Calculus Revisited # 20: Taylor Series and Maclaurin Series

Welcome to Part 20 of 21: Taylor and Maclaurin Series.

The Taylor and Maclaurin series are representation of the function f(x) by using an infinite series. If we use a finite number of terms, the series can (I stress can), but a good approximation f(x).

A Taylor series of f(x) is centered on a focus point x = a. Generally, approximations are best when x is around a, and gets worse the further x gets from a.

Taylor Series:

About the point x = a:

f(x) = f(a) + f'(a) * (x - 1) + f''(a) * (x - a)^2 / 2! + f'''(a) * (x - a)^3 / 3! + ....

If the series is cut off at n terms, the final term of the Taylor series is:

f^(n+1)(t) / (n+1)! * (x - a)^(n+1). This is known as the error term.

Maclaurin Series:

The Taylor series with a = 0.

f(x) = f(0) + f'(0) * x + f''(0) * x^2/2! + f'''(0) *x^3/3! + ...

With error term f^(n+1)(0) / (n+1)! * x^(n+1)

Some famous series:

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + .... + x^n/n! + ....

sin x = x - x^3/3! + x^5/5! - x^7/7! + .... + ( (-1)^n * x^(2n+1) ) / (2n+1)! + ....

cos x = 1 - x^2/2! + x^4/4! - x^6/6! + ... + ( (-1)^n * x^(2n) ) / (2n)! + ....

Problems

1. Find a Macluarin series for

f(x) = ln(x + 1) to four terms

f(x) = ln(x + 1)
f(0) = ln(0 + 1) = 0

f'(x) = 1/(x+1)
f'(0) = 1/1 = 1

f''(x) = -1/(x+1)^2
f''(0) = -1/(1^2) = -1

f'''(x) = 2/(x+1)^3
f'''(0) = 2/(1^3) = 2

Then:

ln(x + 1) = 0 + 1 * x + (-1) * x^2/2! + 2 * x^3/3! + ....
= x - x^2/2! + 2x^3/3! + ....

2. Find the Macluarin series for

f(x) = e^(-x^2) to three nonzero terms. Approximate its integral.

f(x) = e^(-x^2)
f(0) = 1

f'(x) = e^(-x^2) * (-2x)
f'(0) = 0

f''(x) = e^(-x^2) * (4x^2 - 2)
f''(0) = -2

f'''(x) = e^(-x^2) * (-8x^3 + 12x)
f'''(0) = 0

f''''(x) = e^(-x^2) * (16x^4 - 48x^2 + 12)
f''''(0) = 12

Then:

e^(-x^2) = 1 - 2 * x^2/2! + 12 * x^4/4! + ....
= 1 - x^2 + x^4/2 + ...

∫ e^(-x^2) dx = x -x^3/3 + x^5/10 + .... + C

3. Find the Taylor series for cos x at a = π/4 to three nonzero terms.

f(x) = cos x
f(π/4) = √2/2

f'(x) = - sin x
f'(π/4) = -√2/2

f''(x) = -cos x
f''(π/4) = -√2/2

Then:

cos x = √2/2 - √2/2 * (x - π/4) + √2/2 * (x - π/4)^2/2! + ....

That concludes our section on Taylor and Maclaurin series. The next and last entry will be our "catch all" entry.

Thank you as always,

Eddie

This blog is property of Edward Shore. © 2012

Announcement: Calculus Revisited starts 4/23/2012

Blog Entry #72

Starting on April 23, 2012, I will revisit topics of calculus for 21 posts (my goal is to post daily - at least on Monday - Friday). It can be a used for a review, visiting past (hopefully pleasant) memories, or an aid for studying for the finals.

I will cover some of the major topics in calculus including:
* Trigonometric Identities
* Functions
* Limits
* Derivatives
* Integrals
* Series
* Entry 21 will be a "catch all" blog

See you soon!

Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #10: Integration Basics

Welcome to Part 10 of our 21 Calculus Revisited series. Today, we enter the world of Integration.

A function f(x) has an antiderivative F(x) if the following is true:

∫ f(x) dx = F(x) and d/dx F(x) = f(x)

Indefinite Integral: Finding a symbolic integral of f(x) to describe all the antiderivatives of f(x).

In a calculus course, if F(x) is the antiderivative of f(x), then

∫ f(x) dx = F(x) + C

where C is a numerical constant. Calculators with CAS (computer algebraic system) capabilities usually do not include the C when figuring out indefinite integrals. It is not technically wrong, the CAS system assumes that C = 0, most of the time.

Definite Integral: An integral that has a limit from a
to b (usually a < b, but not always) which produces a numerical answer. The most famous example of a definite integral is the area under the curve f(x) from a to b.

By the Fundamental Theorem of Calculus,

b
∫ f(x) dx = F(b) - F(a)
a

Table of Integrals

Addition Rule:
∫ f(x) + g(x) dx = ∫ f(x) dx + ∫ g(x) dx

∫ a dx = a * x (a is a constant)
∫ x^n dx = (x^(n+1)) / (n+1)
∫ 1/x dx = ln x
∫ e^x dx = e^x
∫ sin x dx = -cos x
∫ cos x dx = sin x
∫ tan x dx = -ln |cos x|
∫ asin x dx = x asin x + √(1 - x^2)
∫ acos x dx = x acos x - √(1 - x^2)
∫ atan x dx = x atan x - ln(x^2 + 1)/2
∫ 1/√(1 - x^2) dx = asin x
∫ 1/(1 + x^2) dx = atan x
∫ 1/√(x^2 - 1) dx = ln(√(x^2 - 1) + x)
∫ a^x dx = a^x/(ln a)

Problems
Indefinite Integrals

1. ∫ 2x^2 + x - 1 dx

∫ 2x^2 + x - 1 dx
= ∫ 2x^2 dx + ∫ x dx - ∫ 1 dx
= 2/3 * x^3 + C + x^2/2 + C - x + C
C just symbolizes a general constant, do not worry if it is 3C or just C in this case
= 2/3 * x^3 + x^2/2 - x + C

In the future, for indefinite integral problems, we will just add the C at the end.

2. ∫ 1/x^2 - 1/x dx
= ∫ x^(-2) dx - ∫ 1/x dx
= -1/x - ln x + C

3. ∫ 2 * 3^x dx

Note: ∫ a * f(x) dx = a * ∫ f(x) dx. Know this - it will come in handy a lot.

∫ 2 * 3^x dx
= 2 * ∫ 3^x dx
= 2 * (3^x)/ln 3 + C

Definite Integrals

Hint: When using a calculator to calculate definite integrals - as a general rule, always have your calculator in Radians mode.

4.

