What he said.

:-p

Chappy,

Look at my immediately prior post. Does that equation help? I put it in terms of circle circumference and angle off tangent.

I think the situation for a vehicle with 2 axles can be described with a relatioship among (1) the steering angle of the front wheels, relative to the rear wheels (assuming the front axle is the only steering axle); (2) the vehicle wheelbase; and (3) the circumference of the circular path traced by the midpoint of the vehicle. Note that in a vehicle with only one steerable axle, the different axles will trace different paths, each having its own perimiter. The non-steering axle will follow the smaller perimeter.

The longer the wheelbase, the greater the difference between the travel distances of the different axles, and the greater the circumference of the resulting midpoint circle for a given steering angle.

Okay Rion… Calling it a night. This isn’t over yet. Still, put the math to work. You logic may or may not be correct. Do you have an equation to test. Wanna get rubber-meets-the-road on this.

Good night though.

Hey, no offense Rion…everybody’s got somebody…just making a point…I think…: )

Even single people have to love themselves… ;-)

A circle is a relationship…a single wheel is like a single person…no relationship. ;-)

No. Although the ratio of wheel base to total path length for the same number of 2 degree turns would be the same.

Naturally a Ferd F-teen thousand truck would have a larger turning radius than a teeny McMo-ped. However, for turn-to-turn path lengths greater than the wheel base of the vehicle, the turning radii would indeed be identical.

So your argument is that a moped will travel the same distance turning at a persistant 2’ as a tractor trailer?

What he said.

Not really. If you think about a single point of contact, a unicycle, versus a car, the difference is only one point, since you disregard the outer rear wheel (it freewheels, contributing little in the way of thrust or drag) and the outer front tire moves parallel to the inner front wheel, and also freewheels, thus providing only negligible drag.

For a single tire as on the unicycle, you are correct in assuming intermittent steering corrections to approximate a circle with a polygon, although you must consider the possibility for tilt of the wheel in the vertical plane. For a rounded tire, or even a “flat” profile tire which is inflated with a nonzero gauge pressure, the outermost portion of the tire rides along a larger diameter arc than the innermost portion, thus largely determining the radius of arc as by the ratio of these two effective diameters.

That is a bit obscure for a nonrigid tire, since the tire surface deforms in at least two dimensions under load, making the math a little bit wonky to say the least.

If you assume a rigid, circular cross-section tire, perhaps a tungsten torus (doughnut) riding on a likewise infinitely rigid, perfectly flat surface with finite coefficient of friction, then only gravity and inertia come into play, acting as moments about the center of mass of the assembly and the axle. Tilt then redirects some of the normal force to drive the wheel in a circular path, and it would be possible to trace an arbitrary circular arc or spiral without approximating with short line segments (chords).

for the wheelbase variability, you are correct to assume that a shorter or longer wheelbase would result in a smaller or larger turning radius, as is the moped’s ability to turn its front wheel nearly 90 degrees to the direction of travel at low speeds, resulting in a turning radius less than or only slightly greater than its own length.

To find an infinitely variable solution for a single wheel with no lateral tilt, you would simply set up your equations for the polygonal approximation, then take the limit as the length of the polygon’s face approaches zero, or the number of faces/turns approaches infinity. I think it was mentioned before regarding the circle vs. spiral issue… as long as the distance between course corrections is exactly the same each time, and the angle of the course correction is identical in direction and magnitude, then you will trace a circle. If either the distance between course corrections or the angle of the correction is altered, then you will generate a spiral. The same math applies to the circular solution, since you can indeed transpose the chord and the wheel base.

So then are you wondering how long the chord length is, for a line that forms a 2-degree angle with a tangent line at the point n{a}, and then intersects the circle again at n{b}? Or are you looking for the arc length?

The arc will actually be 4 degrees, because at the distant point, that second tangent line will intersect at -2-degrees (or 178 degrees, based on how you handle the branch cut). To continue in a circle, one would need to turn 4 degrees off the most recent path (which is 2 degrees off the new tangent line). The arc length would then be (4/360)*Circumference = Circumference /90.

If you changed so that the initial deviation from the tangent was 1-degree, and then subsequent turns were 2-degrees, then each arc length would be Circumference/180. Basically, the arc subtends 2*Angle of the circle, where ANgle is defined as the angle relative to the first tangent.

But if you are looking for chord length, that is calculable using sines. The radius of the circle is R = C/(2*pi). The chord is symmeric about a bisector, which is oriented at half of (2*Angle), relative to a line going from the center of the circle to the point of first deviation, which you label n{a}.

