Categories
Uncategorized

Pullups and Logarithmic Decay

So, I was talking to a friend about how pullups need more rest time between sets than other exercises:

Pullups need significant recovery time to not have logarithmic decay in the number of reps you can do.

I suggested waiting about 3 minutes between sets.  Of course, the next time I did them, I wanted to see just how logarithmic the decay is.  I plugged the numbers into Excel and did a log regression.  Wow, I wasn’t kidding, look at that correlation coefficient !!

Categories
Learning Skills Tutoring

Before you hire a private math tutor…

Before you spend your hard earned money on a private tutor, first utilize the resources you are already paying for.  To that end, please encourage the student to help themselves.  (That’s what it’s all about, right?)  The lynchpin of my learning philosophy is to “teach a man to fish instead of giving a man a fish.”  Esoteric mathematical skills aside, use this opportunity to push beyond your comfort zone and improve your ability to learn anything.

  1. Print out my chart below
  2. Keep it in your notebook
  3. Try everything at least once.
  4. Hand it in to me

Please attempt every strategy at least once.  Print my chart, put it in your notebook, and check off each one after you do it once.  I think completing the grid within 2 weeks is reasonable.   (Sure, hide it from your friends so they don’t laugh at you.)  The goal here is to “nudge” and evolve your scholarship repertoire when reacting to a complex academic challenge.  Developing these habits will benefit the remainder of your academic career.

 

Categories
Learning Skills SAT

The Wrong Way to Prepare for the Math SAT

The Math SAT is an aggressively timed exam which hinges upon numeric, algebraic, and geometric foundations. The de facto approach taken by many SAT books, courses, and tutors focuses on doing sample SAT tests.  This is productive when the student’s SAT math foundations are already strong.  The student isn’t reviewing concepts at this stage, but rather, is practicing & optimizing. But, is this approach ideal for students who are just beginning to prepare for the exam?

I feel that effective preparation for the SAT entails a two stage approach. You should first back-fill core foundations before jumping into randomized SAT problems (which students may not yet be sufficiently prepared to tackle). In other words, targeted review “by topic” will prepare him for tackling the comprehensive practice tests. If you do 10 problems of a specific concept in a row, the concept is much more likely to “stick”. In contrast, prematurely doing practice tests can result in the following unproductive cycle: try a problem, get it wrong, move to an entirely different problem.

If you buy into my rationale, the key is to find an SAT book that clusters the practice problems “by topic”.

Categories
Learning Skills

Excerpts from Rapt: Attention and the Focused Life

  • Learning tennis is best done between the ages of 8 and 15.  The longer you delay, the harder it will be, and your ability to play will suffer.  This principle applies to all sorts of skills, both physical and mental, including the ability to concentrate, direct your focus at will, manage your time.  Kids also need to work at developing the capacity for the concentrated, sustained attention required to succeed in many endeavors.
  • The young can get away with IM’ing while playing a computer game, but there’s a risk.  If you grow up assuming that you can pay attention to several things at once, you may not realize that the way you process such things is superficial at best.  When you’re finally forced to confront intellectually demanding situations in high school or college, you may find you’ve traded depth of knowledge for breadth, and stunted your capacity for serious thought.
  • Studies show that through practice, you can expand your capacity to focus.
  • Where big breakthroughs are concerned, getting to “That’s it!!” requires not only the intense focus and explicit learning …but also plenty of (non-conscious) incubation, mind wandering, and implicit learning.
  • Science has determined that multitasking, for most practical purposes, is a myth.  Focusing on 2 demanding activities simultaneously is a skill that requires months of drilling to acquire, and, even then, is confined to just those two tasks!
  • You may think you’re multitasking, but what you’re really doing is switching back and forth between two activities.  The extra effort involved actually makes you less productive.  Your overall performance will be inefficient, error prone, and more time consuming than if you had done one thing at a time.  If your train of thought is interrupted even for a second, you have to go back and say “Where was I?”  There are startup costs each time as you reload everything into memory, and people aren’t as good at is as they think they are.
  • When you focus on a demanding task, your brain’s hippocampus, which is important to memory, is in charge.  However, if you try to work while distracted by instant messaging, or the like, the Striatum, which is involved in rote activities takes over.  As a result, even if you get the job done, your recollection of it will be more fragmented, less adaptable, and harder to retrieve than it would be had you given it your undivided attention.

 

Categories
Learning Skills

So, you’re not happy with your test grade?

