Plots Graphs and Sketches 1 1 Plots or

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Plots, Graphs, and Sketches (1. 1) • Plots or Graphs - Generally the most

Plots, Graphs, and Sketches (1. 1) • Plots or Graphs - Generally the most effective format for displaying and conveying the interrelation of experimental variables. • Sketches - Quick and informal method of sharing ideas with others or clarify concepts for yourself. Free body diagrams (FBDs) are an example.

Plots and Graphs • Means of Communicating Ideas Concisely – Axes ® X-axis (horizontal,

Plots and Graphs • Means of Communicating Ideas Concisely – Axes ® X-axis (horizontal, independent variable) ® Y-axis (vertical, dependent variable) ® Divide major axes into appropriate divisions of 1, 2, 10, 10 X, etc ® Label with words, symbols, and units ® Minor axes should be distributed evenly

Plots and Graphs

Plots and Graphs

Sketc h

Sketc h

Sections 1. 2 ü Functions: ü Independent variables (Cause /Input /Action) ü Dependent variables

Sections 1. 2 ü Functions: ü Independent variables (Cause /Input /Action) ü Dependent variables (Effect /Output / Reaction)

Dependent / Independent Variables Dependent variables (Y-axis) values will depend on the value of

Dependent / Independent Variables Dependent variables (Y-axis) values will depend on the value of independent variables (X-axis). Notation in math and science - parameter name = f ( independent 1, independent 2, …) - example : p = f ( , z) Developing the relationship between the dependent and relationship independent variable Experiment: collect raw data and draw the curve. apply fairing algorithm and interpolate regression analysis. Conservation law or theoretical principles Semi-empirical equation 1) identify the important parameter. 2) develop analytical equations through experiments

Sections 1. 3 ü “Area under the curve” ü “Slope of the curve”

Sections 1. 3 ü “Area under the curve” ü “Slope of the curve”

Area Under and Instantaneous Slope of a Curve

Area Under and Instantaneous Slope of a Curve

Section 1. 4 ü Unit Systems ü S. I. ü Pound force - pound

Section 1. 4 ü Unit Systems ü S. I. ü Pound force - pound mass ü Pound - slug: What we will use ü Carry units through all calculations ü If the units are wrong, the answer is wrong

Units 1 slug is defined as the mass that will be accelerated at a

Units 1 slug is defined as the mass that will be accelerated at a rate of 1 ft/s 2 by a 1 lb force.

Unit Analysis If you get unreasonable values: - check the units in the process

Unit Analysis If you get unreasonable values: - check the units in the process of calculations To avoid mistakes: - know the units of each parameter - understand the relationship between fundamental dimensions and their units Ruled Lines Method : LT = 4000 ft³ | 1. 99 lb-s² | 32. 17 ft | 1 LT | = 114. 32 LT ft^4 | s² | 2240 lb

Section 1. 5 ü Significant figures ü Exact numbers ü Measurements ü Addition/subtraction ü

Section 1. 5 ü Significant figures ü Exact numbers ü Measurements ü Addition/subtraction ü Multiplication/division

Significant Figures ü The number of accurate digits in a number – Example: 2.

Significant Figures ü The number of accurate digits in a number – Example: 2. 65 has 3 significant figures – Example: 10 has 1 or 2 , 10. 0 has 3 ü Multiplication / Division: Use the same # of significant figures as the number with the least # of significant figures Example: 20 x 3. 444 = 69 ü Addition / Subtraction: Use the same # of decimal places as the number with the least # of decimal places Example: 3. 6 + 1. 212 = 4. 8

ü General Problem Solving Technique Write down applicable reference equation which contains the desired

ü General Problem Solving Technique Write down applicable reference equation which contains the desired “answer variable”. ü Solve the reference equation for the “answer variable”. ü Write down additional reference equations and solve for unknown variables in the “answer variable” equation, if needed. ü Draw a quick sketch to show what information is given and needed and identify variables, if applicable. ü Rewrite “answer variable” equation, substituting numeric values with units for variables. ü Simplify this expanded equation, including units, to arrive at the final answer. ü Check the answer: ü Do units match answer? ü Is the answer on the right order of magnitude?