Half of this chord (from n{a} to where it intersects the bisector) is R*sin(Angle). So the total length of the chord, which I will label as S, to denote the side of a polygon is then S = 2*R*sin(Angle).

The total perimeter of the polygon is N*S, where N is the number of sides. N = (360/(2*Angle)). NOTE #1: I am defining Angle by the angle relative to the first tangent line. Subsequent turns will be at an angle of 2*Angle. NOTE #2: The measure of Angle is in degrees, not radians.

As a check, cosider the following:

Perimeter of the resulting straight-edge regular closed polygon, P, is found by:
P = (360/(2*Angle))*2*R*sin(Angle)
P = (360/Angle)*R*sin(Angle)
P = (360/Angle)(C/(2pi))sin(Angle)
(now the final expression)
P = (180/Angle)(C/pi)*sin(Angle)

We can verify using 2 limiting cases (set C=2*pi for simplicity). First, set Angle = 45 degrees. This will make a square inside a circle of radius 1. It’s perimiter will be
P = (180/45)(2pi/pi)sin(45) = 42*0.7071 = 5.657 This is correct.

Now take the limit as Angle goes to zero (approaching an infinite-sided polygon that becomes a circle.

The angles must be expressed in radians, rather than degrees. So (180/Angle) becomes (pi/R_Angle), where R_Angle is Angle expressed in radians. Using the small argument approximation for sine, sine(R_Angle) = R_Angle. If I recall properly, what I will be doing is something like L’Hopital’s rule.

P = (pi/R_Angle)(C/pi)sin(R_Angle)
P = (C/R_Angle)*(R_Angle)
P = C

This result is expected, and basically says that as the chord length goes to zero, the perimeter of the polygon becomes the circumference of the circle.

OK, Chapppy.

Does this answer your question?

A car has too many points touching the ground. The length of the wheelbase being variable gives an inconsistent false-positive for the answer. If a moped travels this course or a semi does. Will a 2’ alteration in course result in a smaller circumference for the moped? You betcha, Therefore a car cannot be a test subject. Use a single point of contact… Still, no answer.

AHA!

This may get convoluted.

oh well.

For the case of a car, here goes;

Assume the front inside tire is steered tangent to your circle. Assume your inside rear tire is also tangent to the circle. Also assume you have an open differential, lest the outer tire slide and slip and do all sorts of noisily nasty things. Also assume the outside front tire has sufficient toe-in to accommodate the turn without doing similarly noisy and nasty things.

Thus, the center of the circle containing the arc which is traversed can be found by using the normal to the tangents formed by the instantaneous path of the front and rear inside tires.

It takes a bit of trig, but if you turn your front tires to the left by two degrees, then you have essentially set up two right triangles. (technically you’ve set up a bunch, but the only two that matter are formed between the axes of the front and rear tires)

Long story short, you’ve got two triangles, sharing a hypoteneuse that essentially bisects the car. The between the two legs leading to the wheels from the center of the circle would be your 2 degrees as set by the front tire, and by simply knowing the wheel base, and the angle, you can fudge a very close approximation to the actual case. The reason for the approximation would be that your wheelbase actually tends to change, since your tires don’t actually rotate about their centers, but have an ever so slight offset since it’s tough to jam all your suspension dohickeys inside the diameter of the wheel – that’s why we have fancy contraptions like ball joints and offset spindles and the like.

So, you’ve got a 88 degree angle at the front and wheel, and a 2 degree apex angle, leaving a 90 degree angle with the rear wheel’s track. Do thy trig, and thou art done.

I uploaded a quickie diagram to my gallery thing. (or will in about ten seconds…)

Well, Kelce @ 2’ there are 180 potential turns in a perfect circle andso momentum is the element that makes the circumference. The rate of change over the base distance enables a trajectory calculation.

Do this:
line n{} is infinite North and South. Circle C1{} is a circumference that meets n{} @ n{a}. C1{} has a diameter of 1’=12". At point n{a} there is an imaginary ray set 2’ (deg) from line n{}. At 2’ What is the length of the chord of that intersect points C1{x} and C1{ZA}. Solve for Arc between x and ZA. Multiplied by 180 is the cirucumference.

This is true with an invariable or static rate of change. Change the rate of change and change teh circumference of the circle and thusly the diameter.

Chappy,

With that clarification, I have a question. How many discrete “turn” events are there?