Categories
Learning Skills

NYMag: Distraction

Distraction

  • Poor retention:  “…when forced to multitask, the overloaded brain shifts its processing from the hippocampus (responsible for memory) to the striatum (responsible for rote tasks), making it hard to learn a task or even recall what you’ve been doing once you’re done.”
  • He sees our distraction as a full-blown epidemic—a cognitive plague that has the potential to wipe out an entire generation of focused and productive thought. He compares it, in fact, to smoking. “People aren’t aware what’s happening to their mental processes,” he says, “in the same way that people years ago couldn’t look into their lungs and see the residual deposits.”
  • If John Lennon had a BlackBerry, do you think he would have done everything he did with the Beatles in less than ten years?”
Categories
Tutoring

Practical Advice For Hiring a Math Tutor

  1. Try to hire an experienced classroom teacher.  The perspective gleaned from correcting 1000’s of student exams over the years can not be understated. From this experience, a classroom teacher instinctively knows what mistakes students typically make, and which topics need extra attention.  Classroom teachers are also very familiar with the content and grading styles of NY State standardized tests.
  2. Pair up with a friend for tutoring sessions. 1 tutor. 2 students. You have cut your tutoring expense by 50% while still maintaining a high level of individual attention. Students also may find the session more enjoyable when doing problems with a friend.

I am a certified, tenured, full time faculty member at Somers High School.  I find time for occasional tutoring in Westchester, NY.   I enjoy doing Math problems with motivated students, and it’s a way to keep abreast of what other schools in the area expect from their students.

I only tutor what I have taught in the classroom:

  • Algebra (9th)
  • Geometry (10th)
  • Algebra 2 / Trigonometry (11th)
  • Precalculus (12th)
  • SAT / ACT / SAT 2 (Math: IC & IIC)

You can contact me for more information

 

Categories
Learning Skills

Procrastination and Setting Smaller Milestones

Here’s an interesting study that confirms what you already know about procrastination. You’re better off spacing out your work with self-imposed deadlines, if you want to avoid waiting until the last minute.  English teachers know this well, and that’s why they explicitly break up phases of a longer-term research paper into smaller, more regular milestones.

 They set up three classes, and each had three weeks to finish three papers. Class A had to turn in all three papers on the last day of class, Class B had to pick three different deadlines and stick to them, and Class C had to turn in one paper a week.

Which class had the better grades?

Class C, the one with three specific deadlines, did the best. Class B, which had to pick deadlines ahead of time but had complete freedom, did the second best, and the group whose only deadline was the last day, Class A, did the worst.

 

Categories
Learning Skills SAT

FAQ: SAT, SAT II

Q: When should I take the SAT II (Level IC)?
A: It can be taken after finishing Math 2 (Geometry)

  • Math 1: Algebra
  • Math 2: Geometry

______________________________________________________________________________

Q: When should I take the SAT II (Level IIC)?
A: It can be taken after finishing Math 3 (Alg2/Trig)

  • Math 1: Algebra
  • Math 2: Geometry
  • Math 3: Selected Algebra2/Trig topics:
    • Imaginary numbers (i)
    • Systems of equations
    • Functions
    • Composition of functions
    • Logs
    • Trigonometry
    • Law of Sin & Law of Cos

__________________________________________________________________________________

Q: Are Math 3 (Algebra2/Trig) topics on the ACT?
A: Yes, about 8-10 problems out of 60.

  • Domain/Range
  • Systems of equations
  • Absolute Value Inequalities
  • Logs
  • Trigonometry
  • SOH|CAH|TOA
  • Radians (pi/180)
  • Sin/Cos graphs (0 to 2pi)
Categories
Learning Skills

To Deal With Obsession, Some Defriend Facebook

 

To Deal With Obsession, Some Defriend Facebook

Categories
Learning Skills

Computers at Home: Educational Hope vs. Teenage Reality

 

 

Computers at Home:  Educational Hope vs. Teenage Reality

The Duke paper reports that the negative effect on test scores was not universal, but was largely confined to lower-income households, in which, the authors hypothesized, parental supervision might be spottier, giving students greater opportunity to use the computer for entertainment unrelated to homework and reducing the amount of time spent studying.

Categories
Geometry

What does pi (3.14159) have to do with taking great photographs?