1
∫ 1/(√(1 - x^2) dx
0

Use the Fundamental Theorem of Calculus to assist us with these type of problems.

f(x) = 1/(√(1 - x^2)
F(x) = asin x

1
∫ 1/(√(1 - x^2) dx
0

= asin 1 - asin 0
= π/2 - 0
= π/2

Hint: Do not worry about the "C" with definite integrals, they cancel out!

b
∫ f(x) dx
a

= F(b) + C - (F(a) + C)
= F(b) + C - F(a) - C
= F(b) - F(a)

5.

2
∫ 2 * e^x dx
0

2
= ∫ e^x dx
0

= 2 (e^2 - e^0)
= 2 (e^2 - 1) ≈ 12.77811

( ∫ e^x dx = e^x )

Limits do not have to be constants. Check out the next problem.

6.

t
∫ cos x dx
0

= sin(t) - sin 0
= sin(t)

( ∫ cos x dx = sin x)


Initial Value Problems - Basically, figure out what C is.

7. Find the antiderivative of f(x) = 2x + 5 if the antiderivative passes the point (0, 1).

We are faced with two tasks:
1. Find F(x) in general
2. Find the specific F(x) knowing F(0) = 1. This is where we solve for C.

∫ f(x) dx = F(x) + C

∫ 2x + 5 dx
= x^2 + 5x + C

With F(0) = 1 (going through point (0,1)): (x = 0, y = 1)

1 = (0)^2 + 5 * (0) + C
1 = 0 + C
1 = C

The final answer is F(x) = x^2 + 5x + 1

8. The slope of the function y(x) is described by the function sin x. It is known that y(x) passes through the point (1/2, 1/2). Find y(x).

The slope of the function y(x) is the derivative of y(x), or y'(x).

Then ∫ y'(x) dx = y(x)

And
∫ sin x dx = - cos x + C
Almost there, we have to find what C is. Use the point (1/2, 1/2) to help.
1/2 = - cos(1/2) + C
1/2 + cos(1/2) = C
We can conclude that
y(x) = cos x + (1/2 + cos(1/2)) ≈ cos x + 1.37758

We have just started with integrals. The next four parts will deal with techniques of integration.

See you soon, Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #8: Related Rates

In our 8th in our 21 blog entry series, we will deal with some problems involving related rates.

Simply put, if you have a function f(x), then f'(x) describes the rate of change at point x.

In physics, you can describe the position of an object given any point in time by the equation x(t). The velocity is found by the first derivative x'(t), and the acceleration is found by the second derivative x''(t).

Position -(derivative)-> Velocity -(derivative)-> Acceleration

Problems
1. A ball is dropped from a building four stories high (48 feet). Let the position equation be x(t) = -1/2 * g * t^2 + 48 where g is the gravitational constant at 32.17405 ft/s^2. When will the ball hit the ground? What will be the velocity and acceleration when the ball hits the ground?

When will the ball hit the ground? How much time will it take for the ball to hit the ground? Solve x(t) = 0.

-1/2 * 32.17405 * t^2 + 48 = 0
-16.087025 * t^2 = -48
t^2 ≈ 2.98377
t ≈ 1.72736 (negative time is nonsensical)

It takes approximately 1.72736 seconds for the ball to reach the ground.

What is the velocity and acceleration when the ball hits the ground?

Velocity: x'(t) = -g * t
x'(1.72736) ≈ -32.17405 * 1.72736 ≈ -55.57617 ft/sec (velocity of the ball moving downward)

Acceleration: x''(t) = -g
The acceleration is 32.17405 ft/sec^2.

2. A rectangular tub 6 feet wide by 4 feet deep is filled with water with its height increasing at 6/5 in/sec. How fast is the volume of water increasing? If the tub is 3 feet high, when will the tub be full?

1 foot = 12 inches

The tub is 72 inches wide by 48 inches deep by 36 inches high. The rate of the increase in height in water is 1.2 in/sec.

Volume of the water = width of the tub * depth of the tub * height of the water

Note that width and depth do not change, only the height.

Then:
V_water(t) = 72 * 48 * h(t)
= 3456 * h(t)
Take the derivative with respect to t:
V_water'(t) = 3456 * h'(t)
The water is rising at 1.2 in/sec. Hence h'(t) = 1.2
V_water'(t) = 3456 * 1.2 = 4147.2

The volume of water is increasing at a rate of 4,147.2 in^3/sec, or 2.4 ft^3/sec.

When will the tub be filled?

The tub will be filled when the height of the water reaches 3 feet, or 36 inches.

distance = rate * time
height of tub = h'(t) * time
36 = 1.2 * t
t = 30

The tub will be full in 30 seconds.

3. Let q(t) = 1/3 * t^3 - t^2 + t coulombs of charge flow through a conducting wire with time (t) being measured in seconds. What is the current after 4 seconds? If there is a 25-amp fuse in the wire, how long will it last?

In electronics, the current of the rate of charge. Current is symbolized by i(t).

Then i(t) = q'(t).

So i(t) = d/dt (1/3*t^3 - t^2 + t) = t^2 - 2t + 1

And i(4) = 16 - 8 + 1 = 9

The current is 9 amp (coulombs/sec).

If there is a 25-amp fuse, how long will it last?

The fuse burns out when the current is 25 amps. So we are solving for t.

i(t) = 25
t^2 - 2t + 1 = 25
t^2 - 2t - 24 = 0
( t - 6 ) ( t + 4 ) = 0

This implies that t = 6 (negative time is nonsensical).

The fuse lasts for 6 seconds.

This concludes Part 8 of this 21 part series. Next time, we will solve non-linear equations.

In good calculation, Eddie.

This blog is property of Edward Shore. © 2012

How Far Scientific Calculators Have Come

Happy Friday the 13th - Blog Entry # 71

How Far Scientific Calculators Have Come

To put it briefly, the Casio fx-115 ES is a Casio 9860g minus the programming and graphing.

The TI-36X Pro is a TI-84 Plus minus the programming and graphing.

Both have definite integrals, numeric derivatives, numeric solvers, polynomial solvers, (some) complex number operations, statistics, base operations, a library of constants and conversions, and Σ (sum of a series).

All for about $20-$25.

40 years ago we were looking at getting just trigonometry and logarithms for $395. (HP 35). (Thank you Hewlett Packard for making the scientific calculator part of our lives. And for bringing back the HP 15C. )

I am excited about the new Casio fx-115ES Plus that is coming soon. I think the model adds product of a series (Π), GCM, LCD, integer part, and fraction part. (partially to catch up to the TI-36X Pro).

Casio has got to consider expanding the programmable fx-3650P, and marketing it in America.

Can a solar graphing calculator not be far behind? Who would do it first?

What about RPN? I have been forever dreaming of a having an RPN solar calculator.