In a perfect circle, there are an infinite number of 0-degree turns, with no distance between them. Thus, the chord becomes a straight tangent line at the circle boundary. Otherwise, with a finite number of turns, we have a situation of a closed polygon with some number of sides, N.

For a polygon with N flat sides, traversing the perimiter in a single direction results in N pivot turns at a vertex. Each turn (in degrees) is 360/N.

Oh, and another thought.

If there is any variation in the average path length between completions of a turn, that is, if E{x} varies as a function of x, then we have the following:

For E{x} decreasing with increasing x, there will be an inward spiral toward a point. For E{x} increasing with increasing x, there will be an outwardly growing spiral. Of course, they’ll be jagged, rather than smooth. The spiral will be sort of like a snail shell.

For E{x} constant, the expected completion point will be co-linear with the initial vector, but could be either inside or outside with equal chance. Perhaps.

For diameter calculation, after estimating path length, maybe divide by pi for an initial estimate. It will likely be slightly greater, because a flat-sided polygon will have a “radius” greater than a circle of the same circumference. And having an end point different than the starting point makes “diameter” and “radius” somewhat subjective terms.

I’m not anywhere close to certain that my suggestion is anywhere close to the answer you were seeking. But hopefully you’ll be able to get it figured eventually.

Okay, Lets turn it inside out. Circumference C1 meets line n{} at point n{a} in vector ZZ. If C1 is a perfect circle with a diameter of 100’; what is the angle at which CC departs from n{}?

Chappy,

I think I must be missing something significant in your question. I have no idea where the other people get speed, “slip friction”, tires versus rails, …

Here is how I see it – from a purely vector analysis, and no real-world mechanical effects.

If each time a turn occurs, it is 2 degrees in the same direction, and all paths are confined to a single flat plane (not airplane), then it takes 180 turns to make a 360-degree turning path. Add in complicating factors, such as the direction or amount of the turn can be different, and none of the rest of my comments will apply.

If each segment is a random length (from the starting point, or else the completion of the prior turn), then the total path length traversed is merely 180 x E{distance between turns + distance required to complete a 2-degree turn}.

Basically, figure out what the average (mean) distance is between when subsequent turns are completed. This mean value is the E{x} expected value. Multiply it by 180. That’s it. The hard part is figuring out how to determine the average. Non-pivot turns (arced curves) will complicate the path length calculation.

The resulting path will likely not form a regular polygon, unless you have some constraint that all distances between turns are identical. So you won’t get anything that looks like a rough “circle” but is made out of a bunch of straight sections (like a STOP sign with with 180 sides instead of 8).

In fact, there is no certainty that you would end up back at the same point or at any point co-linear with the initial vector. Your 181st vector will be parallel to the initial one, but who knows whether it would be inside the radius or outside the radius, relative to center point of the path.

It might be that the average radius or diameter of the range of shapes could be computed according to a polygon that had sides all equal to E{X} (assuming the turns are effectively instantaneous pivots). But I don’t know, without actually trying to derive a relationship between E{x}, the standard deviation, and the expected diameter of the resulting shape.

This is an interesting problem, Chappy. I have no idea what spawned it, but am rather curious now.

Chap,

It is the complete opposite for me…..A few of these posts I thought were a spoof……then I realized they weren’t….now THAT is sad! ;)

For the record, you should never “disregard turbulent flow”…..they make Pepto-Bismol for that….

Greg,
That is funny… I was actually following him until the “7 transistors”, that was just not possible or necessary, the environment is too hot… lol.

Then i realized it was a spoof… I was embarassed. Good one!

Right Rion;
it was also determined, when discussing, length of arcs in a turning vehicle, that the length of a chord would be the distance of the wheel base. Then the conversation transitioned to train tracks and derail speeds as Michael H is an officianado on the subject being that he is a rail road engineer. Then the focus took flight as Frank, a master in aeronautics and open fluid media, added in his expertise. So then I ascerted that it may be a law of inertia that makes the answer then scalar depending on speed and object mass.

So I concluded in my own mind that when talking “Travel” which is the base of the original question the pmethod to solve will always involve inertia and that engineering calculus is for static equations.

Then I submitted my product that incorporated these principals :-)

Now it is just for fun. Still interested in your response.

Yeah…what he said.

;)
http://www.youtube.com/watch?v=rLDgQg6bq7o

Aha.

Yes, and no.

In an incompressible fluid, no. In reality, yes.