The lens in this photo has a focal length of 50mm.  In general, the higher the focal length, the more zoom you’ll have.  For example, a wide angle lens is between 9mm-24mm, where a zoom/telephoto lens can be 150-400m.  However, let’s focus on the list of numbers on the bottom edge of the lens:  1.4, 2.8, 4, 5.6, 8, 11, 16.   At first glance, this is a bizarre sequence of random numbers.  These numbers allow you to choose the f/stop, which is the ratio between the diameter of the lens opening (aperture) and the focal length of the lens.  For example, for an f-stop of 2 (written f/2.0) the diameter would be 25mm while the focal length is 50mm, because 25mm divides into 50mm two times.  Hence, the general equation is:  \(f/stop = \frac{focal\ length}{diameter}\)  Note that when the f/stop is low, it means the aperture is large.  (FYI, the point of a large aperture is to let in a ton of light to improve picture quality).

http://www.http://mathaholic.com/images/ApertureLens.jpg.com/

Next, let’s play around with some numbers and see where this leads us.  The simplest example would be to consider an f/stop ratio of 1 (written f/1.0) on a 50mm lens:  This would mean the diameter of the lens opening (aperture) is 50mm, making the radius equal to 25mm.  Remember the old circle formulas from middle school?  Well, knowing the radius, we can calculate the actual area of the lens opening at f/1.0.

http://commons.wikimedia.org/wiki/File:CIRCLE_1.svg

\(Area = \pi r^2\)

 

\(Circumference = 2 \pi r = \pi d\)

 

\(A = \pi r^2 = \pi (25)^2 = 1963.5\)

 

 

Ok, so let’s try an f/stop that is actually on the lens (f/1.4):

\(1.4 = \frac{50}{diameter}\) … (so d = 35.7 and r = 17.86)

\(A = \pi r^2 = \pi (17.86)^2 = 1001.8\)

An f/1.4 lens has an aperture area of about 1000.  Do you see any relationship between the area for f/1.0 vs. f/1.4?  If not, I plugged in the rest of the f/stop numbers printed on the lens into a spreadsheet.  What do you notice about the area of the circle for each subsequent f/stop?

In fact, the f/stop numbers that initially seemed so random actually do have a very precise relationship to each other.  The area of the circle is being approximately halved for each of these f/stops.  Conversely, for each f/stop you drop down, you are doubling the area of the lens aperture (effectively doubling the amount of light that the lens will let in!) Compare f/16 to f/1.4.  They are 7 stops apart, meaning the amount of light doubles seven times.  That means an f/1.4 lens allows 27 = 128 times as much light as the f/16 lens!!  That makes a huge impact on the kind of pictures you can take when lighting is not optimal (indoors, night, etc).  Most pocket cameras are about f/4.  Even an f/2 lens will allow 4x as much light in (2 stops down)

Categories
Algebra Graphs Life

The older you get, the faster each year passes by. Why?

RE: Birthday Dinner
Yes, life does go by fast.  Strangely, the older you get, the faster it goes.  I do not know why this is.

Ever get an email like this?  Well, as your age varies, the percentage of your life that a single calendar year represents also varies.  As you get older, a year is a smaller percentage of your overall life.  In other words, 1 year represents 50% of a 2 year old’s life.  However, it is only 2% of a 50 year old’s life.  So, perhaps that is why each year seems to go by faster.

Want to see the percentage for every age from 0 to 80?  Let’s make a formula and graph it.  The percentage of your life that a single year represents is just a function of your age:  \(f(age) = \frac{1}{age}\)  If you graph this on a spreadsheet, you’ll get the following:

How would you interpret this graph?  You’ll notice that once you pass the inflection point, the percentage seems to flatten out.  So, at what point can a person legitimately start saying “Wow, this year really flew by?”  Based on the graph, teenagers might feel this almost as much as middle aged people.

Lastly, do you notice how scaling of the y-axis makes the difference between age 15 and 50 look trivial?  In order to properly display percentage changes, I will scale the y-axis logarithmically.  Here is the result:

With this scaling, you can see a year in the life of a teenager (~6%) is quite different than a year in the life of someone in their 50s (~2%)

 

Categories
Statistics

The Wallaby That Roared Across the Wine Industry

The Wallaby That Roared Across the Wine Industry

By the end of 2001, 225,000 cases of Yellow Tail had been sold to retailers. In 2002, 1.2 million cases were sold. The figure climbed to 4.2 million in 2003 — including a million in October alone — and to 6.5 million in 2004. And, last year, sales surpassed 7.5 million — all for a wine that no one had heard of just five years earlier.