Announcements

Coming at the end of April, I am anticipating April 30, 2012, I am going to post a review series on one-variable calculus. It will last about a few weeks.

Have a good weekend everyone!

This blog is property of Edward Shore. © 2012

Taking Some Mystery Out Of Baseball Statistics

Welcome to Blog Entry #70.

A shout to all who are following my blog. Many sincere thanks!

This will probably be my first anniversary post, even though it is still about a week away. I started this blog on April 16, 2011.

If you have any math or calculator questions, please feel free to leave a comment - I will do my best to answer it. Maybe your question becomes a future blog entry. Thanks!




Today's blog entry will cover baseball statistics. Any one who follows baseball, especially in America, will hear tons of statistics for each player, for every single kind of situation possible. I guess this is how a lot of people keep busy watching the sport.

I am going to go over some of the more common statistics. Apparently the only thing complex about statistics is the amount of statistics and abbreviations and not the mathematics itself.

Baseball Stats

BA: Batting Average

BA = H / AB

The higher, the better.

H = the number of hits a player makes. This is each time a player successfully reaches at least one base after hitting a pitch.

AB = number of at-bats, the number of the times a player appears in front of a pitcher. However an at-bat does not include situations when the batter walks (receives four pitches deemed to be "balls"), the batter is hit by a pitch, or hits a sacrifice (intentionally bats out to get a teammate to score a run).

Examples:
Note: All examples use statistics from the 2011 season - provided by MLB and ESPN. All calculations result in three decimal places.

Matt Kemp, Los Angeles Dodgers: 195 hits in 602 at bats
BA = 195 / 602 = .324

Evan Longoria, Tampa Bay Rays: 118 hits in 483 at bats
BA = 118 / 483 = .244

SLG: Slugging Percentage

SLG = TB / AB = (singles + 2 * doubles + 3 * triples + 4 * home runs) / AB

The higher the better.

TB = total bases, a weighted sum of hits

Matt Kemp: 353 total bases in 602 at bats
SLG = 353 / 602 = .586

Evan Longoria: 239 total bases in 483 at bats
SLG = 239 / 483 = .495

ERA: Earned Run Average

ERA = 9 * ER / IP

The lower, the better.

ER = earned runs scored, in this case, by the opponent. This does not include any runs scored by "errors".

IP = innings pitched. Innings are counted by outs. So if a pitcher threw for six innings and got two outs in the seventh inning, the pitcher has pitched for 6 2/3 innings.

Caution: For IP, some statisticians use shortcut notation. For an IP listed as n.1, it means n and 1/3 innings. (n is a number). If the IP is listed as n.2, it means n and 2/3 innings. Example: A listed IP of 233.1, it means 233 1/3.

Clayton Kershaw, Cy Young Award Winner, Los Angeles Dodgers: 59 earned runs allowed in 233 1/3 innings pitched.
ERA = 9 * 59 / (233 1/3) = 2.276

Chris Carpenter of the 2011 World Series Champions St. Louis Cardinals: 91 earned runs in 237 1/3 innings pitched.
ERA = 9 * 91 / (237 1/3) = 3.451

K/9: Strikeouts Per 9 Innings Pitched

K/9 = 9 * K / IP

The higher, the better.

K = strike out. A pitcher gets this when the pitcher gets the batter to swing and miss on a pitch which results in Strike 3 (symbolized by a forward K), or the pitcher throws the ball to a batter's strike zone but the batter does nothing and is thrown out on Strike 3 (symbolized by a backwards K). Any batter who gets Strike 3 is out.

Per ESPN, Clayton Kershaw's K/9 for 2011 was 9.57 while Chris Carpenter's was 7.24. A K/9 above 9 is considered excellent.

RC: Runs Created (Bill James)

RC = (TB * (H + BB)) / (AB + BB)

The higher the better.

BB = base on balls. A BB is earned when a hitter takes a pitched deemed Ball 4 (the pitcher missed).

This is an estimate of runs created and not a count of the number of runs the player scored.

Paul Konerko, Chicago White Sox: 281 total bases, 163 hits, 77 walks in 543 at bats
RC = (281 * (163 + 77)) / (543 + 77) = 108.774 (estimated 109 runs created)

RBI: Runs Batted In

Commonly known as "ribbys", a batter gets an RBI every time the team scores a run that do not result from "errors". A batter can get up to four RBIs every time the batter is at bat.

Example: A batter has teammates at second and third base. The batter hits a double going to second base, allowing both teammates to reach home base. The batter scores 2 RBIs.

WHIP: Walks and Hits Per Innings Pitched

WHIP = (BB + H)/IP

The lower the better, real good of WHIP is below 1.

C.J. Wilson, was with the Texas Rangers in 2011, now with the Los Angeles Angels of Anaheim in 2012: 74 walks (BB), 191 hits allowed, 233 1/3 innings pitched
WHIP = (74 + 191)/(233 1/3) = 1.187


WAR: Wins Above Placement

Thankfully this is not a measurement of a baseball player's ability to use a weapon. WAR is an encompassing statistic of a player's contribution both offensively and defensively. You may have seen WAR graphs on various sports web sites. Generally, if your WAR is above 8, you are MVP material. If your WAR is around 5, you are an All-Star. If your WAR is around 2, you're an average starter. WAR at 0 and... you will most likely soon to be replaced with a minor leaguer.


These are just some of the statistics that baseball uses. Now it's time to go grab a soda and a hot dog - PLAY BALL!

Source:

MLB Clubhouse - Glossary Page

This blog is property of Edward Shore. © 2012

Review: iMathPac App from Mobile Caltronics

Welcome to blog entry # 69: Happy Easter to Everyone!

Review: iMathPac by Mobile Electronics

Platforms: iPhone, iPod Touch, iPad (can be zoomed)
Cost: Free
Link: Mobile Electronics Website
Location of Company: Central Florida

iMathPac is an app that features a scientific calculator and a graphing application. The features iMathPac have are:

* Trigonometry
* Angle Modes included are Degrees and Radians
* Factorial for any number, not just integers
* Permutations and Combinations
* Full complex number support (rectangular notation only)
* Graphing of functions, inequalities, and inverse functions.
* One touch conversion between fractions and decimal approximations

Below are some screen shots of this powerful and free calculator.