Short answer:

If you disregard things like turbulent flow, slip, finite friction, etc., then you can assume you’re moving very slowly through a viscous medium, and that your motion is altered instantaneously when your control surface (or tire) is altered accordingly. Thus, if you travel a finite distance (l), then alter your course and travel another finite distance (l), and so on, then your arc length is approximately the sum of the number of turns you make multiplied by the distance (l).

For real circumstances, however, tires, control surfaces, etc. have a finite ability to support a reaction force. Tires have a limited coefficient of friction and limited contact patch, and your vehicle has a finite mass. As you corner in a vehicle, for example, your mass has inertia which tends to pull you out of the turn onto a straight line path tangent to the turn’s path. Your mass, and the normal (vertical) component of the cornering force tends to hold the tire in better contact with the road surface. Aerodynamic forces can come into play as well, since your vehicle may develop down-force and be able to corner better at higher speeds due to the effective increased weight via the down-force.

So, mathematically, no matter how quickly you make the turn, the actual turn is what determines your arc length. In reality, many factors alter your path from where you’re actually steering. The simple fact of moving a specific distance, then turning, means your speed doesn’t matter. If you are basing your determination on the performance of control inputs, then the aviation examples are spot-on – the faster you go, the more control input you have to apply in order to complete a turn of a given radius.

I can’t post images or links, but there’s a book I’ve got that shows that the center of a slip-free turn for a vehicle depends on the intersection point of the front and rear axles, give or take a few caveats. I’ll see if I can figure a workaround, or simply post an image this evening to my gallery.

You’re also kind of describing in a local polar coordinate form the first and possibly second derivative of a function describing your position in global cartesian coordinates. in rough terms, your first derivative is “2 degrees per step” and your second derivative would likely be a small integer constant, most likely 1, if my thinking is correct, but either way, we’d have to reconcile the two frames of reference.

I think I’ve got a good idea of where to start, but work calls – a few breaks between clients, so I’ll see if I can sketch something up for you.

Cheers!

Actually Rion,

This whole conversation may be right on topic… I will explain that ascertion later. For now the question is, and it has not been absolutely determined to this point, the following:

If traveling on line n in vector zz and change direction at random point n(a) by 2deg thereby creating a new location and new direction in vector zz, {o,b}; if from zz {o,b} again alter direction 2’. Repeat until path of travel is n again. What is the circumference of the circle or solve for diameter.

What the conversation has led to is that it depends on the distance of he chord between zz{n,a) and zz{o,b}. Then the question was presented: “Would speed change that circumference?” with varried conjecture.

Rion,
What say you?

Hi Kelce,
I was eager for you to join this thread.

Thank you in advance!

Chappy!

I think perhaps I’ve localized the source of the offense, so to speak.

There are quite a few crackpots that have attempted to pursue so-called solutions to the effect I mentioned, and they are all lacking in scientific rigor. When someone questions my scientific rigor, or knowledge, out-of-hand, it’s rather perturbing, since I’m generally the skeptic that is shooting down others’ claims of unusual phenomena with my own scientific knowledge and mathematical ability.

Again, I think we’ve got some sort of miscommunication going on, so I’ll attempt to leave it at that. Seems to be happening rather often over the past month or two. Very simple comments yield ridiculously out-of-character reactions from the receiving party.

One minute I’m saying “Hey [random friend], it’s been forever, how you been?” and the next I’m cleaning their drink off my shirt.

Back on topic somewhat – glad your Sunday’s been glorious. Mine’s rather cold and sleet-ridden. shiver

Yay Virginia weather.

We’ve had a humbling experience and I was just offering my blessings on this glorious Sunday!

I’m done killing the weeds. Did I miss anything?

Rion, look back and compare what you wrote with the analysis Kelce gave and assume Kelce is conversing at a post-graduate level. What level would you say you’re trying to communicate at? I often-times find myself humbled by the intellect and inventiveness of other people I associate with, but at my age I’m content at just hoping I can be a good inventor before it’s too late. Believe me when I say I can BS with the best of them when I want to! But I have developed a knack for spotting BS when it happens upon me and yet I still enjoy BS-ing because it keeps my mind young and vigorous. So thank you for the challenge and since this is Sunday, God bless you in all your pseudo-scientific travels!

since I can’t post links in the forum, and can’t send you a message, the means of communicating such proof are inaccessible to me. Further, i cannot upload an image, so scanning my notes and posting the image is also moot.

I am unsure as to why you’ve decided to attack me in any way shape or form, but I am not a second year engineering student, nor am I full of baloney. Never did I presume to put myself forth as anything other than the principal investigator on a research project which yielded unusual results, and in attempting to explain them, I had to re-write certain equations into a form not typically used.

