them besides the first one that is defined)
the visualisation of the example solution
In my task I have started calculated this by that: 7700(total lenght)-100(the free space left side)-100(the free space right side)-54(the lengt of the first defined small object)=7446
So 7446 will be the total distance of the pipe to be filled by the smaller objects. I have checked that for 7446 the divaders are i.a. 438, 1241, 2482, so I chose 1241 - as a distance between them.
And then after divide 7446/1241 it equals 6. So besides the firs defined one, there have to be 6 more of the smaller objects, with distance between themselves of 1241. But it is wrong, because the total of objects would be 7 so the one of them will be exactly in the middle, and it can not be like that, the distance from the middle of the pipe have to be the same for the end of left object and the start from the right object. Not in the middle.
the visualisation of my task solution
How can I deal with that kind of task? How can i calculate them and figure out where to put things? Why does my calculations are wrong and how to change them to get distance from the middle exactly the same between two items on pipe, not to one of them be in the middle. I hope all of this is ok, and there is a good place, for You guys to advice me a little.
In general you have a pipe (long object) of length $L$, small objects of length $\ell$, the desired distance $d$ from the ends and the desired distance $D$ between small objects and you want the small objects uniformly spaced.
These demands may be slightly contradictory, so let's see how good approximation we can achieve. Let $n$ (necessarily an integer) be the number of small objects you place on the pipe. Consider their centers. Then they are $d+\frac \ell 2$ away from the pipe ends. This leaves $L-2d-\ell$ length for $n-1$ intervals between the centers, so the the distance between the centers is $\frac{L-2d-\ell}{n-1}$, which translates into the distance $\frac{L-2d-\ell}{n-1}-\ell$ between the objects. Thus, you want this number to be as close to $D$ as possible.
If the objects are like supports for the pipe, then $D$ may have the physical meaning of the longest allowed free flight for avoiding the collapse, so you want to approximate $D$ from below rather then from above. I assume that it is the case and the task is to minimize the required number of supports without putting the structure into danger (like if you are building a bridge here).
Then we want the smallest $n$ with $\frac{L-2d-\ell}{n-1}-\ell\le D$, i.e., $n\ge 1+\frac{L-2d-\ell}{D+\ell}$. So just perform the division, round up, and get $n$. The expressions in terms of $n$ for everything have been already written above.
In your example $L=7700,\ell=54, d=100, D=1000$ (in cm). So we want $ n\ge 1+\frac{7700-2\times 100-54}{1000+54}=8.0645\dots $ Now you have a tough choice: to round down to 8 and to be potentially just slightly unsafe, or to round up to 9 and to have an extra support and an object in the middle, which you for some reason do not want (so you will have to raise the number to an even one, i.e., 10, and get two more supports than you would like to). Fortunately, you are allowed to modify $d$ a little bit. So, let's try to keep $n=8$ but reduce the distance between the objects to the maximal allowed one of 1000 cm.
Then 8 objects would take the length $8\times 54+7\times 1000=7432$, so you now have $(7700-7432)/2=134$ instead of $100$ for the distances from the ends of the pipe. If that is acceptable, you are done. If not, well, the question becomes how much lack of safety you want to accept for the gain of the cost of two extra supports.
So that is how more or less how it is played. Note that it absolutely does not matter whether the fractions are integer or not, so your idea to take the divisors of 7446 is totally misleading: what would you do if the number were 7451 instead? It may make sense in the end to adjust the numbers a bit to round them to full millimeters (or half-centimeters) to facilitate the actual measurements when the installation takes place (nobody will measure exactly 996.56784321 cm in real life and you can easily get a fraction like that when dividing), but that is the last step in the story.
Hope that helps a bit. Feel free to ask questions if something is unclear :-)