Prima facie, it looks like exponential growth.  But, in the real world, nothing ever grows exponentially in perpetuity (except college tuition, it seems)   I looked up sales figures for other years online.  Let’s plot these numbers in a spreadsheet, and see how they look.  As you can see the growth started to flatten out after a few years. I actually couldn’t find the sales data for 2008, so this calls for a statistical regression (fancy words for “line of best fit”).  A linear regression only yielded r=.88, while a 2nd degree polynomial (quadratic) regression gave an r = .96.  This regression equation is \(f(x) = -.13x^2 + 513x – 515887\)

Do you notice the negative leading coefficient of the x2 term? Remember how this makes the parabola “frown”?  Well, this “inverted parabola” shape clearly reflects the flattening of the sales growth.  

By just looking at the trendline,what’s your estimate for the number of cases sold in 2008? Or, plug 2008 into the equation to get the exact coordinates on the red trendline:  \(f(2008) = -.13(2008)^2 + 513(2008) – 515887 \)

 

Categories
Probability

Why Casinos Don’t Lose Money

First, let’s illustrate the law of large numbers.  If you flipped a coin 10 times, you should expect to get 50% heads.  However, this may not happen.  You could get anything from 0 to 10 heads.  The most likely outcome is flipping 5 heads, and the odds of this outcome is 50%.  Variations from this become increasingly less probable.  For example, the odds of you flipping 4 or 6 heads is 26%, and the odds of you flipping 3 or 7 heads is 10%.  Now, let’s say you flipped a coin 100 times.  The odds of getting exactly 50% heads is still 50%.  But, do you think the odds of getting 60 heads is also 26%?  It’s actually only 2%.  It’s much easier to get 6 out of 10 heads, than it is to get 60 out of 100 heads.  What are the odds of flipping 600 heads out of 1000 flips?  It’s 0.00000001%.  With 1000 flips, you’re pretty much always going to get around 48%-52% heads.  Deviations beyond that range are very improbable.  So, in summary, the law of large numbers states that the more trials you have, the closer your actual outcome will be to the theoretical expected probability (In this case, the more coins you flip, the more you’ll start to approach actually getting 50% heads)

So, how does this tie into casinos?  Let’s take the roulette wheel as our example.  There are 37 total numbers.  18 reds, 18 blacks, and 1 green.  If you guess red or black correctly, you’ll get a 1:1 payout (ie: If you bet $1, you’ll get back $2, thereby winning $1).  If the wheel lands on green (0), both red and black lose.  This is where the casino gets it’s edge in this particular gamble.  Let’s calculate the expected value of a $1 bet on red.

\(E[X] = \frac{18}{37}(\$1)+\frac{18}{37}(-\$1)+\frac{1}{37}(\$-1) = -\$.03\)

 

What this means is you have an 18/37 chance of winning $1 (if it lands on red), and 18/37 chance of losing $1 (if it lands on black), and a 1/37 chance of losing $1 (if it lands on green)  The expected profit for playing this game is negative 2 cents.  Now, sometimes you’ll win, and sometimes you’ll lose, but if you play enough times, you’ll be averaging a loss of 23 cents per round.  This is where the law of large numbers comes into play.  As long as enough people are playing, the house will be averaging a profit of 2 cents for every dollar bet on that roulette table.

Question: What is the expected value of correctly guessing a specific number? There are 37 numbers, but the payout is 35:1 (You get paid $35 for each dollar you bet)  Based on this answer, is it smarter to try guessing the color or guessing the number?

 

Categories
Algebra 2

The Beatles meet Mathematics & Physics

About this sound Listen to the opening chord

 

Mathematics, Physics and A Hard Day’s Night

In this article we shall use mathematics and the physics of sound to unravel one ofthe mysteries of rock ’n’ roll – how did the Beatles play the opening chord of A Hard Day’s Night? The song may never sound the same to you again.

 

I just love this paper.  Professor Jason Brown sampled the famous opening chord of this song and (using Math) separated out all the distinct frequencies in the clip.  From this, he was able to determine each individual note that was played.  From there, he determined exactly what each member of the band played.  He even discovered a surprising element relating to George Martin’s 5th Beatle status.

Professor Brown took each frequency and converted it to a musical note on the Western scale.  Here is the function he used to do this:

\(f(x) = 12 log_2(\frac{x}{220})\)    (…where 220 hertz is the frequency for A natural.)

