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Sep 28, 2011, 4:05:55 AM9/28/11

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A friend posted this on Facebook and thought it was interesting enough

to post here:

--------------------------------------------------------------------

To be published in my Titanic book, due out next month.

( Based on information contained in http://www.rmstitanicremembered.com/?page_id=282

)

Revisionist calculations argue that the Titanic would have been

shredded from bow to stern; the

main proponent of this is author Daniel Butler, and the calculations

from his website

deserve to be quoted in full:

"Working in SI units, calculating power generated by ship movement and

using the figures I gave above, 55000 tonnes (note, I’m using metric

tonnes, 2200lbs, not long tons, 2240 lbs) and velocity of 22 knots = …

22 x 1850 (metres in a nautical mile) / 3600 (seconds in 1 hour) =

55,000,000Kg x 9.81 (newtons per kg) x 11.3 m per sec = 6096,915,000

watts, or 6096 Mw.

The force generated, assuming a dead stop, would be mass(Kg) x

velocity(m/sec) x 9.81(grav force, newtons/kg)/ 9.81 (Newtons/kg), or

55000000 x 11.3 x 9.81/ 9.81 = 651,500,000 kg.= 651,500 tonnes.

By comparison, a 1 ton car travelling at 60 mph (96 km/hr) would be:

1000 kg x 9.81 (g) x (96 x 1000/3600) = 1000 x 9.81 x 26.6 = 261600

watts, or 261.6 Kw, and the force generated in coming to a dead stop

would be 1000 kg x 26.6 m/sec = 26600 Kg, or 26.6 tonnes.

Quite a difference. Carbon steel has a maximum strength of 56,000 psi

in tension and compression but a maximum shear strength of only 42,000

psi so the steel would have an shear strength of 800 bar (42000/15).

If the steel was 1.875 (from http://www.tpub.com/content/construction/14250/css/14250_14.htm)

a 1 metre length could withstand a shear impact of 800 / 1 x 0.018 =

44444 newtons, = 4530 kg, or 4.53 tonnes."

Since this figure of 44444 Newtons is less than the force generated by

the impact, the Titanic, we are told, would have been ripped apart

from prow to stern. Based on the numbers, this conclusion isn't

correct.

The ensuing discussion will revolve around the units used by the

scientific community (Metres and Newtons),

termed as SI units. Sadly, this is mathematically intensive, but this

treatment has been made as simple as possible.

The first point of contention is that the author of this analysis has

calculated incorrectly. He gives an answer of 651,500,000 kg as the

"force" when it should actually be 621,500,000 Newtons. Not only has

he calculated wrongly, but he mixes up "mass" (kg) with

"force" (Newtons) - a mistake that no scientist would make.

He also implies that the weight of the ship was 55000 tonnes. Her

actual displacement weight, a measure

of the actual weight of the ship, is dependant on how heavily loaded a

ship was at the time, but

a figure of 53147 tonnes is generally accepted. This is about 3% less

than the number Butler quotes, but we

shall use his numbers to allow a fair comparison.

Then we are told that steel has a maximum shear force of 42,000 psi,

and that this is less than highest compression and tension values. But

he has mixed up the forces. Shear forces exist when two opposing sides

of a block

are forced in opposing directions, thus transforming a rectangle into

a parallelogram. With the Titanic

racing head-long into an iceberg, we would actually have a compressive

force acting on the plates; the ship

heading forwards, and the iceberg acting to resist this forward

motion, squeezing the plates, rather than

shearing the two sides parallel to the motion in opposing directions.

The value of "1.875" is presumably the thickness of the hull plate.

Ironically, the web reference that Butler uses next says, "therefore,

when dealing with maximum strength, you should always state the type

of loading," ironic considering the confusion.

The power is also computed wrongly. The force on the ship due to

gravity (that is "the weight"), is not acting

in the same direction as the force driving the ship forward, so it is

incorrect to say that the power is equal

to the weight multipled by the velocity. The weight is given by the

mass times the acceleration due to

gravity (9.81 metres per second per second) and 9.81 is NOT the number

of Newtons per kilogramme; this is the conversion between mass and

weight, which is a force acting straight down towards the centre of

the Earth and not in the direction of the Titanic's motion. In fact,

the whole concept of power is a red herring, and should be ignored.

