Isaak B.'s Blog at WyzAnt.com

The twelve hints of hercules

posted by WyzAnt tutor: Isaak B.

Spoiler Alert: This post gives a lot of hints on how to do the problem about the famous bullwhip that I posted on Dec 3, 2012 under the heading “Sine of the times”. If you got stuck on the problem, read a hint and then give it another try. Only read another hint if you are still stuck. You might want to have someone read you the hints so that you do not accidentally read more hints than you need.

Hint #1: Since the problem did not provide a diagram, and there are a lot of confusing details, read through the problem carefully and draw a diagram.

Hint#2: The diagram should include the known quantities and some indication of the unknown quantity (the angle).

Hint#3. The diagram should show that this problem boils down to the solving of a triangle.

Hint#4: The type of triangle being solved is SSA.

Hint#5. In particular, the triangle being solved is a sSA triangle, which means that the side next to the given angle is bigger than the other provided side.

Hint#6. There is a special consideration you must be aware of when solving sSA triangles.

Hint#7: The special consideration has to do with there being more than one possible triangle that can be considered a valid solution when given the inputs in the sSA form.

Hint#8: The triangle being solved is a sSA triangle, where the bottom of the triangle (pointing east from the original location of the priceless artifact, in the direction of our travel, is 20 yards, or 60 feet). The other known side length is the length of the bullwhip plus the length of your arm (42 feet). The provided angle is from the original location of the priceless artifact, an angle of 30 degrees north of our direction of triangle (60 degrees east of north, or in other words, clockwise from ninety degrees to the left of our direction of our escape route).

Hint#9: To find one solution of a sSA triangle, use the sine law to solve for the angle between the 42 foot outstretched bullwhip/arm and the unknown length of laser between its source and the place where it will be interrupted.

Hint#10: If you find the simplest solution to the sine law, and if you have read the problem carefully, you will find that the acute angle you have solved for results in a triangle which indicates that you should project the bullwhip obstruction in a direction that the problem stated was not allowed, due to the obstructions from falling debris in that direction. If you showed this restriction in your diagram, you probably remembered this fact and spotted this conflict.

Hint#10: To find the other solution, realize that for any angle theta satisfying the relation: sine theta equals x, there are two angles theta that satisfy the equation, unless x = 1. The first angle solution is the arcsine of x, and the other is 180 degrees (or pi, if using radians) minus the first solution.

Hint#11: You should choose the second solution, so that your triangle solution will give you an answer for your angle of “throw” that does not get blocked by falling debris.

Hint#12: The angle you need to solve for can be found by using the rule about what the sum of the interior angles of a triangle must always add up to.

If you still need more hints, let me know. Or if you simply enjoyed solving this problem, please let me know that too, and I will gladly generate more adventure problems like them. Let me know if there is a particular topic of math, physics, or chemistry you would like to be challenged on or challenge your friends with.

Sine of the times

posted by WyzAnt tutor: Isaak B.

High school trigonometry students who are learning to solve triangles for missing sides and missing angles, here is a problem I made up for you:

You and I and a famous adventurer with a hat and a bullwhip are in a bind. 20 yards straight west and behind him and two fellow explorers (me and you), across a gaping hole in the floor that has since opened up, a trap that he accidentally triggered after grabbing a priceless artifact is emanating a brilliant, 1 foot in diameter red laser beam in a direction 60 degrees east of north. The beam is shining on a crystal which is heating up and contains enough poison gas to flood the cavern we are currently in and all of the escape routes we can see several times over in only two or three seconds. -- we must break the beam entirely, even if only momentarily, to break the feedback loop and save us all. Because of the locations of the gaps in the floor all around us, we can't move closer or farther from the beam, and because of the way the bullwhip works, and the limited tools available, the best plan is to block the beam by attaching something to the end of the 40 foot bullwhip. You are currently the only adventurer undamaged enough to do the job, but don't know how to use the bullwhip well enough to control it at partial extension. In fact, the only way to block a 1 foot beam exactly 40 feet away from you, by swinging the bullwhip in such a way that it becomes completely outstretched by the time it travels through the same horizontal plane that you and the laser beam are in.

You look and see that unfortunately, swinging this apparatus in any direction between due east and due north is not possible because of falling debris.