I cited this process as an example, since making assumptions, or heading down the incorrect path mathematically can have ridiculously large negative consequences on one’s results.

Somehow, this example became the target for your comments regarding my abilities, intelligence, or my understanding of physics and math, namely by your citing Maxwell’s equations, which directly deal with electrodynamics. The device in question is electrostatic, and unless it is used as some form of propulsion system at high speed, electromagnetic effects can be neglected. The device is further a simple embodiment of a capacitor, and the “lack” of reaction mass is really a masking of that same mass. The device allows for acceleration of an arbitrary mass without the carrying of propellant which is exhausted in order to provide a reaction force. It is in no way a stargate, a warp drive, or any other form of science fiction fantasy fare.

I constructed the first of a series of devices to investigate claims by hobbyists, as well as a study performed by the Army Research Laboratory, all of which came up with either scientifically dubious claims of “overunity” cringes, or left certain key questions unanswered regarding the nature of the device. My assumption was that a mass transfer WAS occurring, but that it was being overlooked. In a sense, I was right, but in a way I had not previously considered.

Solving the aforementioned equations describing classical electrostatics, for field instead of charge, one arrives at a solution which allows for a semiconservative, unbalanced field geometry as evaluated with a gaussian surface distal to the device. Taking the limit of the resultant field vector as the radius of the evaluation surface approaches infinity gives a nonzero net field vector. Most other monopole or dipole arrangements would result in a zero net electric field distal to the device.

It turns out that the purely classical approach will give the exact same result, but in taking the “long way around” I was able to confirm that this was not a simple mathematical error, and that it is inherent in the nature of electrostatics.

I really despise that a thread of this nature, attempting to help someone with a mathematical question, was turned around into a personal attack and resulted in such a long-winded defense. If you wish to discuss my intelligence or scientific rigor further, please continue via private message.

Even then I wouldn’t…;-)

Don’t assume my assumptions are correct unless you can’t prove they aren’t.

Yes, I bow to the unimaginable depth and breadth of your assumptions.

Rion, you’ve become a bit humbled I see. We can all use a bit of humbling once in a while. Take care.

Ah, so I was correct in assuming your “second year” comment was indeed smack-talking.

Why?

You seem to have jumped to the conclusion that I’m promoting myself as some sort of mathematical demigod, when in fact I simply did some creative algebra, and a bit of calculus, only to find out that everything I did simplified back down to facepalm algebra.

So, add me or not, pm or not. Just don’t be an ass for no reason.

Hey Rion, meet Kelce! No talking smack there!

Chappy,

I only just now saw this. Sorry I wasn’t helpful earlier. Fortunately, it appears you have your answer.

That is a really beautiful color on the optical illusion spiral. I’ve seen that spiral before. Looking at the center makes the reversal zones look like they’re wiggling. But I’ve never seen it in that color. That hue of magenta is really striking in that pattern. Does anyone know whether it is just equal amounts of red and blue?

My professors forced me to spend a lot of time with pi, when I was in grad school. It just made me gain weight.

Standard deviation is a probability measurement (sort of an average amount of variation from the average value). At first, I thought your question was about something like a spiral version of a “random walk.” Apparently, I completely misinterpreted, because you appeared happy with the answer you received. So I guess I’m glad that I didn’t answer earlier, because that might have led to an initial bout of unnecessary confusion.

As a side note, I haven’t ever seen any way to calculate the expected value of a radius for a circular (spiral) random walk to complete its first 360 degree rotation from an initial vector. I bet the probability distribution function would be something akin to a weibull.

If you have any interest in the subject, look at equiangular spiral antennas. I did some calculations on them in grad school, and actually worked with them some. The Air Force uses them in radar warning receivers, due to their broad bandwidth.

You’ve got me curious now. I bet there are a lot of issues that could be solved by developing expected value calculations for certain parameters of spiral random walks. I’ve only seen random walk analyses in 2-D cartesian coordinates, and none in polar. The 3-D version in spherical coordinates could be useful in chemistry or physics. But, alas, my curiousity will end in a few minutes, when I go out to the garage to pour a bag of weed killer into my lawn spreader.

Any time. Got some provolone?

I can’t seem to PM you without adding as a friend, so in the meantime, I’d suggest a quick google search for “Z machine”, in reference to your coin shrinking/nuclear fission/fusion comment a couple months ago.