"Quite a difference" between a car and a huge ocean liner? Of course;

one wonders why this stated,

it is obvious.

Then the equations are used incorrectly. Force is the rate of change

of momentum (which itself is the multiple of mass and velocity), not

just the change of momentum. One needs to know over how long this

change happened.

A perfectly linear decelaration to a perfect stop over, say, 10

seconds, would enable us say that

force is the momentum divided by 10. It isn't quite as simple as this;

with the iceberg tearing through

the hull, resistive force would be applied, hastening the

deceleration. This resistance could be from

steel plates accumulating in front of the 'berg, or from internal

structures within the ship, hastening

the slowing down process. Then Mr.Butler says that the figures assume

"a dead stop." The ship simply wouldn't

(possibly couldn't) stop instantaneously. Incidentally, his "force

generated" implies a duration for this

of 1 second, or change of momentum/1 second. If a dead stop were to

occur immediately, the forces involved

would be so astronomical as to be unimagineable. It would, succinctly

put, be the same as the Titanic running

at full speed into an unyielding object, like a cliff face. The

iceberg, while massive, would be subject to the usual laws of motion,

including the conservation of momentum, and would not be an object

incapable to movement.

The next set of calculations are also used incorrectly. Mr.Butler says

that 42000 psi is the equivalent of 800

bar. He assumes that dividing by 15 converts these numbers from one

set of units to another (the actual

conversion is about 14.7 since this is atmospheric pressure). At any

rate, using 14.7 - or 15 - gives a figure

of about 2800. No doubt Mr.Butler made a typographical mistake here,

as he says "800", or a factor of 3.5 out,

but we must be dubious as this works in his favour.

Next, we come to the coup de grace: calculating the force from the

stress. To do this, one needs

to know the surface area to which the force is applied. For a shear

stress, this area is parallel to

the direction in which the force is applied. This area is actually the

length of the plate multiplied

by its height since this is where the force is applied, and not, as he

implies, edge-on.

Then Mr.Butler mixes up his non-SI and SI units when he says "800 / 1

x 0.018"; 800 (an inaccurate

figure) is non-SI, and the next two, and the answer are in SI. This is

not trivia. As the old adage goes, its

like comparing apples and oranges. In this case, we end up with

bananas.

So, calculating the numbers correctly, we find that 42000 psi is 290

Million Newtons per square metre.

So, now we have the pressure, or the force acting on an area. To find

the force, we multiply

the area on which this pressure is applied - the surface area of the

hull plate. We do NOT, as Mr.Butler

has done, divide the pressure by the area. This leads to a meaningless

number, as pressure is already

force divided by area.

Estimating the area of the hull plate one uses figures gleaned from

"Titanic: The Ship Magnificent";

we use a length of 36 feet (11 metres) and a height of 6 feet (1.8

metres). The maximum force that could

be applied until failure occurs is therefore 290 Million Newtons x 11

x 1.8 = 5700 Million (Mega) Newtons.

Compare that to the 622 Million Newtons that Mr.Butler would tell us

is imparted by the impact. Using these

figures, it is clear that the plates would deform, but not break.

The danger of using Mr.Butler's figures is that he has taken one plate

and extrapolated to form

a conclusion upon which lesser scientific figures have fallen.

In practise trying to prove a failure scenario would need a model that

had an interior spaces containing beams,

columns, bulkheads and corridors would add resistive strength to the

iceberg crushing through the hull,

slowing its progress as it proceeded.

The plates would also buckle.

It is not unlikely the berg would be stopped well before 216 feet

which is a "worst case scenario." We also

do not know exactly how the rivets holding the seams would fare. Would

they pop before a siseable fraction

of compressive damage was reached, or would they only yield at the

last second? And how easily would the

compressive force be transferred from plate to plate? It is not a

simple matter of taking one plate and guessing;

rigourous and intensive computation modelling is necessary.