Notes:

* The numeric keypad is in the layout of a phone, which is opposite of the key layout on most calculators.
* There is no shift key, which I believe is by the company's design.
* The e key enters the constant e = 2.17828182845905 (to 16 decimal places). To use the exponential function, just press the e key then the ^ key.
* If you press the 1/x, "^(-1)" will be attached to the end of expression.
* Complex number calculations include trigonometric (watch the angle mode), logarithmic, powers and roots, and exponential - which is really nice!
* There is no "EE" key, so you have to enter large exponents as such: nnnn*10^(xxx)
* The minus key, -, acts as both the negation key and subtraction. In calculators with algebraic systems, I prefer this to having separate keys for negation and subtraction.
* The gamma function can be calculated using the factorial: Γ(x) = (x-1)!
* The absolute value (abs) and floor function (||x||) are available in graph mode only. I do believe the graph mode operates in Radians mode only.
* Tap on the graph to find the value of all functions. Pinch to zoom (not perfect).

What I Love about this App
* Full complex number support - and yes I am a big fan of this
* The factorial function is not restricted to integers - I am also a big fan
* The screen looks sweet, well use of color
* The ability to graph inequalities and inverse functions
* Who can beat free? A lot of features for a free calculator.

What Could be Improved
* No memory, store, and recall. This could be helpful in complex calculations.
* There is no fixed decimal mode settings.
* You can copy expressions, but where to paste them?

Recommendation
iMathPac is a free, powerful calculator, with a lot of powerful features. It is a nice scientific calculator with great graphing capabilities. Definitely worth the download.

Rating: * * * * * (out of 5)

This blog is property of Edward Shore. © 2012

Pies in the Face and Inelastic Collisions

Welcome to blog entry # 68 of my math and calculator blog!

Before I start, I dedicate this blog entry to pranksters, professional clowns (the ones whose job is it to bring laughs, not the annoying politician kind), Shane Jensen, Chris Brame, A.J. Burnett, the Texas Rangers, and the New York Yankees.

Today I am on the tail end of a nasty cold. The saying "laughter is the best medicine" and the beginning of the 2012 MLB baseball season reminds me of the good old "pie in the face" prank. Yes, I like baseball for a lot of reasons and I have grown to like the pieing prank as part of it.

This blog entry, I attempt to bring math, physics, and cream pies together. For those of you who expect a serious subject, my apologizes, but I am going for the silly side today.

Here is what typically happens when someone throws a pie in... a face. Disclaimer: I do not claim to be an artist by any means.

So I break the pie in the face into two stages:

1. Before to Splat
2. Splat to Aftermath

Since the pie sticks to the receiver's head, it is safe to say that the pie in the face row is an example of an inelastic collision. Momentum will be preserved but not kinetic energy.

Variables:

m_pie = mass of a cream pie, about 1 to 2 pounds

m_arm = mass of the thrower's arm, which is about 6.5% of a human's weight according to Shirley A Kindrick, PhD of The Ohio State University.

m_head = mass of the receiver's head. It is approximately 8 to 12 pounds according to The Physics Factbook edited by Glenn Elert.

v_t = velocity of the arm throwing the pie

v_f = velocity of the arm, pie, and head after the contact

v_a = velocity of the pie covered head after contact

Assumptions:

1. The entire pie in the face motion is done in one-dimension. That is, back and forth. The receiver is getting the pie head on at no angle.

2. In the Splat to Aftermath stage, the thrower's arm "falls" out of the picture, allowing the reciever's head to come back to neutral.

3. The amount of "pie" lost is negligible when the contact occurs.

4. To keep is problem simple, the thrower does not "smear" the pie over the receiver.

5. Both the thrower and receiver are standing still. The receiver is not trying to duck, avoid, or do anything to avoid being pied. The receiver exerts no effort after being pied.

Momentum Conservation

Before to Splat Stage:

(m_pie + m_arm) * v_t = (m_pie + m_arm + m_head) * v_f

Splat to Aftermath Stage:

(m_pie + m_arm + m_head) * v_f = (m_pie + m_head) * -v_a

v_a is negative since the head is moving "back" after the thrower releases his grip.

Let's try a Numeric Example

I am going to use US Customary units here:

* pounds for mass (lb) [the SI unit is kg]
* feet/second for velocity (ft/s) [the SI unit is m/s]
* foot-pound (lb_f) for a unit of energy [the SI unit is J (Joule)]

Conversion: 1 mi/hr = 22/15 ft/s. (approximately 1.466667)

A thrower, weighing 210 pounds is holding a 1.5 pound pie at 0.6 mph. He launches it on to his friend's face. His friends weights 220 pounds, assuming his head weights 10 pounds. What is the velocity of friends's head at contact and after contact? What is the kinetic energy at the competition of the two stages?

Data:
m_pie = 1.5 lb
m_arm = 6.5% * 210 = 13.65 lbs
m_head = 10 lbs
v_t = 0.6 mph = 0.88 ft/s

Before to Splat Stage:
v_f:
(1.5 + 13.65) * 0.88 = (1.5 + 13.65 + 10) * v_f
15.15 * 0.88 = 25.15 * v_f
13.33202 = 25.15 v_f
0.53010 = v_f

The arm, pie, and head are moving approximately 0.53010 ft/s (0.77748 mph) at Splat.

Kinetic Energy (KE = 1/2 * mass * velocity^2)
KE before Splat = 1/2 * 15.15 * 0.88^2 ≈ 5.86608 ft_lb
KE at Splat = 1/2 * 25.15 * .53010^2 ≈ 3.53365 ft_lb
(a loss of approximately 39.8%)

Splat to Aftermath Stage:
v_a:
25.15 * .53010 = (1.5 + 10) * -v_a
13.33202 = 11.5 * -v_a
v_a ≈ -1.15931

At the aftermath stage, the velocity of the pied head is 1.15931 ft/s (0.79044 mph) in the other direction, a small whiplash after pushing the friend back with that pie.

Kinetic Energy:
At Aftermath: KE = 1/2 * 11.5 * 1.15931^2 ≈ 7.728 lb_ft
(more than double after release due to loss of mass)

So this is an attempt to explain the physics of the pie in the face.

Thanks for reading as always. To Shane Jansen, Chris Brame and those who like pies in the face, I splat you with a virtual pie.

This blog is property of Edward Shore. © 2012

Follow the Bouncing Ball

Blog Entry #67


Introduction

The goal is to find the velocity of a ball, and the angle it travels, given the initial velocity and angle.

The next two tables are for reference. Thanks to my cousin Gina for helping me with the Coefficient of Restitution! You are the best!

Variables:

Given:
v_i = initial velocity
θ = angle of impact before the bounce

Want to find:
v_f = final velocity
Φ = angle of travel after the ball bounces

Assumptions:
1. Gravity has negligible impact. We are talking about the moment just before the ball bounces to just after the ball bounces.
2. The coefficient of resistution (COR) is measured experimentally (see table). I thank my awesome cousin Gina Ramirez helped me with getting data. (see table above)
3. The x-direction is "forwards/backwards" and the y-direction is "up towards the sky/down towards the earth"
4. The coefficeint of friction only affects the y-direction (is this correct?)
5. Kinetic energy is not conserved.
6. The ground does not move and maintains a zero velocity.