Section 1. 6 Output ü Linear Interpolation y 2 y 3=? ü Assume a

Section 1. 6 Output ü Linear Interpolation y 2 y 3=? ü Assume a straight line y 1 ü Use anchor point, output range, and domain fraction x 1 x 3 Input from anchor point to input value. ü Output=anchor+{range*fraction} ü y 3= y 1+ {(y 2 -y 1)*(x 3 -x 1)/(x 2 -x 1)} Output y 2 y 3 y 1=Anchor x 1 x 3 x 2 Input

Linear Interpolation Example Temperature (°F) Density (lb-s 2/ft 4) 61 1. 9381 62 1.

Linear Interpolation Example Temperature (°F) Density (lb-s 2/ft 4) 61 1. 9381 62 1. 9379 63 1. 9377 Solve for the density at 62. 7°F as follows:

Section 1. 7 ü Scalars & Vectors ü Forces & Weight üPoint forces (centroid

Section 1. 7 ü Scalars & Vectors ü Forces & Weight üPoint forces (centroid of a force distribution) üDistributed forces (such as Pressure) ü Moments & Couples: Force × Distance ü Static Equilibrium: ü Hydrostatic Pressure = Phyd = gz ü Fhyd = Phyd×Area

Rectangular Coordinate System z y x

Rectangular Coordinate System z y x

Scalars and Vectors Scalar has only magnitude. - Example : mass, speed, work, energy

Scalars and Vectors Scalar has only magnitude. - Example : mass, speed, work, energy Vector has its magnitude and direction. - Example : velocity, force, momentum, acceleration - Way of presentation : boldface, symbol with a half/full arrow ( notation should be consistent in the text. ) F Fy Vectors can be resolved into scalar components in each principal dimension. F Fx

Vectors - Vector symbols often have subscripts to better clarify. resultant buoyant force resultant

Vectors - Vector symbols often have subscripts to better clarify. resultant buoyant force resultant weight of ship = displacement line of action - Naval Engineers uses to represent a change in a property. Ex ; : change in a draft due to parallel sinkage

Forces, Moments, and Couples • FORCE - a vector quantity with magnitude and a

Forces, Moments, and Couples • FORCE - a vector quantity with magnitude and a direction. • MOMENT - force times a distance with respect to a given origin, i. e. • COUPLE - a special case a of moment that results in pure rotation and no translation.

Forces 1000 lb F 20 ft 100 lb/ft Point Force 10 ft Distributed force

Forces 1000 lb F 20 ft 100 lb/ft Point Force 10 ft Distributed force and its resultant point force.

Moments F d Example of a moment applied with a wrench Increasing F or

Moments F d Example of a moment applied with a wrench Increasing F or d will increase the magnitude of the Moment about the nut Moment = Force x Distance (M = F x d) Units: foot - pounds

Couples Two forces creating a couple F ü Couple = Force distance C =

Couples Two forces creating a couple F ü Couple = Force distance C = F d Units: foot-pounds d F The magnitude of a couple is calculated by multiplying the magnitude of one force by the distance separating the two forces. The two forces must be equal in magnitude and opposite in direction for pure rotation without translation.

Static Equilibrium If an object is neither accelerating or decelerating then it is because

Static Equilibrium If an object is neither accelerating or decelerating then it is because the following 2 conditions are met: ü Sum of the forces = 0 ü Sum of the moments = 0

Hydrostatic Pressure ü “Pressure” is the amount of force applied to a given area

Hydrostatic Pressure ü “Pressure” is the amount of force applied to a given area (p=F/A) ü In English units it is pounds/sq. ft. or pounds/sq. in. , or “psi” Air pressure is ~ 15 psi. At 440 ft below sea level it is ~ 195 psi!