It should also be noted that the 56,000 and 42,000 psi values no doubt

apply to modern steel, and not the

material used ten decades ago.

Free from any revisionist posturing, we must remember what Edward

Wilding, a naval architect at Harland

and Wolff said regarding the stem-on collision proposal: "...she would

have crumpled up in stopping herself. The momentum of the ship would

have crushed in the bows for 80 or perhaps 100 feet. [The impact would

not be] a shock, it is a pressure that lasts three or four seconds,

five seconds perhaps, and whilst it is a big pressure it is not in the

nature of a sharp blow."

It should be pointed out, in 1879, a ship named the Arizona collided

head-on with an iceberg at a speed of 15 knots, and she survived, with

her bows "broken and twisted." Reportedly, she was left with a hole

"20 feet broad by 30 feet deep."

It may be incorrect to compare the Arizona and the Titanic; the afore

named ship was only of some 5600 tons, but it

is not clear what tonnage this refers to, as ships had gross,

displacement, deadweight etc. tonnage. Interested

readers will find more information on David Gittin's titanicebook.com

website.

Other examples of stove-in bows exist but their numbers do not provide

any meaningful comparison: the Stockholm, after the Andrea Doria

Collision had a huge segment of her bow missing but only her speed

(about 18 knots)

is known at the time of the collision; her gross tonnage of about

12200 tonnes does not allow meaningful comparsion.

While it is clear that Wilding's discussion that the Titanic might

have survived was conjecture, with no

calculations to reinforce his argument, the simple argument is that,

spurious physics aside, we do not know.

to post here:

--------------------------------------------------------------------

To be published in my Titanic book, due out next month.

( Based on information contained in http://www.rmstitanicremembered.com/?page_id=282

)

Revisionist calculations argue that the Titanic would have been

shredded from bow to stern; the

main proponent of this is author Daniel Butler, and the calculations

from his website

deserve to be quoted in full:

"Working in SI units, calculating power generated by ship movement and

using the figures I gave above, 55000 tonnes (note, I’m using metric

tonnes, 2200lbs, not long tons, 2240 lbs) and velocity of 22 knots = …

22 x 1850 (metres in a nautical mile) / 3600 (seconds in 1 hour) =

55,000,000Kg x 9.81 (newtons per kg) x 11.3 m per sec = 6096,915,000

watts, or 6096 Mw.

The force generated, assuming a dead stop, would be mass(Kg) x

velocity(m/sec) x 9.81(grav force, newtons/kg)/ 9.81 (Newtons/kg), or

55000000 x 11.3 x 9.81/ 9.81 = 651,500,000 kg.= 651,500 tonnes.

By comparison, a 1 ton car travelling at 60 mph (96 km/hr) would be:

1000 kg x 9.81 (g) x (96 x 1000/3600) = 1000 x 9.81 x 26.6 = 261600

watts, or 261.6 Kw, and the force generated in coming to a dead stop

would be 1000 kg x 26.6 m/sec = 26600 Kg, or 26.6 tonnes.

Quite a difference. Carbon steel has a maximum strength of 56,000 psi

in tension and compression but a maximum shear strength of only 42,000

psi so the steel would have an shear strength of 800 bar (42000/15).

If the steel was 1.875 (from http://www.tpub.com/content/construction/14250/css/14250_14.htm)

a 1 metre length could withstand a shear impact of 800 / 1 x 0.018 =

44444 newtons, = 4530 kg, or 4.53 tonnes."

Since this figure of 44444 Newtons is less than the force generated by

the impact, the Titanic, we are told, would have been ripped apart

from prow to stern. Based on the numbers, this conclusion isn't

correct.

The ensuing discussion will revolve around the units used by the

scientific community (Metres and Newtons),

termed as SI units. Sadly, this is mathematically intensive, but this

treatment has been made as simple as possible.

The first point of contention is that the author of this analysis has

calculated incorrectly. He gives an answer of 651,500,000 kg as the

"force" when it should actually be 621,500,000 Newtons. Not only has

he calculated wrongly, but he mixes up "mass" (kg) with

"force" (Newtons) - a mistake that no scientist would make.

