Oddball sort of question - but basic

K

kirkm

New member
Joined
Feb 19, 2012
Messages
8
Hi,

I've spent weeks trying to work this out, as my maths skills are almost non existent. It could be hard even describing this... hope you can help?

There's an area that 557cm wide.
I want to put 5 objects in that area that are equal distant apart and equal distance between first object and start, and last object and end.
Each object is 90 cm wide and there's six gaps between in total

What I'm after is the left value for each object.

The answer is something like 19,124,230,335,and 440 but that's not right.

I'm hoping there may be a formula I can apply to different cases, where the widths and number of objects vary.

Many thanks for any help !

Regards, Kirk
 
Some information is confusing. There are not six gaps between five objects. There are four.

One piece at a time.

557 is the total.

5 objects at 90 is 450, leaving 557 - 450 = 107

Now we have a problem. You stated the first space is equal to the last space. You also stated that the spaces between the objects are equal. You did NOT state that the "between" spaces are equal to the start and stop spaces. Is this your intent?

If so, we have six spaces (four between and two outer), so 107/6 = 17.833... You WILL have trouble measuring that.

Start: 0

End of first space: 17.8333...
End of first object: 107.8333...

End of second space: 125.6666...
End of second object: 215.6666...

And so forth. If you are careful, you should get

End of last space: 517
 
tkhunny said:
Some information is confusing. There are not six gaps between five objects. There are four.
Click to expand...

Yes, but +2 at the start and end

>One piece at a time.

>557 is the total.

>5 objects at 90 is 450, leaving 557 - 450 = 107

>Now we have a problem. You stated the first space is equal to the last space. You also stated that the spaces between the objects are equal. You did NOT state that the "between" spaces are equal to the start and stop spaces. Is this your intent?

Yes, to have all spaces the same between each object

>If so, we have six spaces (four between and two outer), so 107/6 = 17.833... You WILL have trouble measuring that.

I'll say :D

>Start: 0

>End of first space: 17.8333...
>End of first object: 107.8333...

>End of second space: 125.6666...
>End of second object: 215.6666...

>And so forth. If you are careful, you should get

>End of last space: 517

Not following you 100%... but working on it!
 
Last edited:
Hello, kirkm!

The problem is neater if that width is 558 cm.
I don't suppose you made a sketch . . .

There's an area that 558 cm wide. I want to put 5 objects in that area that are equal distant
apart and equal distance between first object and start, and last object and end.
Each object is 90 cm wide and there are six gaps between in total.

What I'm after is the left value for each object.

The answer is something like 19, 124, 230, 335,and 440, but that's not right.

I'm hoping there may be a formula I can apply to different cases,
. . where the widths and number of objects vary.
Click to expand...

I'm guessing that the situation looks like this:

\(\displaystyle | - - - \bigcirc - - - \bigcirc - - - \bigcirc - - - \bigcirc - - - \bigcirc - - - | \)

The total width is 558 cm.
The objects are 90 cm across; they take up 450 cm.
That leaves: \(\displaystyle 558 - 450 \,=\,108\) cm for the six gaps.
Then each gap is: \(\displaystyle 108 \div 6 \,=\,18\text{ cm wide.}\)

The "left values" are: 18, 126, 234, 342, 450.


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We can create a formula for this problem.

Let \(\displaystyle W\) = width of the entire area (cm).

There are \(\displaystyle n\) objects with a diameter of \(\displaystyle d\) (cm).

Then there are: .\(\displaystyle W - nd\) cm left for the gaps.

Each gap will be: .\(\displaystyle g \:=\:\dfrac{W-nd}{n+1}\) cm wide.

The "left values" will be:
. . . . \(\displaystyle g,\;d+2g,\;3d+3g,\;3d+4g,\;\hdots\;(n-1)d + ng \)
 
Soroban, that's amzing ! That formula makes me feel so dumb - and that's just looking at it! :eek:

I think - from schooldays years ago - that 'nd' is actaully n x d. - and 3g is 3 x g ?

If so, I may be able to figure out the left values....

Denis, your example was spot on - but you never said how to get X to try it out. Or did I miss something...

Everyones help and responces are much appreciated. I'll report back how I got on.
 
Hi Soroban

I have it ! Well, you have it.... but I can emulate it !

Code: 
w = 558
d = 90
n = 5

tmp = w - (n * d)
g = tmp / (n + 1)

Debug.Print g
Debug.Print d + (2 * g)
Debug.Print (3 * d) + (3 * g)
Debug.Print (3 * d) + (4 * g)
Debug.Print (n - 1) * d + (n * g)

My answers are (as you said)
18
126
324
342
450

and if I use 557
17.8333333333333
125.666666666667
323.5
341.333333333333
449.166666666667

How did you work out to change the 557 to 558 ?
 
You must log in or register to reply here.
Top