...tpc / Tpi...

[fradrigo]

Hi all, I need a little help in deriving ratio between metric and imperial unit immersions ( Tonsmetricpercentimeter / Longtonsperinch ) Obviously we need less mass to immerse a boat one centimeter then one inch ( 1 longTon correspond to 1.016 metricTon ), but whatever trick I pull from the bag I keep geting larger than one ratio ( 2.5 to be precise ) What am I doing wrong

[Guillermo]

1 lt/inch = 1.016/2.54 mt/cm = 0.4 mt/cm

1 mt/cm = 2.5 lt/inch

[fradrigo]

Thanks Guillermo, I'm geting the same result...but the common sense tells me that TPC should be less then TPI ( ratio TPC / TPI less than one ) ?

[Guillermo]

....TPC should be less then TPI....
And they are!

0.4 < 1

[fradrigo]

...1 lt/inch = 1.016/2.54 mt/cm = 0.4 mt/cm

1 mt/cm = 2.5 lt/inch...



mt/cm = TPC
lt/inch = TPI

according to your second line:

TPC = 2.5 TPI

TPC / TPI =2.5 > 1

...still confused

[fradrigo]

...I've checked in one of Hydromax's sample models ( varying units for same waterplane area ) and got :

TPC = 1.097 tm/cm

TPI = 2.743 lt / inch
-----------------------

So TPC/TPI = 0.4 ,

Inverse what we derived for the same ratio ( 2.5 )...I'm not very good in dimension analysis..Anyone out there ???

[Guillermo]

1 mt/cm = 2,5 lt/inch
(1/2,5) mt/cm = 1 lt/inch
0,4 mt/cm = 1 lt/inch

This implies you need less weight to sink a boat 1 cm than to sink it 1 inch (as is evident).

An example:
If you take a parallelpipedic hull, of 1 sqm section, floating in fresh water (density = 1), you need 10 kg to sink it 1 cm but you need 25,4 kg kg to sink it 1 inch.

Check: 10 kg = (25,4*0,4)/1,016

What's the doubt....

[fradrigo]

It's clear now, thanks !

' ...open mind for a different view... '

[Guillermo]

Veritas lucet omnibus....

[fradrigo]

Tempus omnia revelat