Cutting corner

[TimeSpaceLightForce]

How can I fabricate (cut,bend,weld) the 1m x 1 m metal sheet into an open top box to hold 75 liters of liquid?

[witzar]

  Reveal hidden contents

1. Cut the sheet into 5 congruent rectangles.
2. Cut&weld each rectangle into a square.
3. Weld 5 squares together into an open top box that can hold (1/5)^(3/2) = 0.0894 m^2 of liquid (more than 89 liters).

[TimeSpaceLightForce]

  On 8/7/2013 at 10:48 AM, witzar said:

  Reveal hidden contents

1. Cut the sheet into 5 congruent rectangles.

2. Cut&weld each rectangle into a square.

3. Weld 5 squares together into an open top box that can hold (1/5)^(3/2) = 0.0894 m^2 of liquid (more than 89 liters).

That is right! Nice 1

  Reveal hidden contents

But the maxima solution and the rectangle to square method are under and over design for fabrication.

The 9 grids solution is enough for the problem.

For the best.. i got the idea from witzars solution..

Edited August 7, 2013 by TimeSpaceLightForce

[TimeSpaceLightForce]

  Reveal hidden contents

Im trying to edit the weld value of Greek cross solution to 3.788m , this make it cheaper for 9 grids solution to fabricate

note: pls hide last figure

Edited August 7, 2013 by TimeSpaceLightForce

[kedes]

  Reveal hidden contents

Roll sheet into a cylindar that will have the volume of 1 meter times area of opening (Circumference of 1 M)

[Anza Power]

  Reveal hidden contents

My instincts tell me that half a sphere would give you best results.

In this case if you can make half a sphere with area of 1 M² you'd have space for 94 liters.

But since you can't make a perfect sphere by cutting and welding, you can try to make it close to a sphere by making a polygon of the highest order as you can, the higher the order the more pieces you'll have to cut and weld

[TimeSpaceLightForce]

  On 8/7/2013 at 7:22 PM, kedes said:

  Reveal hidden contents

Roll sheet into a cylindar that will have the volume of 1 meter times area of opening (Circumference of 1 M)

  Reveal hidden contents

1)For height=circumference:no base

V=h (a) volume

h=C=1 height

C=piD=1 circumference

D=2r=(1/pi) diameter

r=1/2/pi radius

a=pi(r )^2 circle area

V=(1)(pi)(1/2/pi)^2

V= 0.07958

2) For height=diameter :no base

a=pi( r)^2 circle area

D= 2r=(C/pi) diameter

r=C/pi/2 raduis

A=1=2rC cylinder area

C=1/2/r circumference

r=(1/2/r)/pi/2

r=0.28209

V=2r(pi)r^2 volume

V=0.14105

good for holding marbles

[TimeSpaceLightForce]

  On 8/7/2013 at 7:43 PM, Anza Power said:

  Reveal hidden contents

My instincts tell me that half a sphere would give you best results.

In this case if you can make half a sphere with area of 1 M² you'd have space for 94 liters.

But since you can't make a perfect sphere by cutting and welding, you can try to make it close to a sphere by making a polygon of the highest order as you can, the higher the order the more pieces you'll have to cut and weld

  Reveal hidden contents

1) volume of sphere=4/3 (Pi) rrr

area of sphere=4(pi)rr

1=4(pi)rr

r=0.28209

V=0.09403

2) volume of half sphere=2/3 (Pi) rrr

area of half sphere=2(pi)rr

1=2(pi)rr

r=0.39894

V=0.13298

[Anza Power]

  On 8/9/2013 at 7:37 AM, TimeSpaceLightForce said:

  On 8/7/2013 at 7:43 PM, Anza Power said:

  Reveal hidden contents

My instincts tell me that half a sphere would give you best results.

In this case if you can make half a sphere with area of 1 M² you'd have space for 94 liters.

But since you can't make a perfect sphere by cutting and welding, you can try to make it close to a sphere by making a polygon of the highest order as you can, the higher the order the more pieces you'll have to cut and weld

  Reveal hidden contents

1) volume of sphere=4/3 (Pi) rrr

area of sphere=4(pi)rr

1=4(pi)rr

r=0.28209

V=0.09403

2) volume of half sphere=2/3 (Pi) rrr

area of half sphere=2(pi)rr

1=2(pi)rr

r=0.39894

V=0.13298

Yes, the volume is in units of meter³, 1 m³ = 1000 liters.