He also implies that the weight of the ship was 55000 tonnes. Her

actual displacement weight, a measure

of the actual weight of the ship, is dependant on how heavily loaded a

ship was at the time, but

a figure of 53147 tonnes is generally accepted. This is about 3% less

than the number Butler quotes, but we

shall use his numbers to allow a fair comparison.

Then we are told that steel has a maximum shear force of 42,000 psi,

and that this is less than highest compression and tension values. But

he has mixed up the forces. Shear forces exist when two opposing sides

of a block

are forced in opposing directions, thus transforming a rectangle into

a parallelogram. With the Titanic

racing head-long into an iceberg, we would actually have a compressive

force acting on the plates; the ship

heading forwards, and the iceberg acting to resist this forward

motion, squeezing the plates, rather than

shearing the two sides parallel to the motion in opposing directions.

The value of "1.875" is presumably the thickness of the hull plate.

Ironically, the web reference that Butler uses next says, "therefore,

when dealing with maximum strength, you should always state the type

of loading," ironic considering the confusion.

The power is also computed wrongly. The force on the ship due to

gravity (that is "the weight"), is not acting

in the same direction as the force driving the ship forward, so it is

incorrect to say that the power is equal

to the weight multipled by the velocity. The weight is given by the

mass times the acceleration due to

gravity (9.81 metres per second per second) and 9.81 is NOT the number

of Newtons per kilogramme; this is the conversion between mass and

weight, which is a force acting straight down towards the centre of

the Earth and not in the direction of the Titanic's motion. In fact,

the whole concept of power is a red herring, and should be ignored.

"Quite a difference" between a car and a huge ocean liner? Of course;

one wonders why this stated,

it is obvious.

Then the equations are used incorrectly. Force is the rate of change

of momentum (which itself is the multiple of mass and velocity), not

just the change of momentum. One needs to know over how long this

change happened.

A perfectly linear decelaration to a perfect stop over, say, 10

seconds, would enable us say that

force is the momentum divided by 10. It isn't quite as simple as this;

with the iceberg tearing through

the hull, resistive force would be applied, hastening the

deceleration. This resistance could be from

steel plates accumulating in front of the 'berg, or from internal

structures within the ship, hastening

the slowing down process. Then Mr.Butler says that the figures assume

"a dead stop." The ship simply wouldn't

(possibly couldn't) stop instantaneously. Incidentally, his "force

generated" implies a duration for this

of 1 second, or change of momentum/1 second. If a dead stop were to

occur immediately, the forces involved

would be so astronomical as to be unimagineable. It would, succinctly

put, be the same as the Titanic running

at full speed into an unyielding object, like a cliff face. The

iceberg, while massive, would be subject to the usual laws of motion,

including the conservation of momentum, and would not be an object

incapable to movement.

The next set of calculations are also used incorrectly. Mr.Butler says

that 42000 psi is the equivalent of 800

bar. He assumes that dividing by 15 converts these numbers from one

set of units to another (the actual

conversion is about 14.7 since this is atmospheric pressure). At any

rate, using 14.7 - or 15 - gives a figure

of about 2800. No doubt Mr.Butler made a typographical mistake here,

as he says "800", or a factor of 3.5 out,

but we must be dubious as this works in his favour.

Next, we come to the coup de grace: calculating the force from the

stress. To do this, one needs

to know the surface area to which the force is applied. For a shear

stress, this area is parallel to

the direction in which the force is applied. This area is actually the

length of the plate multiplied

by its height since this is where the force is applied, and not, as he

implies, edge-on.

Then Mr.Butler mixes up his non-SI and SI units when he says "800 / 1

x 0.018"; 800 (an inaccurate

figure) is non-SI, and the next two, and the answer are in SI. This is

not trivia. As the old adage goes, its

like comparing apples and oranges. In this case, we end up with

bananas.

So, calculating the numbers correctly, we find that 42000 psi is 290

Million Newtons per square metre.

