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how much hose?
- Thread starter rfcomm2k
- Start date
Assume you have a hose 150' long 1" diameter.
Next, assume a reel 18" diameter outside and 4" diameter inside.
How wide would this reel need to be for all 150' of hose to fit?
Assuming there were no 'winding waste', the first layer would contain x\(\displaystyle \pi\)d length of hose where d is the initial inside diameter and x is the width of the reel(diameter and x in inches). The next layer would contain x\(\displaystyle \pi\)(d+1) [the 1 in d+1 represents the width of the hose] and the next would contain ... Add all of these up and set it greater than or equal to the length of the hose (don't forget those unit measurements).
BTW, in real life there is always 'winding waste' so if you were doing this to actually build something the answer would be different.
Edit: For a little more clarity
Last edited:
re how much hose
So in this case x = the 1" hose width? Or does x = the width of reel, which is the unknown I am trying to figure out to begin with?
Assuming there were no 'winding waste', the first layer would contain x\(\displaystyle \pi\)d length of hose where d is the initial inside diameter and x is the width (diameter and x in inches). The next layer would contain x\(\displaystyle \pi\)(d+1) and the next would contain ... Add all of these up and set it greater than or equal to the length of the hose (don't forget those unit measurements).
BTW, in real life there is always 'winding waste' so if you were doing this to actually build something the answer would be different.
So in this case x = the 1" hose width? Or does x = the width of reel, which is the unknown I am trying to figure out to begin with?
D
Deleted member 4993
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So in this case x = the 1" hose width? Or does x = the width of reel, which is the unknown I am trying to figure out to begin with?
What do you think? Read Ishuda's response again - and draw a sketch for the first layer of the "winding". If you still cannot get it - please ask back.
What do you think? Read Ishuda's response again - and draw a sketch for the first layer of the "winding". If you still cannot get it - please ask back.
I think Ishuda and you should both go back and read my original post. I did not ask for how much hose a given length of 4" cylinder would hold. I KNOW how much hose I have. I asked HOW LONG THE CYLINDER NEEDS TO BE TO ACCOMODATE THE KNOWN LENGTH OF HOSE.
Using the formula provided I know I need 5 layers with a 16" wide reel. But whatif I wanted to use a 10" wide reel. Again, the solution provided got me to the answer needed, just not the way it was asked.
I think Ishuda and you should both go back and read my original post. I did not ask for how much hose a given length of 4" cylinder would hold. I KNOW how much hose I have. I asked HOW LONG THE CYLINDER NEEDS TO BE TO ACCOMODATE THE KNOWN LENGTH OF HOSE.
Using the formula provided I know I need 5 layers with a 16" wide reel. But whatif I wanted to use a 10" wide reel. Again, the solution provided got me to the answer needed, just not the way it was asked.
I did modify my original post to indicate that x was the width of the reel and that the 1 in d+1 was the diameter of the hose. This was to answer your question about x. Sorry it wasn't clear to you.
I'm also sorry you can't follow the hint given you and feel you must attack the people trying to help. I did not answer the question of how much hose a given length of 4" cylinder would hold except in the case of providing step one in the answer to the question you asked.
I will restate my original post (hopefully) more plainly so that you might understand. Assuming you wanted the complete reel used, all the way out to the outside diameter of eighteen , you would have 14 wrappings, each 1 inch (the diameter of the hose) in diameter larger than the previous one. Thus the total length L of the wrapping would be (remember the add them up in the original post of mine?)
L = x \(\displaystyle \pi\) [d + (d+1) + (d+2) + ... + (d+13)]
where I have used d=5 for the first winding (the outside of the hose rather than the inside). If you do compute the value of the series (as a function of x of course), you will have a minimum width for the reel to contain the complete hose.
Several other comments: If windings 7 through 14 were all of the length of winding 7, the total of just those eight windings would be about 238" and you could store a 150' (1800") hose with a reel width of less than about 8". So, although your 16" reel width would work, but it is bigger than needed. But then you changed the problem, didn't you? Only using 5 layers instead of the 14 as I have above. Or maybe we should have said something like the question isn't answerable as it is asked?
If you don't like 5 for d, pick your own number but you might have to adjust that last value depending on the value you use. If you want to use three layers or a 10" width reel or a different length hose or whatever within the frame work of this problem, I hope the example is sufficiently clear so that you could work those problems. But, if not, you might not get anymore help until you can read the posts that try to help you and ask intelligent questions rather than blaming someone else for your short comings.