I have the distinct feeling there’s a miscommunication going on somewhere. I look forward to clearing it up.

Thanks, Rion, you cleared up a lot of my suspicions.

Do a google search for “Maxwell’s Equations”.

Virtually everything listed has to do with time-varying magnetic or electric fields.

I’d be happy to discuss further in a PM, since this is suddenly and dramatically off-topic!

Also, I am neither. Einstein didn’t have access to cloning technology, and I prefer Salami.

one quick caveat – Gauss’ law does in fact deal with the relationship between a charge and a field, but does not go so far as to describe in and of itself the nature of two interacting charges.

Also, I’m not saying I came up with anything new. I just re-solved for field geometry instead of charge. Winds up simplifying back down to a few constants and two variables, and has very much the same form as the original.

Alas, I must get back to laundry, changing light bulbs, and replacing a light fixture on my porch. I’ll be sure to upload a photo if my hair’s standing on end this evening.

Sounds like pseudo-science as believed and distributed by a second-year college engineering student. Please direct me to the exact equations you are referring to so I’ll know for certain if you’re full of baloney or else Einstein’s ‘Mini-Me’.

I think there’s a little something lost in translation.

We’re talking electrostatics, and not implying any tie to gravitation or “warp drive”. Just a means to move something without spitting out a stream of matter behind you, a-la-rockets-and-jets.

It’s just a subset of classical electrostatics that provides for an apparently reactionless propulsion method. No relativistic concerns or time-varying fields were employed. Just straight up DC electric fields.

maxwell’s equations either break, or simplify to classical electrostatics, when applied to non-varying fields.

As a non-insider I can’t post a link to maxwell’s equations on wikipedia. phoey.

Just as an aside, my dog is named Max after James Clerk Maxwell. It’s my cryptically geeky way of calling him “sparky”.

;-)

Rion, I believe Maxwell provided field equations long before you found a solution for ‘field effect propulsion’ and Lorentz discovered a transformation (equation) to make those findings universally transferable to more sensible applications! However, I’m sure the old and proven action-reaction theory of Newton will rule until you can prove there exists anti-gravity propulsion systems or a new quantum-physics based intergalactic human-transporter device. I would be really excited if you invented one of those!

Are you familiar with the principles of calculus?

re: square peg/round hole/pi…

pi is actually the result of calculating the surface area of a circle, or the ratio of circumference to diameter.

I think you’d be well served in taking a look at derivatives, taylor polynomials/taylor sums, and integrals. The drawing you posted regarding the straight line and the “turned” line on a circle…

what you drew were a chord and a tangent. There are a variety of equations describing both of these items, and when solved for one another, can yield an angle. This, however, is a very different problem from a vehicle with one set of fixed and one set of steerable wheels, attempting to go in a circle, especially when considering toe in, caster and camber, along with weight transfer and other suspension changes that alter the tire’s contact patch.

Just depends on how deep you’d like to go down the rabbit hole. I can recommend a few books if you’re thinking about vehicle suspensions, or some further links/reading if you’re just interested in a funky mathematical phenomenon.

I had to teach myself a little math in order to arrive at a solution for classical electrostatics that allows for field effect propulsion. Gotta be careful where you take it without guidance, though, since a small incorrect assumption can lead to vastly inaccurate results later.

Thanks Jane…I just try to play along the best I can. ;-)

True James, you put your heart and soul into your thoughts…which in turn, gets us all thinking…thanks!

Michael, I love reading this and your first posts with the “uneducated”, self-depricating comments… and then you pull out your “Rollarule” patent! I can’t get this grin off my face!! great stuff. And I’d love to see that work… looks like something that would help to teach the concept to truly “uneducated” math dummies like me.
Chappy, another interesting thread!

Lichtenberg figures are common on the bodies of people struck by lightning. Check out pulsed-power coin-shrinking. Shrunken coins can actually be purchased too. Amazing how they’re shrunk in all three dimensions. Perhaps a way to slam fissible-matter together to achieve critical mass and start a nuclear chain-reaction?

Jim, I’ve actually seen that on electrocution victims I’ve treated over the years…
One guy in Richmond who got the Electric Chair, looked almost like he had tree branch
tatoos on his back, arms, and legs… kind scary

Howz it! Shoulda put that one in the acme name search:) Merry Christmas, and enjoy now!

Thanks Jim, that is some pretty cool stuff…and that’s what I’m talking about, guys like Benoit Mandelbrot…where do they come from?
Brilliant People! Or might I say, Fascinating People!