So, now we have the pressure, or the force acting on an area. To find

the force, we multiply

the area on which this pressure is applied - the surface area of the

hull plate. We do NOT, as Mr.Butler

has done, divide the pressure by the area. This leads to a meaningless

number, as pressure is already

force divided by area.

Estimating the area of the hull plate one uses figures gleaned from

"Titanic: The Ship Magnificent";

we use a length of 36 feet (11 metres) and a height of 6 feet (1.8

metres). The maximum force that could

be applied until failure occurs is therefore 290 Million Newtons x 11

x 1.8 = 5700 Million (Mega) Newtons.

Compare that to the 622 Million Newtons that Mr.Butler would tell us

is imparted by the impact. Using these

figures, it is clear that the plates would deform, but not break.

The danger of using Mr.Butler's figures is that he has taken one plate

and extrapolated to form

a conclusion upon which lesser scientific figures have fallen.

In practise trying to prove a failure scenario would need a model that

had an interior spaces containing beams,

columns, bulkheads and corridors would add resistive strength to the

iceberg crushing through the hull,

slowing its progress as it proceeded.

The plates would also buckle.

It is not unlikely the berg would be stopped well before 216 feet

which is a "worst case scenario." We also

do not know exactly how the rivets holding the seams would fare. Would

they pop before a siseable fraction

of compressive damage was reached, or would they only yield at the

last second? And how easily would the

compressive force be transferred from plate to plate? It is not a

simple matter of taking one plate and guessing;

rigourous and intensive computation modelling is necessary.

It should also be noted that the 56,000 and 42,000 psi values no doubt

apply to modern steel, and not the

material used ten decades ago.

Free from any revisionist posturing, we must remember what Edward

Wilding, a naval architect at Harland

and Wolff said regarding the stem-on collision proposal: "...she would

have crumpled up in stopping herself. The momentum of the ship would

have crushed in the bows for 80 or perhaps 100 feet. [The impact would

not be] a shock, it is a pressure that lasts three or four seconds,

five seconds perhaps, and whilst it is a big pressure it is not in the

nature of a sharp blow."

It should be pointed out, in 1879, a ship named the Arizona collided

head-on with an iceberg at a speed of 15 knots, and she survived, with

her bows "broken and twisted." Reportedly, she was left with a hole

"20 feet broad by 30 feet deep."

It may be incorrect to compare the Arizona and the Titanic; the afore

named ship was only of some 5600 tons, but it

is not clear what tonnage this refers to, as ships had gross,

displacement, deadweight etc. tonnage. Interested

readers will find more information on David Gittin's titanicebook.com

website.

Other examples of stove-in bows exist but their numbers do not provide

any meaningful comparison: the Stockholm, after the Andrea Doria

Collision had a huge segment of her bow missing but only her speed

(about 18 knots)

is known at the time of the collision; her gross tonnage of about

12200 tonnes does not allow meaningful comparsion.

While it is clear that Wilding's discussion that the Titanic might

have survived was conjecture, with no

calculations to reinforce his argument, the simple argument is that,

spurious physics aside, we do not know.

Sep 28, 2011, 1:50:59 PM9/28/11

to

--

Cheers, SDM -- a 21st Century Schizoid Man

Systems Theory project website: http://systemstheory.net

find us on MySpace, GarageBand, Reverb Nation, Last FM, CDBaby

free MP3s of Systems Theory, Mike Dickson & Greg Amov music at

http://mikedickson.org.uk

Nov 2, 2011, 12:44:12 PM11/2/11

to

Dec 14, 2011, 7:00:38 PM12/14/11

to

On Nov 2, 9:44 am, "Paul.Lee.1...@gmail.com" <paul.lee.

newsgroup has their own "special" Dan Butler story. ;-)

-Mike

1...@googlemail.com> wrote:

> Here are some more musings on Mr.Butler:

>

> http://www.paullee.com/bandb/butler.php#Nov22011

Paul, I think just about everyone who's been a poster on this
> Here are some more musings on Mr.Butler:

>

> http://www.paullee.com/bandb/butler.php#Nov22011

newsgroup has their own "special" Dan Butler story. ;-)

-Mike

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