Assume the center of the hat attaches to the end of the whip, the distance from the center of your arm's rotation to the nearest end of the bullwhip is 2 feet. Assume that the hat size is sufficient to block the beam if the center of the hat passes through the center of the beam, and that you can swing the whip such that it will be at its full length in the same plane as the beam and your path of travel.

Provide the compass heading of the direction in which you must swing the bullwhip to block the laser beam using its full length, accurate to within two decimal places.

Kicker physics

posted by WyzAnt tutor: Isaak B.

I was working with a physics student the other day and came across a football trajectory problem which was very timely because Superbowl Sunday fast approaching and the Pittsburgh Steelers are still in the mix. Yeah!

Most trajectory problems in physics problems meant to be worked “by hand” have to make some assumptions so that the equations can be solved without using a computer. The biggest and most common assumption is that the effect of the air resistance on the object is negligible. “Negligible” means that we consider (or at least are willing to temporarily pretend) that the effect is small enough that if we neglect accounting for it in our calculations our answer could still be useful in some way. For example, if:

a) we know our kicker can usually kick with an certain initial speed v, and if

b) we know how to use basic trajectory physics calculations to tell us the maximum possible range R for a kick at a given angle from the horizontal (let’s call it theta), and if

c) we know how to use calculus to optimize our choice of initial conditions for various purposes, such as maximizing range, or height above the cross bar, and if

d) we are willing to accept an answer whose mileage may vary depending on how the ball is kicked (a tight spiral is best, but end over end is next best) due to the variety of magnitudes of air resistance, and if

e) we are willing to recognize that our theoretically calculated range is in practice greatly reduced due to air resistance forces (drag) by an additive quantity which is some factor times the square of the result of taking the ball’s forward speed minus the wind’s forward speed, with both speeds being measured with respect to the ground, and

f) point e) applies to a greater extent as for initial velocity and final range increase because drag forces and the resulting decelerations factor in more heavily as the speed and its square increase (therefore drag typically especially decelerates the ball during the initial fastest moments of the trajectory)

Note that in the coordinate system I am defining for these problems, a headwind (kicking into the direction the wind is coming from) will be considered a negative wind speed while a tailwind (kicking with the wind) could be considered a positive wind speed. That way, the difference between the ball’s (also defined as positive in the forward-going direction) forward speed and the wind’s forward speed increases to a value higher than the ball’s speed relative to the ground in a headwind, and decreases to a value less than the ball’s speed relative to the ground if kicked in a tailwind.

We can also see that in a very unusual special situation where the tailwind is roughly constant, in the same direction of the kick and equal to the ball’s initial horizontal ground speed, the assumption of negligible air resistance is much more reasonable, especially for fairly low angles of kick (drag forces would still be acting in the vertical equation of motion, even in this situation). Although highly unlikely this is theoretically possible. In fact, to get an idea of how much wind would be reasonable to hypothesize, I searched the web for "windiest NFL game" and found a hilarious video of some field goal attempts gone wild during an NFL game that was played in winds gusting to 45mph, which is fairly close to a typical kickoff’s initial horizontal velocity. Such weather could happen again, and if the wind’s direction and speed happens to line up with the kicker’s targeted direction, we can expect negligible drag predictions to actually become fairly realistic. In such ideal conditions, for example one might expect a kick with an initial velocity of 30m/s to fly around 30 yards further (between 80 and 85 yards) before returning to earth than the identically launched kick would travel under calm conditions. I would guess that wind conditions were favorable when the records for longest field goal (63 yards for the NFL, 62 yards CFL) and longest punts (98 yards NFL, 108 CFL) were set. Note that with the longer field (110 yards) and the rules for recording a punt, the difference in best punt likely has more to do with how the statistics for punting are kept than the actual distance the ball traveled -- there may have also been some lucky bounces involved).

However for typical football weather, we would simply expect our answer to over-estimate the range because in realty the football simply does not have enough mass to avoid letting the small forces due to air resistance (also known as drag) apply a negative horizontal acceleration (or a slowing effect) to the ball’s trajectory, especially for long punts, field goal attempts or hail mary passes.