Variables:
COR = coefficient of restitution, depending on the type of ball and surface the ball is being bounced on
μ = coefficient of friction, dependent on the surface the ball is being bounced on
v_ix, v_iy = initial velocity of the ball in the x-direction and y-direction, respectively
v_fx, v_fy = final velocity of the ball in the x-direction and y-direction, respectively

COR = √ ( (height of ball after the bounce) / (height of the ball before the bounce) )

Since momentum must be conserved in both the x and y directions:

m * v_i * cos θ = m * v_f * cos Φ
m * v_i * sin θ = m * v_f * sin Φ

m * v_i * cos θ - m * v_f * cos Φ = 0
m * v_i * sin θ - m * v_f * sin Φ = 0

m * v_i * cos θ - m * v_f * cos Φ = m * v_i * sin θ - m * v_f * sin Φ

m * ( v_i * cos θ - v_f * cos Φ ) = m * ( v_i * sin θ - v_f * sin Φ )

Note the definition of Impulse, I = m * Δv = Δp

Section A: Coefficient for Restitution

COR = v_fy / v_iy

v_fy = v_iy * COR

Hence: v_fy - v_iy = v_iy * COR - v_iy = v_iy * (COR - 1)

Section B: Angle After Bounce

tan Φ = v_fy / v_fx
tan Φ = (v_iy * COR) / v_fx
cot Φ = v_fx / (v_iy * COR)
COR * cot Φ = v_fx / v_iy

Section C: Angle Before Bounce

tan (-θ) = -v_iy / v_ix
- tan θ = - v_iy / v_ix
tan θ = v_iy / v_ix

Main:

Finding Final Angle After Bounce

Impulse, taking friction ( μ ) into consideration:

m * μ * (v_fy - v_iy) = m * (v_fx - v_ix)
μ * (v_fy - v_fi) = v_fx - v_ix

From Section A:

μ * v_iy * (COR - 1) = v_fx - v_ix
μ * (COR - 1) = v_fx/v_iy - v_ix/v_iy

From Section B:

μ * (COR - 1) = COR * cot Φ - v_ix/v_iy

From Section C:

μ * (COR - 1) = COR * cot Φ - tan θ
μ * (COR - 1) + tan θ = COR * cot Φ
(μ * (COR - 1) + tan θ) / COR = cot Φ
COR / (μ * (COR - 1) + tan θ) = tan Φ

which implies that:

Φ = arctan ( COR / (μ * (COR - 1) + tan θ) )

Finding Final Velocity

From Section B:

v_fx = (v_iy * COR) / tan Φ

We also know that:

cos Φ = v_fx / v_f

Then:

v_f = v_fx / cos Φ
v_f = (v_iy * COR) / (tan Φ * cos Φ)
v_f = (v_iy * COR) / sin Φ
v_f = (v_i * sin θ * COR) / sin Φ

Results

Final Angle:

Φ = arctan ( COR / (μ * (COR - 1) + tan θ) )

Final Velcoity:

v_f = (v_i * sin θ * COR) / sin Φ

Sources:

"Friction" http://physics.info/friction Retrieved 3/25/2012

"Coefficient of Restitution" Wikipedia. http://en.wikipedia.org/wiki/Coefficient_of_restitution Retrieved 3/22/2012

H. Brody "That's how the ball bounces" from "The Physics of Sports" edited by Armenti Angelo. American Institute of Physics: New York. 1992


This blog is property of Edward Shore. © 2012

HP 50g: Elastic Collisions

Welcome to Blog Entry #66

This is a program for some solving elastic collision problems for the Hewlett Packard 50g calculator.

Variables (refer to the diagrams above):

MA = mass of object A
VA = initial velocity of object A
θA = initial angle of A

MB = mass of object B
VB = initial velocity of object B
θB = initial angle of B

FA = final velocity of object A
αA = final angle of object A

FB = final velocity of object B
αB = final angle of object B

To type θ, press ΑLPHA right-shift T

To type α, press ALPHA right-shift A

Program:

PROGRAM NAME: ELASTIC
(spaces added for readability)

<< DEG
{ ' MA * VA * COS(θA) + MB * VB * COS(θB) = MA * FA * COS(αA) + MB * FB * COS(αB) '
' MA * VA * SIN(θA) + MB * VB * SIN(θB) = MA * FA * SIN(αA) + MB * FB * SIN(αB) '
' .5 * MA * VA^2 + .5 * MB * VB^2 = .5 * MA * FA^2 + .5 * MB * FB^2 ' }
'EQ' STO
MINIT
"Elastic Collisions"
{ ' MA ' ' VA ' ' θA ' ' MB ' ' VB ' ' θB ' ' FA ' ' αA ' ' FB ' ' αB ' }
MITM MSOLVR >>

Angle Guide

Examples: (FIX 4)

Example 1:

Known:

MA = 1.2450
VA = 40.0000
θA = 30.0000

MB = 0.9780
VB = 45.0000
θB = 225.000

FA = 25.0000
αA = 170.0000

Solved:

FB = 57.1505
αB = 40.2485

Example 2:

Known:

MA = 4.0000
VA = 3.4565
θA = 0.0000

MB = 4.0000
VB = 4.5000
θB = 270.0000

FA = 2.8750
αA = 170.0000
αB = 80.0000

Solved:

FB = 10.2957


This blog is property of Edward Shore. © 2012

Collisions: Elastic and Inelastic

Welcome to blog entry #65! This is the first blog entry using the new Apple iPad (3rd generation, white with 64 GB, so excited)

Introduction

In physics when two objects collide, in general, one of two things can happen:

* They can stick together, producing an inelastic collision, or

* They can bounce off of each other, producing an elastic collision.

In realty, each collision lies somewhere in between inelastic and perfectly elastic, but we will deal with the general cases here.

As the diagram above shows objects A and B are headed on a collision course; each one with their own mass, velocity, and direction. If A and B combine, let C represent the object formed from both A and B.

Define the following variables as:

MA = mass of object A
MB = mass of object B
MC = combined mass of objects A and B = mass of object C

Prior to Collision
VA = initial velocity of object A
VB = initial velocity of object B
θA = initial angle of direction of object A
θB = initial angle of direction of object B

After Collision - Inelastic Collision
FC = final velocity of the aggregate object after objects A and B collide
ΦC = final angle of the aggregate object after objects A and B collide

After Collision - Elastic Collision
FA = final velocity of object A
FB = final velocity of object B
ΦA = final angle of object A
ΦB = final angle of object B

In this blog entry, we will assume we know all the initial conditions VA, VB, θA, and θB. Additionally, assume that friction is negligible, and not a factor in calculation.

