"Thrust Vector" and other ideas

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Jim K

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Hi, Tuner and guys,

Yes, I decided a new thread might be a good idea.

First, let me say that I have a lot of respect for Jerry Kuhnhausen, although I have never met him. I have, I think, all of his books, and generally the information can be taken to the bank. In fact, some of the information I post here is from those books; I try to give credit when the information is available only in a specific book (that is, is not general knowledge), and do so with Mr. Kuhnhausen's books as well. When I have a technical problem, or a question about specifications or fitting parts, Kuhnhausen is my guide. There is none better. My only regret is that when I was doing the work, I did not have all the nice tools he has.

All that being said, his explanation of the way a 1911 type pistol operates is, to put it mildly, somewhat flawed. In his illustrations, and in the accompanying text, he states that the barrel/slide unit does not begin to move until the bullet is out of the barrel. That is simply not true; if it were, the pistol would not work. The barrel begins to move the instant the bullet begins to move. The barrel remains locked to the slide until the bullet is out the barrel, at which point it has recoiled about 1/10 inch. This has been shown repeatedly in high speed films.

His explanation of a "thrust vector" (Vol. II, Page 42, Figure 4 caption) is nonsense, as is a lot of Para 5, Page 38.

So what really happens? And how does the pistol work? First, if you don't believe in the laws of physics, go to another thread.

The answer is recoil. This is sometimes couched in terms of Newton's Third Law, "for every action there is an equal and opposite reaction." That law is now called the Law of Conservation of Momentum, but the idea is the same. When the bullet moves forward, the part of the gun in contact with the bullet, the barrel, moves back. Is pressure involved? Of course, pressure is what moves the bullet. But pressure does not directly cause the gun to function in a recoil operated pistol as it does in a blowback pistol. (No, the pistol is not kept locked by the bullet pushing forward on the barrel; that force is essentially negligible.)

The bullet's forward motion causes the barrel's rearward motion, and it is that barrel motion that shoves the slide back (through the locking lugs and the hood) and operates the pistol. Now here is where it gets interesting. If the bullet does not move, the barrel does not recoil and the gun does not cycle, period. If the barrel is blocked directly ahead of the bullet, so the bullet cannot move, the gun will not operate. It remains locked, the pressure leaks away, and nothing happens. No, the gun does not "blow up"; the cartridge case does not explode. Nothing happens. (Yes, I have done it, but under tightly controlled conditions; please do not destroy a pistol or hurt yourself trying this.)

So, if the barrel/slide begins to move when the bullet does, when does the barrel unlock from the slide? That happens after the bullet has left the barrel and pressure has dropped to a safe level. Because the bullet is much lighter (has less mass) than the slide/barrel combination, it moves 4 inches or so, while the barrel/slide has moved only 1/10 inch or so.

That movement is the distance the barrel moves before the barrel foot slips off the slide stop pin, the barrel engages the link and begins linkdown. Until that happens, the barrel and slide are locked to withstand the high pressure in the barrel. By the time the unlocking takes place, the bullet has left the barrel and pressure has dropped. Note that the barrel is (or should be) kept locked by the barrel foot riding on the slide stop pin, not by the link.

OK, how does that affect some other fond ideas? Doesn't the recoil spring keep the barrel and slide locked? Well, no. A strong recoil spring can slow down the slide after it unlocks, but no reasonable spring can keep the recoil from shoving the slide back, only mass can do that, and no spring of reasonable weight is massive enough. In fact, the pistol will function and lock for the same amount of time whether a recoil spring is present or not. (Without one, however, the slide will batter the frame, since there is nothing to slow it down. (Yes, I have done this, too. Same warning!)

Does a recoil buffer help? It helps cushion the contact between the slide and the recoil spring guide/frame, and gives some people peace of mind that the frame is not being pounded. But steel is highly elastic, and the pistol is designed to allow the slide to bounce off the frame and use some of the recovered recoil energy in the forward part of the cycle. That is why installing a recoil buffer often results in failures to feed; not enough of the slide's energy is returned to assist the forward part of the cycle. (Remember those little steel balls in the rack that bounced for a long time once started? Put a buffer between them and the bouncing will soon stop.)

