DimensionalReasoning：空间推理.ppt

Dimensions 500.101 Dimensional Reasoning Dimensions 500.101 Dimensions and Measurements “Dimension” is characteristic of the object, condition, or event and is described quantitatively in terms of defined “units”. A physical quantity is equal to the product of two elements: A quality or dimension A quantity expressed in terms of “units” Dimensions Physical things are measurable in terms of three primitive qualities (Maxwell 1871) Mass (M) Length (L) Time (T) Note: (Temperature, electrical charge, chemical quantity, and luminosity were added as “primitives” some years later.) Dimensions 500.101 Examples Length (L) Velocity (L/T) Force (ML/T2) Units Measurements systems-cgs, MKS, SI-define units SI units are now the international standard (although many engineers continue to use Imperial or U.S.) Dimensions and Measurements (cont.) Dimensions 500.101 SI Primitives DIMENSIONUNITSYMBOL for UNIT Lengthmeterm Masskilogramkg Timeseconds Elec. CurrentampereA luminous intensitycandelacd amount of substancemolemol Dimensions 500.101 SI Derived units DESCRIPTIONDERIVED UNITSYMBOLDIMENSION ForcenewtonNmkg/s2 EnergyjouleJm2kg/s2 PressurepascalPakg/(ms2) PowerwattWm2kg/s3 Dimensions 500.101 Dimensional analysis Fundamental rules: All terms in an equation must reduce to identical primitive dimensions Dimensions can be algebraically manipulated, e.g. Example: Uses: Check consistency of equations Deduce expression for physical phenomenon Dimensions 500.101 Dimensional analysis distance s = s0 +vt2 + 0.5at3 constant = p + gh +v2/2 volume of a torus = 22(Rr)2 Dimensions 500.101 Deduce expressions Example: What is the period of oscillation for a pendulum? Possible variables: length l L, mass m M, gravity g i.e. period P = f(l , m, g) Period = T , so combinations of variables must be equivalent to T . m l g Dimensions 500.101 Dimensional analysis (cont.) Example: What is the period of oscillation for a pendulum? Possible variables: length l L, mass m M, gravity g i.e. period P = f(l , m, g) Period = T , so combinations of variables must be equivalent to T . Only possible combination is Note: mass is not involved m l g Dimensions 500.101 Quantitative considerations Each measurement carries a unit of measurement Example: it is meaningless to say that a board is “3” long. “3” what? Perhaps “3 meters” long. Units can be algebraically manipulated (like dimensions) Conversions between measurement systems can be accommodated, e.g., 1 m = 100 cm, or or Example: Dimensions 500.101 Quantitative considerations (cont.) Arithmetic manipulations between terms can take place only with identical units. Example: but, Dimensions 500.101 Buckingham Pi Theorem (1915) Pi theorem tells how many dimensionless groups define a problem. Theorem: If n variables are involved in a problem and these are expressed using k primitive dimensions, then (n-k) dimensionless groups are required to characterize the problem. Example: in the pendulum, the variables were time T, gravity L/T2, length L, mass M . So, n = 4 k = 3. So, only one dimensionless group describes the system. Dimensions 500.101 Buckingham Pi Theorem (cont.) How to find the dimensional groups: Pendulum example: where a,b,c,d are coefficients to be determined. In terms of dimensions: Dimensions 500.101 Buckingham Pi Theorem (cont.) Therefore: a - 2c = 0 b + c = 0 d = 0 Arbitrarily choose a = 1. Then b = -1/2, c = 1/2, d = 0. This yields Dimensions 500.101 A star undergoes some mode of oscillation. How does the frequency of oscillation depend upon the properties of the star? Certainly the density and the radius R are important; well also need the gravitational constant G which appears in Newtons law of universal gravitation. We could add the mass m to the list, but if we assume that the density is constant, then m = (4R3/3) and the mass is redundant. Therefore, is the governed parameter, with dimensions = T-1, and (; R; G) are the governing parameters, with dimensions = ML-3, R = L, and G = M-1L3T-2 (check the last one). You can easily check that (; R;G) have independent dimensions; therefore, n = 3; k = 3, so the function is simply a constant in this case. Next, determine the exponents: = T-1 = aRbGc = Ma-cL-3a+b+3cT-2c Equating exponents on both sides, we have a - c = 0; -3a + b + 3c = 0; -2c = -1 Solving, we find a = c = 1/2, b = 0, so that = C(G)1/2, with C a constant. We see that the frequency of oscillation is proportional to the square root of the density, and independent of the radius. Buckingham Pi Theorem (cont.) Oscillations of a star Dimensions 500.101 “Dimensionless” Quantities Dimensional quantities can be made “dimensionless” by “normalizing” them with respect to another dimensional quantity of the same dimensionality. Example: speed V (m/s) can be made “dimensionless“ by dividing by the velocity of sound c (m/s) to obtain M = V/c, a dimensionless speed known as the Mach number. M1 is faster than the speed of sound; M1 is slower than the speed of sound. Other examples: percent, relative humidity, efficiency Equations and variables can be made dimensionless, e.g., Cd = 2D/(v2A) Useful properties: Dimensionless equations and variable are independent of units. Relative importance of terms can be easily estimated. Scale (battleship or model ship) is automatically built into the dimensionless expression. Dimensions 500.101 Dimensionless quantities (cont.) Reduces many problems to a single problem through normalization. Example: Convert a dimensional stochastic variable x to a non-dimensional variable to represent its position with respect to a Gaussian curve-N(0,1), e.g., grades on an exam Dimensions 500.101 The area of any triangle depends on its size and shape, which can be unambiguously identified by the length of one of its edges (for example, the largest) and by any two of its angles (the third being determined by the fact that the sum of all three is ). Thus, recalling that an area has the dimensions of a length squared, we can write: area = largest edge2 f (angle1, angle2), where f is an nondimensional function of the angles. Now, referring to the figure at right, if we divide a right triangle in two smaller ones by tracing the segment perpendicular to its hypotenuse and passing by the opposite vertex, and express the obvious fact that the total area is the sum of the two smaller ones, by applying the previous equation we have: c2 f (, /2) = a2 f (, /2) + b2 f (, /2). And, eliminating f: c2 = a2 + b2, Q.E.D. Proof of the Pythagorean Theorem Dimensions 500.101 Scaling, modeling, similarity Types of “similarity” between two objects/processes. Geometric similarity linear dimensions are proportional; angles are the same. Kinematic similarity includes proportional time scales, i.e., velocity, which are similar. Dynamic similarity includes force scale similarity, i.e., equality of Reynolds number (inertial/viscous), Froud number (inertial/buoyancy), Rossby number (inertial/Coriolis), Euler number (inertial/surface tension). Dimensions 500.101 Distorted models Sometimes its necessary to violate geometric similarity: A 1/1000 scale model of the Chesapeake Bay is ten times as deep as it should be, becau