Hydrostatic Pressure & Force If an object is floating in water and the object

Hydrostatic Pressure & Force If an object is floating in water and the object and water are both at rest, then the pressure exerted by the water on the object is referred to as hydrostatic pressure and is defined as: Phyd = gz where = water density (lb-s 2/ft 4) g = acceleration of gravity (ft/s 2) z = depth of object below the water’s surface (ft) Hydrostatic pressure acting over an area (A) results in a Hydrostatic Force where Fhyd = Phyd A.

The Mathematical First, Second and Third Moments • These integrals are used in mathematical

The Mathematical First, Second and Third Moments • These integrals are used in mathematical descriptions of physical problems Where: s = some distance db = some differential property = summation

The Mathematical First and Second Moments • In Naval Architecture: – “b” could represent

The Mathematical First and Second Moments • In Naval Architecture: – “b” could represent length, area, volume, or mass – “s” is a length or distance ® First Moment of Mass ® Second Moment of Area

Weighted Averages In Naval Architecture, we use the simplified form: Used to find: Longitudinal

Weighted Averages In Naval Architecture, we use the simplified form: Used to find: Longitudinal Center of Flotation (LCF), Longitudinal Center of Buoyancy (LCB) Centers of Gravity (LCG, TCG, VCG)

Mathematical Moments and the Parallel Axis Theorem y ü First Moment=òsdm ü Mx=òyd. A;

Mathematical Moments and the Parallel Axis Theorem y ü First Moment=òsdm ü Mx=òyd. A; My=òxd. A; AT=òd. A x ü Centroids: y = Mx/AT : x = My/AT y ü Second Moment=òs²dm ü Ix=òy²d. A; Iy=òx²d. A x y yc ü Ix= Ixc+Ad 12; Iy= Iyc+Ad 22 d 2 ü Use Appendix C xc C A d 1 x

Translational vs. Rotational Motion ü Ship freely floating is subject to 6 degrees of

Translational vs. Rotational Motion ü Ship freely floating is subject to 6 degrees of freedom. ü Three are Translational 1. Heave (z) 2. Sway (y) 3. Surge (x) ü Three are Rotational 1. Yaw (z) 2. Pitch (y) 3. Roll (x)

Ship’s Axes and Degrees of Freedom z Pitch/Trim x Heave Roll/List/Heel y Sway Yaw

Ship’s Axes and Degrees of Freedom z Pitch/Trim x Heave Roll/List/Heel y Sway Yaw Surge

Section 1. 9 Bernoulli’s Equation: Along a line of equal energy (a streamline) in

Section 1. 9 Bernoulli’s Equation: Along a line of equal energy (a streamline) in a fluid, the above is a constant. Flow Energy + Kinetic Energy + Gravitational Potential Energy=Available Energy (Assumes steady, frictionless, incompressible flow along a streamline) p=hydrostatic pressure at any point in the stream V=speed of the fluid at any point on the stream g=acceleration of gravity z=height of the streamline point above reference =fluid density

Bernoulli Equation • Total pressure is constant in a fluid, if: ® inviscid flow

Bernoulli Equation • Total pressure is constant in a fluid, if: ® inviscid flow (no viscosity) ® incompressible flow ® steady flow This gives us hydrostatic and hydrodynamic pressure. These are the water loads on the vessel.

Pressure Prediction ü Vertical pressure supports the vessel (lift versus weight) ü Horizontal pressure

Pressure Prediction ü Vertical pressure supports the vessel (lift versus weight) ü Horizontal pressure is thrust and drag These are the same as an aircraft!

Example Problem ü A 50 ft wide, 100 ft long barge on an even

Example Problem ü A 50 ft wide, 100 ft long barge on an even keel is divided into forward and aft compartments by a transverse bulkhead 60 ft aft of the bow. The barge weighs 100 LT empty with an even weight distribution. The forward tank is then filled to a height of 5 ft and the aft tank to a height of 4 ft with oil ( =1. 80 lb-s²/ft 4) ü What is the total weight of the barge? ü At what point (fore and aft) would this weight be applied? ü Which degrees of freedom for the barge were affected by loading the oil and in what direction?

