Calculating the tilt of a floating box?

1. Apr 21, 2004

DanishStudent2

Does anyone know how to calculate the angle at which the box will tilt, if the center of mass is above the center of buoyancy?

Suppose we have a box with the dimensions:
W:15.4 inches
L:27.4 inches
H:27 inches
With the weight 2796,5 grams
Center of mass is 8 inches from the bottom of the box.

I´ve calculated the depth of the box to 6,6407 inches

I can´t figure out how to calculate the angle at which the box will tilt.
Any ideas

2. Apr 21, 2004

Gokul43201

Staff Emeritus
The tilt angle is that which makes the CoB and the CoM in the same vertical line. However, the position of the CoB depends on the tilt, so this is not so easy, in general. In this particular case though, it's not hard.

Before you figure this out, you only know the volume fraction that is submerged, not the actual depth. But some of your numbers don't check out...the CoM should be 7.8 inches from the bottom - okay that's close enough to 8. But the submerged volume fraction (which is equal to the density in g/cc) I get is about 1.5 %, a lot less than your depth suggests.

When objects are so light (density << 0.5g/cc) that most of the volume is above water, they like to lie flat on the largest face, so that the potential energy of the box is minimised.

So there is no tilt in this case, but there would have been a tilt if the box were dense enough that most of it was submerged.

3. Apr 21, 2004

jepe

For floating objects you need to know another point to check if the object has a positive stability (floats without list). this is the Metacentric height.
Here is the solution in metric units:
W = width = 39.1 cm
L = length = 69.6 cm
H = height = 68.6 cm
T = depth of the box in water = weight / (W*L) = 2796.5 / 2721.4 = 1.0 cm
(and not 6.6407 inches!!!)
assume vertical line through middle of the box with the following points:
K = bottom of the box
B = centre of buoyancy, distance KB = 0.5 cm (half of depth)
G = centre of mass = 20.3 cm above K
M = Metracentric height
distance MB = I/V, where I = waterline inertia, V = volume of box in water
with I = W^3*L/12 and V = L*W*T
MB = W^2 / (12*T) = 127.4 cm
distance MG = MB+MK-GK = 127.4+0.5-20.3 = 107.6 cm
Metracentre is above centre of mass, hence box is floting stable,
hence NO LIST.

4. Apr 23, 2004

DanishStudent2

Thanks for the answers, I have just realized that the measures that I gave you earlier, wasn´t in inches but in metric units.

So my box, would be:
W: 15.4 cm
L:27.4 cm
H:27 cm

and the weigth 2796.5 grams

The CoB was measured, to approximately 8 cm from the bottom of the box.

I calculated the depth to 6,6407 cm.

Please would you try to help me with this, I need some kind of generel formulae to calculate, the tilt degree. Can anyone explain this to me, by using this example?

5. Apr 23, 2004

jepe

OK, let's do it with the new numbers:
For floating objects you need to know another point to check if the object has a positive stability (floats without tilt). this is the Metacentric height.
Here is the solution:
W = width = 15.4 cm
L = length = 27.4 cm
H = height = 27.0 cm
T = draft of the box in water = weight / (W*L) = 2796.5 / 421.96 = 6.62 cm