A Word About Direction Angle
Angle is used to dictate the direction of the object. 0° is considered going due right (east) (postitive in the x-direction only). If the direction angle is 90°, the object is going due up (north), positive in the y direction only. Refer to the diagram below.

Inelastic Collisions

In inelastic collisions, when objects collide, they tend to "stick" together. Examples include a car wreck or a when football (rugby) defensive end tackles a wide receiver.

Only momentum is conserved in inelastic collisions. Kinetic energy that was given by the two objects is lost. Some energy is either vibrations or heat given by the combined object.

The equation for momentum is: P = m * v, where m is mass and v is velocity.

When solving for the general case, we have to deal with the total momentum in both the x-direction and the y-direction.

The initial momentum for each direction respectively is:

Px = MA * VA * cos θA + MB * VB * cos θB (1)
Py = MA* VA *sin θA + MB * VB * sin θB (2)

Since momentum is conserved, the total initial momentum must be equal to the total final momentum, in the x- and y-direction. Hence:

Px = MC * FC * cos ΦC (3)
Py = MC * FC * sin ΦC (4)

where MC = MA + MB (5)

As a result, we have two equations of two unknowns: VC and ΦC. A calculator (such as HP 50G, TI 89, TI nSpire) or a sufficient mathematics computer program can solve the set of equations easily.

Since the equations are relatively simple, we can find explicit solutions by the following strategy.

First, solve for ΦC. Either equation (3) or (4) can be used. For this blog entry, I will use equation (3).

Px = MC * FC * cos ΦC
Px / (MC * FC) = cos ΦC

Then ΦC = arccos (Px / (MC * FC)) (6)

Sub into (4) to get:

Py = MC * FC * sin (arccos (Px / (MC * FC)))

Note that sin(arccos x) = √ (1 - x^2). Also cos(arcsin x) = √ (1 - x^2)

Then:

Py = MC * FC * √ ( 1 - (Px / (MC * FC))^2 )
Py = MC * FC * √ ( ( MC^2 FC^2 - Px^2 ) / (MC^2 * FC^2) )
Py = (MC * FC * √ ( MC^2 FC^2 - Px^2 ) ) / (MC * FC)
Py = √ ( MC^2 * FC^2 - Px^2 )

Square both sides and rearrange terms:

MC^2 * FC^2 = Py^2 + Px^2

Solving for FC gives:

FC = √ ( (Py^2 + Px^2) / MC^2 ) (7)

(We can just take the principal root, since the ΦC will take care of the direction of object C)

In Summary:
MC = MA + MB
FC = √ ( (Py^2 + Px^2) / MC^2 )
ΦC = arccos (Px / (MC * FC))

This is a solution for an inelastic collision for any two objects of any mass, velocity, and direction angle.

Elastic Collisions

In perfectly elastic collisions, when objects collide, they tend to "bounce off" each other. Examples include a sub-atomic gas particle collision or two billiard balls hitting each other.

Like inelastic collisions, momentum is conserved in inelastic collisions. Unlike inelastic collisions, kinetic energy is also conserved.

The equation for momentum is: P = m * v, where m is mass and v is velocity.

When solving for the general case, we have to deal with the total momentum in both the x-direction and the y-direction.

The initial momentum for each direction respectively is:

Px = MA * VA * cos θA + MB * VB * cos θB (1)
Py = MA* VA *sin θA + MB * VB * sin θB (2)

The equation for kinetic energy is: KE = 1/2 * m * v^2. Kinetic energy does not take separate compenent directions into account.

The initial kinetic energy is:

KE = 1/2 * MA * VA^2 + 1/2 * MB * VB^2 (8)

Since momentum and kinetic energy are conserved, the total initial momentum must be equal to the total final momentum, in the x- and y-direction; and the total initial kinetic energy must be equal to the total final kinetic energy. Hence:

Px = MA * FA * cos ΦA + MB * FB * cos ΦB (9)
Py = MA * FA * sin ΦA + MB * FB * sin ΦB (10)
KE = 1/2 * MA * FA^2 + 1/2 * MB * FB^2 (11)

Note if we do not know anything about the final conditions, then we are dealing with a system of three equations with four unknowns: FA, ΦA, FB, and ΦB.

In order for this system to be solvable, we must know one of the final conditions. Then the system becomes three equations and three unknowns. A calculator (such as HP 50G, TI 89, TI nSpire) or mathematics computer program can assist in solving the set of equations.

What if one of the velocities is known?

Let's assume that FA is known. We can simplify the system of equations (9), (10), and (11). A similar strategy can be used if FB is known.

Start with (11). With FA known, solve for FB.

KE = 1/2 * MA * FA^2 + 1/2 * MB * FB^2
1/2 * MB * FB^2 = KE - 1/2 * MA * FA^2
FB^2 = 2 * KE / MB - MA * FA^2 / MB

Then:

FB = √ ( ( 2 * KE - MA * FA^2 ) / MB ) (12)

Now we are left with the system of two equations with two unknowns: ΦA and ΦB. Use your favorite calculating tool to solve for them.

Px = MA * FA * cos ΦA + MB * FB * cos ΦB (9)
Py = MA * FA * sin ΦA + MB * FB * sin ΦB (10)


For more information of elastic and inelastic collisions, click here (Hyperphysics Page). . Thanks Mime Grigsby, a very good friend of mine and fellow mathematician.


Until next time,

Eddie

Diagrams: drawn using Whiteboard

This blog entry is dedicated to Uncle Ralph (born March 21). Me and my family miss you and love you. Thank you for touching our lives.

This blog entry is property of Edward Shore. © 2012

Billiards - Determining Final Velocity (Program for TI 84+ and HP 50g too!)

Welcome to blog entry #64!

HAPPY ST PATRICK'S DAY (WEEKEND)!

Billiards and Math

Ever wondered what happens with you take the cue stick, strike the cue ball, hit your target ball and the target ball sails right into a pocket? If you played billiards, commonly known as pool, you have experienced this many times.

We can use mathematics and physics to describe what happens when the cue ball hits the target ball. Since the balls do not "stick" together like steamed rice, the collision is considered to be nearly elastic.

In an elastic collision, two properties are conserved: momentum and kinetic energy. Momentum is the product of mass and it's velocity. Kinetic energy is the energy of movement. The equations for momentum and kinetic energy are, respectively:

P = m * v

KE = 1/2 * m * v^2

Note that direction is not considered in calculating kinetic energy.

Problem:

You hit a cue ball at a certain velocity VI, and hit a target ball at angle θ. We want to find out the direction of the cue ball after collision and the velocity of both balls.

Assume the following:

1. The masses of the cue ball and the target ball are the same. This will allow a great simplicafication in calculation, as we will see.

