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How do I do this (Math)

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BUDD

05-20-2006 08:13:29




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I have a suit case weight that weighs 68 lbs. I want it to weigh 50 lbs. How much do I cut off? This weight is 65.125" around the outside, if I take 65.125 / 4 = 16.28125 x 16.28125 = 265.0791 sq in. (is that rite) Now I take 68 lbs / 265.0791 = .2565272 lbs per sq in (is that rite) Now I take .2565272 x 18 lbs (amount I want to cut off) = 4.6174896 sq in to cut off (this just does not sound rite) How do I take this number and put it on the weight so I know how much to cut off? I want to cut this part off the bottom.
BUDD

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Bob and Tom

05-20-2006 19:26:30




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
There are 16 oz to the lb. You want to remove 18 lbs worth of weight.
18*16=288
Take off 288 oz.
Hope this helps.



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T_Bone

05-20-2006 18:22:31




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
Hi Budd,

Easy. Say your weight measures 2" thick x 12" wide at the bottom where you want to cut to be.
Weight of steel .28lbs/ci

2 x 12 = 24ci
24 x .28= 6.72lbs for every 1" of material at the bottom or 1"x 2"x 12" existing

18/6.72= 2.68" to cut off or the drop off piece would be 2.68"x 2"x 12".

math check:
2.68x 2 x 12= 64.32ci x .28lb/ci = 18lbs - 68lbs = 50lbs

Only if you wanted to make a equal cut all the way around would your 65.125" measurement be needed, so it's confusing you, get rid of it.

T_Bone

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old art

05-20-2006 18:04:29




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
maybe you could use a hole saw and cut holes and weight it after each hole you make.wont get any smaller around. or slice it thinner. have fun .
old art



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WVhillbilly

05-20-2006 16:12:12




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
Just start cutting small pieces or whatever you are doing and keep weighing it till you get down to 50 pounds



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BUDD

05-20-2006 15:28:19




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
Ok thanks to all I need to cut off 26 % ish. This is a cast iron suit case weight that is not sq. What I did to get the sq. inches is run a string around the outer edge. The string was 65.125" long. I divided that number by 4 x its self to get the 265.0791 sq in. I think I did this part rite!

Ok now I should cut off 68 sq in to get to the 50 lbs I want. But because this weight has angles on it is not that simple to figure how to get the 68 sq in. What I need to know is at 68 sq in how long would my string be so I can lay it out on the weight? I hope that makes some sense
BUDD

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RAB

05-21-2006 01:10:28




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 15:28:19  
Think about this. If the weight was very shallow, it"s perimeter could be very long indeed!
Circle has a minimum perimeter for a given area. Of quadrilaterals, square has least perimeter for a given area and any rectangular area of that size will have a longer perimeter.
RAB



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BUDD

05-20-2006 15:36:37




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 15:28:19  
Ok I think I just got it. If I take 68 sq in squared that gives me 8.246211 x its self will give me a string length of 32.9848. Is this rite?

BUDD



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bil b va

05-20-2006 15:22:36




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  

a lot simpler to lay the weight on a set of scales and start grinding until you get what you want .



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Not the Genius

05-20-2006 11:56:34




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
Find Area of the weight. 50/68=.735.(weight you want left).735xArea= how much Area you want left, or roughly 3/4 of original. As Cowman says cut 1/4 of the weight off.



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frankiee

05-20-2006 11:39:48




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
I take it that it is an irregular shape. I think that it is cast iron I think cast iron weighs between 6800Kg and 7800Kg
Sorry, my mind only works in metric.

I would put the weight in a container and have the weight attached to a string.
I would mark how high the water level is with the weight submerged.
I would pull the weight out.
I would carefully fill the container back up with water which is measured by liters or gallons or what ever you are use to.
From that you can find how much it weighs per unit (which ever you chose)
For instance, I would use liters being I am a metric person.
I would find exactly how many liters and milliliters it took to fill the container to its original height with the weight in it.
From that I would multiply by 61 to get cubic inches.
That will be how big the weight is in cid.

Now take the weight (68lbs) divided by the cubic inches.
That will tell you how much each cubic inch weights.
With that knowledge you can figure out how much to cut off the square end.

I would still cut twice. Leave some for error.

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RAB

05-20-2006 11:10:40




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
Budd,

You start with:

Mass = volume*density,

so, Volume = Mass/Density.

Your mass is 18lb, so your volume is:

18lbs/density

Now you know the volume you can work out from

Volume = length*width*thickness

to find the unknown (you measure the thickness and the length (from front to back)) you rearrange to find width, so:

Volume/(length*thickness) = width

Now you know from the first equation that the Volume is 18/density, so

(18/density)/(length*thickness) = width

Easy when you do it in simple stages, but make sure all your units are the same (don"t mix feet with inches or metres or anything else)

Regards, RAB

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Stan(PA)

05-20-2006 10:34:13




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
Your calculations don"t look bad, except that you would DIVIDE the 18 pounds by the .2565272 to get square inches. You should cut of 70 square inches. If you cut straight across the bottom, you would cut 1.0625 inches from the bottom, and straight across. This will equate to removing 69.195 square inches, which equates to removing 17.75 pounds. (69.195 X .2565272)



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Stan(PA)

05-20-2006 10:38:58




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 Re: How do I do this (Math) in reply to Stan(PA), 05-20-2006 10:34:13  
Disregard what I said, as I was thinking it was 65.125 along the bottom. Looks like you were talking around the outside of the whole. Makes sense now, as 65 would be a LONG weight :)



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Dachshund

05-20-2006 10:11:00




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
I would start by cutting a "rough" quarter of it off. if you are still a little heavy, then get a good drill bit and "tweak" the weight.



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Bob - MI

05-20-2006 09:52:50




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
Density of iron is around 0.29 lbs/cubic inch.

10 cubic inches= 2.9#

62 cubic inches = 18#



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Midwest redneck

05-20-2006 09:51:58




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
Steel weighs in at .2833 pounds per cubic inch. Knowing this then you can determine how much to remove.



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Richard H.

05-20-2006 08:39:30




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
Budd, If it twas me I would sector it out by marking it in squares and rectangles, then multiply side A to side B = sq. inches and add them all together for your total.



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J Schwiebert

05-20-2006 08:27:47




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
Budd: I did the rough calculation in my head. 50 pounds is close to 70% of 68 pounds. So if the dentisty of the weight is the same No handles bolt holes or any thing you want to cut off close to 30% so that should be between 75 and 80 square inches. It is 16 inches wide at the bottom you want to cut about 4-1/2 to five inches off. Later tonight I get set down with paper and pencil to do this. Good luck!

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Nebraska Cowman

05-20-2006 08:25:45




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 Re: How do I do this (Math) in reply to BUDD, 05-20-2006 08:13:29  
No, you will need to remove approx 25% or 1/4 of the weight. Most guys cut weights out of flat plate and will have an assortment of weights in the 10 pound range. I assume you are pulling? and every scale will weigh different, even the same scale will weigh different on another day.

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