assume vertical line through middle of the box with the following points:
K = bottom of the box
B = centre of buoyancy, distance BK = 3.31 cm (half of draft)
G = centre of mass = 8.0 cm above K
M = Metracentric height
MB = I/V, where I = waterline inertia, V = volume of box in water
with I = W^3*L/12 and V = L*W*T
MB = W^2 / (12*T) = 2.99 cm
MG = MB+BK-GK = 2.99+3.31-8.0 = -1.70 cm
Negative MG means metacentre below centre of gravity, which means instable condition.
This means that box will float under a tilt angle
This tilt angle can be calculated using Scribanti's law.
If the box floats under a list, the metacentre point M moves upward to a point N, where N is called the false metacentre.
The false metacentre depends on the tilt angle phi in the following way:
NB = MB*(1+0.5*tan^2(phi)) (Scribanti's law)
List angle phi will assume a value so that N and G fall together.
Knowing this, phi can be calculated as follows:
NB = MB+1.70 = 2.99+1.70 = 4.69 cm
4.69 = 2.99*(1+0.5*tan^2(phi))
hence: tan^2(phi) = 1.14
tan(phi) = sqrt(1.14) = 1.07
phi = atan(1.07) = 46.9 degrees
Scribanti' law is only valid for vertical sides of the box (which is the case) and as long as the bottom corner remains below water. Since it is a large angle we have to check this!!!
Bottom corner comes above water if tilt angle is larger than atan(T/(0.5*W), which is 40.8 degrees.
So, unfortunately for you, Scribanti's law give not the right answer, but it is an approximation. If the corner comes above water, the mathematics to determine the correct tilt angle are far more complicated.

6. Apr 24, 2004

DanishStudent2

Thanks to everyone!

Thanks Jepe, that´s exactly what I needed..

Super, know I can get on with my project.

7. Apr 28, 2004

DanishStudent2

By the way...

Do anyone know where I can get some sort of information about Scribanti´s law? I can´t figure out, why this formulae works??

8. Apr 28, 2004

jepe

The basics of the Scibanti Formula can be found in:
http://www.ocp.tudelft.nl/mt/journee/Files/Lectures/OffshoreHydromechanics_Intro.pdf [Broken]
Note that in this pdf file: B = breadth which corresponds to W = width in your thread
B is also centre of buoyancy (bit confusing, but should be clear in formulaes)

Using the nomenclature in figure 5.2 in the .pdf file the Scibanti Formula is explaned as follows:
The vertical shift zi' zi causes a vertical displacement of the centre of buoyancy from BΦ' to BΦ which is equal to the distance from M to the false metacentre NΦ. Hence:
M NΦ = BΦ' BΦ
The vertical buoyancy force, ρgV, intersects the vertical buoancy force at angle Φ= 0 degrees at the metacentre NΦ
The first moment of volumes with respect to a plane parallel to the not-heeled water plane through the centre of buoyancy B of the barge in the not-heeled condition is given by:
{LBT}.{ M NΦ} = 2.{(L/2).(B/2).(B/2).tan Φ} . {(1/3).(B/2).tan Φ}

(refer also equation 2.8 in pdf file)

so that:

M NΦ = {(1/12).L.B^3)}/{LBT} . {(1/2). Tan^2 Φ}

With {(1/12).L.B^3)}/{LBT} = I/V = MB

B NΦ = BM + MNΦ

B NΦ = BM {1+(1/2). Tan^2 Φ} (Scribanti's Law)

Last edited by a moderator: May 1, 2017
9. Apr 28, 2004

jepe

sorry, figure 5.2 should be figure 2.5

10. Apr 29, 2004

DanishStudent2

Thanks

Thanks again Jepe.

I think I´ve got it now..

11. May 6, 2004

DanishStudent2

Okay, so far so good.

I think I´ve got the Scribanti law covered now, but I can´t seem to understand how you conclude that the new metacentric height is
NB = MB+1.70 = 2.99+1.70 = 4.69 cm

where 1.70 = is MG = MB+BK-GK

If I draw a box with the points, M, G, B, K and of course the new metacentric height, I get the NB to be the distance between G and B.

Can you explain this..?

Soren

12. May 6, 2004

jepe

Referring to the previous pdf file:http://www.ocp.tudelft.nl/mt/journee/Files/Lectures/OffshoreHydromechanics_Intro.pdf [Broken]
The stability chapter is referring to ships and barges which are designed to always have a positive stability. A positive stability means that the metacenter M is always vertical ABOVE the centre of gravity G. (refer figure 2.5.) For ships and barges it is interesting to know what the tilt is when a large moment is working on the vessel, e.g. a lateral wind force, or a large weight hoisted by a vessel (refer figure 2.6.). If this external moment is Mh, the tilt will be given by equation 2.7, assuming the tilt angle is small. If the tilt angle is large, the Scribanti formula is required.
In your case of the box, the metacentre M is BELOW the centre of gravity G, hence there is no initial stability. If you put the box in the water with zero tilt and you release the box, the tilt of the box will increase. With increasing tilt, the metacentre M will move up to the false metacentre N. This will continue until stability is reached, hence until N has reached the centre of gravity G.
Only then the box floats stable in the water.
Therefore, you quite rightly conclude that in that case NB is also the distance between G and B.
Hope this explanation will do?