2. The Intial direction of the ball is 0° (due right). As a visual, you are on the left hitting the ball to due right.

Note the diagrams below:

3. The angles created after the hit add up to 90°.

From the diagram above, θ + Φ = 90°.

4. The cue ball does not hit any balls other than the target ball. If more than two balls were involved, calculation can get hairy, quick.

5. The cue ball is hit at it's sweet spot, which eliminates any friction developing between the ball and the felt of the pool table. According to The Real World Physics Problems web site, the sweet spot is 1.4 times the ball's radius.

An American billiard ball has a diameter of about 2.25 in, or 5.715 cm. The ball has a typical weight of 5.5 to 6 oz, or 156 to 170 grams. This link has more information on billiard balls.

Solution, Assuming All Masses Are Equal

Let:

VI = initial velocity of the cue ball, going due right. In the x-direction, VI cos 0° = VI. In the y-direction, VI sin 0° = 0, as expected.

VCF = final velocity of the cue ball

VBF = final velocity of the target ball. Note since the target ball was at rest, it's initial velocity is zero.

Φ = Angle where the cue ball goes after collision. Simply put, Φ = 90° - θ.

MC = mass of the cue ball

MB = mass of the target ball

Note: We assumed the masses were equal, MB = MC.

Momentum and kinetic energy has to be preserved. So for momentum in the x-direction:

MC * VI = MB * VBF cos θ + MC * VCF cos Φ

and momentum in the y-direction:

0 = MB * VBF sin θ - MC * VCF sin Φ

Note: Φ is a angle in the clockwise direction and sin (-Φ) = -sin Φ

and for kinetic energy:

1/2 * MC * VI^2 = 1/2 * MB * VBF^2 + 1/2 * MC * VCF^2

Note: Since mass is the same for all balls (MC = MB), mass "cancels" out in all three equations.

To Recap:
VI = VBF cos θ + VCF cos Φ
0 = VBF sin θ - VCF sin Φ
1/2 * VI^2 = 1/2 * VBF^2 + 1/2 * VCF^2

Making use of the trigonometric identities:

sin (90° - x) = cos x
cos (90° - x) = sin x

The equations simplify to:
VI = VBF cos θ + VCF sin θ
0 = VBF sin θ - VCF cos θ
1/2 * VI^2 = 1/2 * VBF^2 + 1/2 * VCF^2

Observe that if we solve for VBF in the second equation, we get:
VBF = VCF * (cos θ / sin θ)

Then:
VI = VBF cos θ + VCF sin θ
VI = VCF * (cos θ / sin θ) *cos θ + VCF sin θ
VI = VCF * (cos^2 θ / sin θ ) + VCF sin θ
VI = (VCF * cos^2 θ + VCF sin^2 θ) / sin θ

Note: sin^2 x + cos^2 x = 1 for an angle x.

And:
VI = VCF / sin θ

Solving for VCF yields:
VCF = VI sin θ

Let's get VBF in terms of VI:
VBF = VCF * (cos θ / sin θ)
VBF = VI sin θ* (cos θ / sin θ)
VBF = VI cos θ

Before we can celebrate, we need to satisfy the conservation of kinetic energy, not just momentum.
1/2 * VI^2 = 1/2 * VBF^2 + 1/2 * VCF^2

First, let's just multiply everything by 2:
VI^2 = VBF^2 + VCF^2
VI^2 = (VI cos θ )^2 + (VI sin θ)^2
VI^2 = VI^2 * (cos^2 θ + sin^2 θ )
VI = VI

So this checks out because we assumed the masses of both the cue ball and target ball were equal. .

Final solution - All Masses being Equal:

VBF = VI cos θ
VCF = VI sin θ
Φ = 90° - θ

What if the Mass of the Cue Ball and Target Ball are not equal?

Let:

MC = mass of the cue ball

MB = mass of the target ball

If I use the strategy of solving for VCF and VBF above, I get:

VCF = VI * sin θ.
VBF = (MC / MB) * VI * cos θ

We are working with elastic collisions, so conservation of kinetic energy must be considered:

1/2 * MC * VI^2 = 1/2 * MB * VBF^2 + 1/2 * MC * VCF^2

But:
1/2 * MB * VBF^2 + 1/2 * MC * VCF^2
= 1/2 * (MB * (VI * cos θ * MC / MB)^2 + MC * VI^2 * sin^2 θ))
= 1/2 * MC * VI^2 * (cos^2 θ / MB + sin^2 θ)
≠ 1/2 * MC * VI^2

Yep. No go. Confession, I thought the above way was good to go when MC ≠ MB too.

Here is a real solution if MC ≠ MB:

VBF = (MC * VCF) / (MB * tan θ)

And we have to solve the quadratic equation for VCF:

0 = VCF^2 * (MB * MC^2 * sin θ + MC^2 * cos^2 θ)
+ VCF * (MB * MC)
+ (MB * sin θ * (MC * VI + MC * VI^2))

The following calculator programs solve for Φ, VCF, and VBF with the assumption that the mass of both the cue ball and target ball are the same.

Variables for the programs:
C = VI
A = θ

TI 84 Plus:

NAME: BILLIARD

Degree
Disp "Deg MODE"
Input "INIT. VEL.",C
Input "ANGLE HIT",A
C*SIN(A) → S
C*COS(A) → T
Disp "ANGLE CUE=", 90 - A
Pause
Disp "FINAL VEL. CUE=",S
Disp "FINAL VEL. BALL=",T

HP 50g:

NAME: BILLIARD

<< DEG
"Degrees Mode" MSGBOX
"Initial Velocity, Cue" PROMPT
"Hit Angle" PROMPT
→ C A
<< "Finals:"
90 A - →NUM "Angle, Cue" →TAG
C A SIN * →NUM "Vel., Cue" →TAG
C A COS * →NUM "Vel., Ball" →TAG
>> >>

Test data:
VI = 5, θ = 15°; Results: VCF ≈ 1.2941, VBF ≈ 4.8296

VI = 2, θ = 30°; Results: VCF = 1, VBF = √3 ≈ 1.7321


Thank you, hope you enjoyed this blog entry. Next time I probably be blogging on a third generation iPad. For the first 64 entries, I have been using the Blogger+ App by ThinkTek Co. Ltd. on my iPad 2. Thanks everyone!

Eddie


This blog is property of Edward Shore. © 2012

Projectiles - A Realistic Point of View (Linear Drag)

Projectile Motion

This blog is about projectile motion. Most resources give an introductory treatment of projectile motion. And yes, I will present the non-air resistance model in this section. But I am also going to present a more realistic model, taking into account air resistance, size of the projectile, and temperature.