I thought I’d try his calculation myself, because it’s a good opportunity to use the change of base formula for logarithms.  If you look in the original white paper, the first frequency in the table is 110.34   Let’s plug this into the function:

\(f(110.34) = 12 log_2(\frac{110.34}{220}) = 12 log_2(.5015)\)

How do you evaluate the above expression?  There is no \(log_2()\) button on most calculators.  Well, here’s where the change of base formula for logs comes in:  \(log_b(x) = \frac{log_d(x)}{log_d(b)}\)   So, let’s choose \(log_{10}\) (since calculators do have this) and continue:

\(12 log_2(.5015)=12(\frac{log_{10}.5015}{log_{10}2}) = 12(-.9957) = -11.9466\)

 

In other words, 110.34hz is -11.9466 semi-tones below the note of A.  It should really be 12, but as Professor Brown noted, the Beatles’ instruments weren’t in perfect tune, so the values are not whole numbers!

So, which note is 12 semi-tones below A?  Actually, 12 semi-tones makes an octave, so the answer an A note.  Using this method, he determined every note that was played.  The rest of the paper describes how he deduced which groups of notes were played by which instrument/band member.  Fascinating.

 

Categories
Statistics

Does taking LSD prevent crime?

http://en.wikipedia.org/wiki/History_of_LSD

Dr. Leary began conducting experiments with psilocybin in 1960 on himself and a number of Harvard graduate students after trying hallucinogenic mushrooms used in Native American religious rituals while visiting Mexico. His group began conducting experiments on state prisoners, where they claimed a 90% success rate preventing repeat offenses. Later reexamination of Leary’s data reveals his results to be skewed, whether intentionally or not; the percent of men in the study who ended up back in prison later in life was approximately 2% lower than the usual rate.

Well, the question is this:  Was the drop from 92% down to 90% explained by random chance, or did the LSD really have a statistically significant impact on reducing the crime rate?  Since the text does not provide a sample size, I will just use n=100 to do the math.

The calculations:

H0:  LSD takers had no difference in their repeat offense rates.
HALSD takers did have a difference in their repeat offense rates.

 

First, take stock of the given information:

\(n = 100 \\\\ p = .92 \\\\ \hat{p}=.90\)

 

Next, you calculate the standard deviation of samples of this size.

\(SD(\hat{p})= \sqrt{\frac{(.92)(.08)}{100}}=.03\)

 

To determine how unlikely your sampling result was, you calculate how many standard deviations away from the expected proportion it was (Z-score).

\(Z(\hat{p})= \frac{\hat{p}-p_0}{SD(\hat{p})}=\frac{.90-.92}{.03}=-.67\)

 

Then, you calculate the odds of getting this Z-score via the normal cumulative distribution function.  (What are the odds of this happening randomly?)  If it’s under 5%, then you reject the null hypothesis, because it’s unlikely this variation can be attributed to random chance.  ie: Odds are, the hair is indeed different.

\(p(Z \le -.67) = .25 = 25\% \)

 

Conclusion:  If the odds of being a repeat offender is 92%, then the odds of having 90% (or less) repeat offenders in a random sample of 100 men is quite likely.  The math shows that the odds of this reduction simply happening by chance (random variations) is 25%.  This is large enough (over 5%), that we can not assume the LSD had any true effect on reducing crime rate.  ie:  The 2% reduction was probably due to chance.  So, we accept the null hypothesis (H0):  In a sample of 100 test subjects, the LSD had no effect if it only reduced the repeat offender rate to 90%.

 So, do you have the same lingering question that I did?  How large would the sample size have to be in order for the 2% drop to not be an accident? (Recall, I just made up n=100).  Well, some simple algebra should answer this for us:

First, let’s determine the Z-score at the 5th percentile:

\(invNorm(.05) = -1.64 \)

 

Let’s use that in the Z-score calculation to figure out what standard deviation we’d need

\(-1.64 = \frac{.90-.92}{SD}\)     (…SD = .012)

 

Backing this into the SD formula will help us solve for the sample size (n)
\(.012= \sqrt{\frac{(.92)(.08)}{n}}\)    (…n = 495)

So, if Timothy Leary showed a repeat offender drop of 2% with a sample size of 495, then we could say the LSD did have an effect.  Why?  Because that much of a drop only has a 5% chance of happening randomly.

Categories
Statistics

Average hours of sleep (normal distribution)

How Little Sleep Can You Get Away With?