Last edited by a moderator: May 1, 2017
13. May 10, 2004

DanishStudent2

Thanks

All right, my bad, didn´t realize the thing about positive stability.. Thanks

14. May 13, 2004

DanishStudent2

Problems with the formulae, in case the weight increases..

Hi again..

Allright, now I checked the angle 47, degrees, in a wave Laboratory, this seems correct. I wrote a MathLab program, that should be able to calculate this angle. The program gives me the right angle, but as I tried adding more weight at the exact point of gravity the formula, gave me some sort of complex angle...???

Here´s the data:

INPUT:
Height 27.4 cm
Width 15.4 cm
Length 27 cm
Center of gravity: 8 cm.
Weight: 5687 grams

My OUTPUT:
The depth: 13.70 cm
Metacenter Height: 0.294 cm
Distance BM = 1.4421 cm
And the angle which puzzles me the most: 0+43.3432i degrees.

I don´t know If you have any experience in programming, but heres the code:

Phi = (atan(sqrt((((BNO/MO)-1)/0.5))))*(180/pi);

where BNO = The distance From gravity center to center of buoancy.
And MO = I/V

IT´s obvious that as long as BNO/MO<1, the angle isn´t defined because of the sqrt.

Can you figure out, what I have done wrong??

15. May 13, 2004

jepe

Scribanti's formula, as you want to use it, only works if you have a negative stability, hence if the metacentre is below the centre of gravity.
If you add weight to the exact point of the centre of gravity until you have a depth of 13.7 cm, the metacentre is above the centre of gravity, in which case, by definition, you have a positive stability, and the tilt angle is than always zero degrees (assuming no external moment on the box)
So, in your nomenclature, a positive stability means that BNO/MO<1.
In that case the tilt angle is always zero, and your formula to calculate the tilt angle is no longer valid. The formula you wrote to calculate phi, by the way, looks OK to me.

16. May 18, 2004

DanishStudent2

But....

I tried to use the exact specifications of the box, and when I used it in the lab., it was obvious that it had a tilt degree of approx. 20 degrees.

But according to you, It would have no list. But isn´t it so that the metacenter height, becomes more and more stable the longer the distance? And my metacenter height is no more than 0.29 cm, which probably isn´t the most stable condition?

17. May 18, 2004

jepe

You are absolutely right, a metacenter height of just 0.29 cm is certainly not the most stable condition. I understand that you have a box with the dimensions you specified and you actually put it in the water.
If everything is absolutely correct, the positive metacenter should result in zero tilt. What in your case might have happened is that the centre of gravity is not exactly on the vertical centerline.
Refer Figure 2.5 of the .pdf file, in your case the "centre of gravity" might be located in point z. But because of the very small metacentre height point z is only 0.29*sin 20 deg = 0.10 cm off the centerline, which is only 1 mm. This is very difficult to measure on your box, a think.
Another check to verify if this is the case is that the box will always tilt 20 degrees to the SAME side. Even if you force the box to give a small angle to the other side, it goes back to the original side.
If the tilt is caused by instability (i.e. a negative metacenter) the box will find an equilibrium tilt at either side of zero degrees.
I am interesting to know what the problem was.

18. May 19, 2004

DanishStudent2

All-right...

The problem in our case was that the center of gravity, was measured with some insecurity. This affected, the point, at which we placed the weight, and therefore the metacenterheight became negative. Therfore the tilt angle became complex. The box had absolutely no problem with finding a stable condition on each side of 0.

Thanks for the help.
Soren