Let’s take a brief side trip to examine why the football’s relatively low mass means neglecting air resistance throws off our answer more than if the football were heavier. We will compare and contrast the flight of a shot put and a football with identical initial speeds and launch angles in air and in a vacuum. In a vacuum, with no air resistance one would expect both objects to travel identical trajectories, but in air, we would expect both objects, especially the football, to slow down and fall short of that ideal. Let’s apply Newton’s second law of motion, F = m*a (where F is the force acting on an object, m = the mass of that object, and a = the acceleration, or rate of change of its speed, of an object) to examine this more closely. I will assume that the drag force on the shot put and the football are about equal when the football is thrown in a tight spiral because they have roughly the same cross-sectional area when viewed from the front, and aerodynamicists say that drag forces on an object moving through a fluid are typically proportional to cross-sectional area, fluid density, and the square of v_fluid, where v_fluid represents the velocity of the object with respect to the fluid. There is also another factor called the drag coefficient that accounts for the effect on drag of an object’s shape but for the purposes of this discussion we will consider the drag coefficients for a tightly spiraling football and a shot put to be close enough not to worry about. To simplify comparing the acceleration due to air’s drag on a football with the acceleration due to air’s drag on a men’s Olympic shotput we can solve F = ma for a (by dividing both sides by m) to write the same law in a different form:

a = F/m.

This expression for acceleration shows that the resulting accelerations of the objects due to drag will increase in direct proportion to the drag forces acting on the objects. In our case we are assuming that the drag forces on the two objects are roughly equal, so this is of not much concern. The equation also shows that the resulting accelerations of the objects due to drag are inversely proportional to the masses of the objects. In other words, if the weight of a football ranges around 14-15 oz | 0.9 lb |0.4 kg, and the weight of a (men’s Olympic size) shot put is 16 lb | 7.26 kg (women’s is 8.8 lb or 4 kg), then the roughly 16 times more massive shotput will experience roughly one sixteenth as much acceleration due to drag as the football. I chose not to use the word deceleration here despite recognizing that these drag forces, and therefore the accelerations due to drag, always oppose the motion of the objects relative to the fluid, because in the case of a tail wind, “drag” forces actually impart a positive acceleration to the object; more on that later. The main point of all this is that we can conclude that neglecting drag always gives erroneous results when calculating trajectories of object’s moving through fluids, but the magnitude of the error is inversely proportional to the object’s mass. We would expect drag to shorten a football’s flight trajectory somewhere between ten and twenty times more than a men’s Olympic shot put thrown with the same initial velocity.

Having given a little justification for the common practice of assuming negligible air resistance, in my next blog I will discuss how to determine the general formula for the expected range when air resistance is neglected. I'm planning to follow that up with another blog outlining the procedure for determining the optimal launch angle is (assuming negligible air resistance). Later, if anyone expresses interest, I’m prepared to discuss the general procedure necessary to take wind and drag into account; practical basic computing techniques will be involved in developing a simple physics simulation. If there is any interest I may even eventually write a companion entry touching on ordinary differential equations and their digital corollary, difference equations, and how they apply to this kind of simulation.

If you enjoyed this post and would like to see more on this or similar topics, please let me know by sending me a message or leave a comment for all to share.

Smart students

posted by WyzAnt tutor: Isaak B.

A wise man I know, a first-year engineering student at the time and now a working registered professional electrical engineer in training, once told me that the key to success is not to focus so much on "working hard" but "working smart". Meaning that the person who keeps their wits about them and constantly evaluates whether they are spending their time wisely is far more likely to succeed than the person who expends a lot of wasted effort hammering away at something using the same technique even when the results need to be improved.

In that spirit of self-examination, how has your semester gone so far this year? Are you ready for finals and on top of everything your course has discussed or are there a couple topics here and there that you are struggling with? At this time of year (exams approaching), many of you reading this are probably hoping that getting a tutor now can help you deal with the final exam, and at least compared to not getting a tutor and trying to cram alone for an exam on material you haven't quite understood yet. Cramming with a tutor is still an excellent idea.

Even better, though, is to deal with problem areas as they arise. Dealing with difficult concepts as they arise (whether that means dedicating extra study time, soliciting extra help from the teacher, finding a classmate willing to help, or acquiring professional tutoring services) as you go along can benefit your marks throughout the semester and ensures that the time you spent in class is more productive. Facing a final exam about material you know you have recently learned is a much better prospect than facing a final exam which will partly test material you know you don't quite understand and have little time to catch up on. Why not resolve to change your approach from working hard to working smart, by making the best use of all available resources to maximize your investment in your education, and schedule a weekly or semi-weekly review today?