Answer: 60 ft 5 ft 4 ft deep Weightbow compt= g. V= (1. 80

Answer: 60 ft 5 ft 4 ft deep Weightbow compt= g. V= (1. 80 lb-s²/ft 4) (32. 17 ft/s²)(50 ft× 60 ft× 5 ft)(1 LT/2240 lb)= 388 LT Weightaft compt= g. V= (1. 80 lb-s²/ft 4) (32. 17 ft/s²)(50 ft× 4 ft)(1 LT/2240 lb) =207 LT ü Weighttotal=388 LT+207 LT+100 LT=695 LT 40 ft

40 ft 5 ft deep 4 ft deep 388 LT@30 ft 50 ft Answer:

40 ft 5 ft deep 4 ft deep 388 [email protected] ft 50 ft Answer: 60 ft 207 LT @80 ft 100 [email protected] ft Application point= = [(100 LT× 50 ft)+(388 LT× 30 ft)+(207 LT× 80 ft)] 695 LT = 47. 8 ft Barge became heavier, draft increased, barge sank in the heave direction. More weight was placed forward, so barge pitched forward.

Example Problem A Landing Craft Medium (LCM 8) boat has an empty draft of

Example Problem A Landing Craft Medium (LCM 8) boat has an empty draft of 1 ft. A 60 LT tank is loaded into the boat. How many 250 lb combat ready Marines can board the LCM and still be able to cross a 4 foot shoal with 1 foot to spare on the way to the beach? Data: ü LCM dimensions: 75 ft long × 21 ft wide × 4. 5 ft high ü Propulsion: 1300 HP from 2 diesels on separate shafts d. T=dw/TPI (T=draft; w=weight; TPI= long tons/inch) 1 LT=2240 lb gsw=64 lb/ft 3

Example Answer ü Displacement = Density x Submerged Volume Vessel Submerged Volume = (75

Example Answer ü Displacement = Density x Submerged Volume Vessel Submerged Volume = (75 ft)(21 ft )(1 ft) = 1575 ft 3 (62. 4 lb/ ft 3)(1575 ft 3)=98280 lb (98280 lb)/(2240 lb/LT)= 43. 875 LT (Empty) 4 foot shoal Tons per Inch (TPI) = 43. 875 LT /12 in= 3. 656 LT/ in 4 foot (24 in)(3. 656 LT/ in)= 87. 74 LT-60 LT=27. 74 LT is allowable shoal 60 LTs + # Marines (27. 74 LT)(2240 LT/lb)/(250 lb/Marine)=248 Marines Max change in draft = 4 foot shoal – 60 LT + 1 foot draft empty – # Marines 1 foot margin = 2 feet 1 foot draft empty 1 foot margin 1 foot draft empty Max change in draft = 4’ shoal – 1’draft empty – 1’ margin = 2’

Example Problem A 100 ft Barge with an empty weight distribution of 2 LT/ft

Example Problem A 100 ft Barge with an empty weight distribution of 2 LT/ft is loaded with the following additional loads: Y <-30 ft-> 2 LT/ft <-40 ft-> 4 LT/ft <-30 ft-> 1 LT/ft Empty Barge=2 LT/ft 0 100 ft X What is the total weight of the barge and where, along the length, is the center of gravity?

Example Answer ü Weight =(2 LT/ft× 100 ft)+(2 LT/ft× 30 ft)+(4 LT/ft× 40 ft)+(1

Example Answer ü Weight =(2 LT/ft× 100 ft)+(2 LT/ft× 30 ft)+(4 LT/ft× 40 ft)+(1 LT/ft× 30 ft) = 200 LT+60 LT+160 LT+30 LT=450 LT ü Weighted Average of each weight: 200 LT × 50 ft= 10, 000 LT-ft 60 LT × 15 ft= 900 LT-ft 160 LT × 50 ft= 8000 LT-ft 30 LT × 85 ft= 2550 LT-ft Sum=10, 000+900+8000+2550 = 21450 LT-ft Weighted Average=21450 LT-ft/450 LT=47. 7 ft ü Center of Gravity along X axis=47. 7 ft 450 LT Y 47. 7 ft Loaded Barge 0 100 ft X