All amounts are presented in SI (kilograms, meters, seconds) units.

Quick Conversions
1 ft = 3.280839895 m
1 lb = 0.4535924 kg

Simple Projectile Motion

A perfect projectile can be described by the following equations:

x(t) = v0 t cos θ
y(t) = v0 t sin θ - 0.5 g t^2

Where
v0 = initial velocity
θ = Intial angle, in degrees
g = gravitational acceleration constant = 9.80665 m/s^2 = 32.17404856 ft/s^2
And we assume the starting height is zero (on the ground)

Note that air resistence, features of the object, and other outside objects are ignored.

Example: Hitting a Golf Ball

Use the simplistic model to plot a trajectory with an initial velocity of 150 mph (67.056 m/s) at a 35° angle.

The trajectory lasts for approximately 7.844 seconds, with a range of approximately 430.86 meters and a height of approximately 75.4 meters.

A More Realisitc View of Projectile Motion

The equations above describe simple projectile motion. However, if we want a realistic picture, we have to consider factors such as temperature, air pressure, and the size of the projectile.

So of the variables we will consider are the drag coefficient, the density of air, and the object's terminal velocity.

Note: Linear drag is assumed. This is best for objects moving at slow speeds. No lift is used.

Drag Coefficient

The drag coefficient is a dimensionless quantity that quantifies the resistance of an objects's movement, taking into consideration factors such as the object's size, shape, and smoothness.

Approximate Drag Coefficients:
Perfect Smooth Baseball: 0.1
Baseball: 0.3
Golf Ball: 0.4
Tennis Ball: 0.6

Density of Air

The density of air can be determined using the Ideal Gas Law:

P V = N R T

where
P = absolute pressure = 101325 kPa
R = Specific Gas Constant = 287.058 J/kg K
T = temperature in Kelvins
N = number of moles (mass)
V = volume of the gas
ρ = air density = N/V in kg/m^3

Solving for ρ gives:

ρ = P / (R T) ≈ 352.9774471/T

Note: To convert degrees Celsius to Kevlins, add 273.15.

Terminal Velocity

When an object falls, it reaches a terminal velocity when the drag becomes equal to the object's weight. At terminal velocity, the object no longer accelerates.

Terminal Velocity is measured by:

V = √ ( ( 2 × mass of object × g ) / ( drag coefficient × ρ × surface area of object ) )

TROJECT
Calculator: CASIO fx-CG10 (Prizm)

Note: Let [triangle] be the symbol for the "stop and display triangle"


{.3, .6, .4} → List "C"
{.145, .057, .045} → List "M"
{.041043, .003318, .001385} → List "S"
Deg
ClrText
"Init Velocity="?→ V
"Init Angle="? → θ
9.80665 → G
Menu "Object","Baseball",1,"Tennis Ball",2,"Golf Ball",3
Lbl 1: 1 → I : Goto Y
Lbl 2 : 2 → I : Goto Y
Lbl 3 : 3 → I : Goto Y
Lbl Y
Menu "Temperature","104°F/40°C",A,"95°F/35°C",B,
"86°F/30°C",C,"77°F/25°C",D,
"68°F/20°C",E,"59°F/15°C",F,
"50°F/10°C",G,"41°F/5°C",H,
"32°F/0°C",I
Lbl A : 1.1839 → P : Goto Z
Lbl B : 1.2041 → P : Goto Z
Lbl C : 1.2250 → P : Goto Z
Lbl D : 1.2466 → P : Goto Z
Lbl E : 1.2690 → P : Goto Z
Lbl F : 1.2920 → P : Goto Z
Lbl G : 1.3163 → P : Goto Z
Lbl H : 1.3413 → P : Goto Z
Lbl I : 1.3943 → P : Goto Z
Lbl Z
List "M"[I] → M
√ (( 2 × M × G) ÷ (List "C"[I] × P × List "S"[I])) → W
M × G ÷ W → K
Solve(-MGX ÷ K + M ÷ K × ( 1 - e^(-KX ÷ M)),100) → E
MV cos θ ÷ K × ( 1 - e^(-KE ÷ M)) → R
"Range="
R [triangle]
-M ÷ K × ln ((MG) ÷ K × (V sin θ + MG ÷ K)^-1) → F
-MGF ÷ K + M ÷ K × (V sin θ + MG ÷ K) × (1 - e^(-KF ÷ M)) → H
"Height="
H [triangle]
ClrGraph
ViewWindow -.5, R + .5, 1, -1, H+1, 1, 0, E, .05
θ → A
ParamType
G SelOff
G SelOn 1
"MV cos A ÷ K × (1 - e^(-KT ÷ M))" → Xt1
"-MGT ÷ K + M ÷ K × (V sin A + MG ÷ K) × (1 - e^(-KT ÷ M))" → Yt1
DrawGraph

-----------

Variables:

M = mass of object, in grams
V = initial velocity, in metes/seconds
A = θ = initial angle, in degrees
C = drag coefficient
S = radius of the object
P = gas density
K = terminal velocity

Revisiting the Example: The Golf Ball

Use the proposed model to plot a trajectory with an initial velocity of 150 mph (67.056 m/s) at a 35° angle. Assume it is 77°F outside (25°C).

The trajectory has a range of approximately 163.46 meters and a height of approximately 45.02 meters. It lasts about 6.177 seconds.

References:

Article: Erlichson, Herman. "Maximum Projectile range with drag and lift, with particular application to golf." The College of Staten Island, 1982

From:

Armenti, Angelo Jr. (editor) The Physics of Sports American Institute of Physics 1983 pg. 71-78

Reference Links:

Terminal Velocity (NASA.gov)

Drag Coefficient (NASA.gov)

Drag on a Baseball (NASA.gov)

Air Pressure

My Favorite Number: π

Why π is My Favorite Number

The number π describes a beautiful relationship between a circle's circumference, area, and radius.

The π is irrational, which means the decimal expansion goes without end.

π ≈ 3.14159265358979323846... (this is from memory). Mathematicians from ancient Greece on forward have worked on obtaining more digits of π. I think the current count is a trillion decimal places.

The gamma of 1/2 is the square root of π.

π is involved in the normal distribution function: f(x) = e^(-2x)/√ (2 π)

Unlike pie, enjoying π doesn't make anyone fat. Furthermore, unlike pie, π has no calories.

The (basic) sine and cosine functions have a period of 2 π radians. (360°)

The number π is involved in infinite series. For example:
π /4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 - 1/15 + 1/17 - 1/19 + 1/21...

or π = 4 Σ ((-1)^n / (2n + 1), n = 0 to infinity)

This series converges slowly - at 10,000 terms I only get 3 decimal points (3.14169264367).

I was born on March 14 - and π = 3.14 to two decimal places.