 

Nice example of a real life phenomena closely modelling a Gaussian normal distribution.  The average hours of sleep on a weeknight (for males) was 6.9 hours with a standard deviation of 1.5 hours.  Using this data, let’s calculate what percentage of men get a good night’s sleep.  The diagram indicates 27%.

\(Z = \frac{8 – 6.9}{1.5} = .73\)

 

\(normalCDF(.73,99) = .23 = 23\%\)

 

 

Categories
Statistics

The Statistics of Gaydar

The Science of Gaydar

Lippa had gathered survey data from more than 50 short-haired men and photographed their pates (women were excluded because their hairstyles, even at the pride festival, were too long for simple determination; crewcuts are the ideal Rorschach, he explains). About 23 percent had counterclockwise hair whorls. In the general population, that figure is 8 percent.

Well, just how meaningful is this 23% discrepancy from the norm of 8%?  Maybe it’s just randomness, right?  Well, try the omitted calculations for yourself.  This is an example of a “hypothesis test” in Statistics.  The Null Hypothesis (H0) says that there is no difference in the groups.  The Alternative Hypothesis (HA) says there is a statistically significant difference in the groups.  In a hypothesis test, the essential question is this:  What are the odds that a sample varies this much from the expected percentage (proportion) simply due to natural random variation?  (For example, if you flip a coin 10 times, you usually get 5 heads.  Sometimes, however, you might get 6.  In fact, that should happen 26% of the time.  Nothing to be alarmed about.  However, the odds of getting 8 heads is only about 3%.  If you do get 8 heads, that’s rare enough to indicate the coin might be rigged.  Odds are you won’t do it again!)

So, for this hair test, we need to ask, “What are the odds of taking a sample of 50 guys and seeing that 23% having a counterclockwise whorl?”  We should expect to get 8%, as per the broad population.  If it’s very very rare to get 23%, then we might suspect there is a connection, and gay men do have different hair swirls than the broad population.  In Statistics, we define “very very rare” as under 5%.  In other words, if the odds that 23% of a sample of 50 have a counterclockwise whorl is under 5%, then it is statistically significant.

 

The calculations:

H0:  Gay men have no difference in their hair whorl orientation.
HA: Gay men do have a difference in their hair whorl orientation.

 

First, take stock of the given information:

\(n = 50 \\\\ p = .08 \\\\ \hat{p}=.23\)

 

Next, you calculate the standard deviation of samples of this size.

\(SD(\hat{p})= \sqrt{\frac{(.08)(.92)}{50}}=.04\)

 

To determine how unlikely your sampling result was, you calculate how many standard deviations away from the expected proportion it was (Z-score).

\(Z(\hat{p})= \frac{\hat{p}-p_0}{SD(\hat{p})}=\frac{.23-.08}{.04}=3.75\)

 

Then, you calculate the odds of getting this Z-score via the normal cumulative distribution function.  (What are the odds of this happening randomly?)  If it’s under 5%, then you reject the null hypothesis, because it’s unlikely this variation can be attributed to random chance.  ie: Odds are, the hair is indeed different.

\(p(Z \ge 3.75) = .000088 = 0\% \)

 

Conclusion:  If the odds of having counterclockwise hair whorl is 8%, then the odds of having 23% of 50 random men exhibit this trait is unlikely.  The odds of this happening by chance (random variations) is basically 0%.  So, we reject the null hypothesis (H0), and accept the alternative hypothesis (HA)

 

 

Categories
Algebra

Optimal Snowboard Length Formula

http://www.livestrong.com/article/87496-size-snowboards/

Evaluate your height. This is the best way to determine snowboard length. One typical formula used by professional snowboards is: rider height (in inches) x 2.54 x 0.88 = suggested snowboard length. This will help you to start narrowing down your snowboard choices.

When I saw this formula, I wondered what it meant.  Note that snowboards are measured in centimeter units.  Well, to convert inches to centimeters, you multiply by 2.54.  So, we’re converting height to centimeters and then taking 88% of that.  I have no idea where the 88% rule comes from.  The point being, the “formula” is just saying to get a snowboard that is 88% of your height.  This lines up with my mouth, and I am pretty sure I am well-proportioned.  So, maybe it’s just easier to say that a snowboard should come up to your mouth.

This is also an example of intentionally not simplifying an expression because you lose some inherent meaning (explicit unit conversion, etc).  Otherwise, the equation could be simplified to length = 2.235 * rider height (in inches)