Doesn't the link lock the barrel back to the slide? No. The barrel foot is made to cam up on the slide stop pin. The gun can be fired without the link; it will not unlock consistently, since gravity is not that reliable in the circumstances, but return to battery is no problem. (Same again, same warning!)

So, Tuner and guys, have at it. I know there will be tons of argument about what I have written, but your significant others will thank me for distracting you and letting them get some sleep.

Jim
 
Keunhausen's book on Muasers drove me crazy when I was learning to sporterize. I am a scatter brain, and I can't read a scatter brain and get orgainized.




I did a calculation on how far cases fly.
It is interesting to me that the velocity of the bullet and gas, the mass of the barrel and slide, the travel of the slide, the spring force of the recoil spring, all do not add up to how fare the cases fly. It would if the pistol were in a heavy vice, but the hand moves back and keeps the slide from getting all the potential side velocity relative to the frame. So we need less spring than one would calculate.

link to my calculaion on far cases fly from my 1911
 
Vectored!

Hey Jim! I was wonderin' where you were hidin'.

I used to have a problem with Kuhnhausen's description of the recoil cycle too...but the more I read it and think about it, I don't think his twist is flawed so much as it is the way he describes it. Aybody who has read his manuals can attest to the fact that he's a little hard to follow. He's a great
pistolsmith, but he ain't no Hemmingway.

His description leads the reader to believe that the bullet exits BEFORE the recoil cycle starts...which is decidedly false. I'm more inclined to believe that he just didn't present it correctly. The bullet is out of the barrel before the cycle is completed, to be sure...but it's still there when the cycle BEGINS...and the momentum needed to complete the cycle is established during that brief instant that the bullet is moving from the point of origin to
the open air.

The thrust vector exists. I believe that the barrel is pulled forward by the frictional effect of the bullet, while the thrust is acting on the slide...As the
bullet moves, the rearward thrust begins to overcome the forward movement of the barrel, and the slide begins to pull the barrel backward with it. The thrust is also acting on the expanded case and the case helps
the barrel to move rearward against the slide...and it all comes together
at the right time...which helps to explain why powders with a slow burning rate won't cycle the slide reliably. The bullet accelerates more gradually
and exerts the forward pull on the barrel for a nanosecond too long...
and the whole thing comes to a halt....or nearly so.

Standin' by...
 
I"ve told myself not to get involved in discussions like this, but here goes . . .

I've read both this post and the one started by Tuner, and you guys are making this way more complicated than necessary. (Now that I've finished writing my post, maybe I am, too). I respect the hell out of both Tuner and Mr. Keenan. They both have oodles more experience with firearms than me, and I've learned much from reading their posts in the past. However, I do believe in the laws of physics (I do, I do), and I do believe they go against Mr. Keenan's version of events and in favor of Tuner. (I've never read Kuhnhausen, so I'll not comment on that. I'll stick to Mr. Keenan's post)

It's simple. For a recoil operated pistol, such as JMB's 1911, the following is the basic sequence: A fired round in a chamber creates a lot of gas pressure very quickly. In one direction that gas pressure pushes the bullet out of the cartridge and down the barrel. Simultaneously, in the other direction that same pressure pushes back against the cartridge base, which pushes against the breech that is part of the slide. The result in that direction is the slide is accelerated rearward - and, oh yeah, the slide is locked to the barrel by the locking lugs, so the slide pulls the barrel to the rear with it, until the rear of the barrel is pulled down and out of engagement with the slide. The slide then continues on its merry way due to momentum to finish the cycling sequence. In essence: One big pressure, pushing in two equal and opposite directions simultaneously achieving two results - bullet accelerated down barrel and slide accelerated rearward. Bullet travels farther, faster than the slide/barrel because the former has much less mass than the latter and therefore is accelerated much quicker. (See Newton's 2d law: F=ma. Here the F is the same, only the m's differ, leading to different a's). Period, end of story.

The barrel does not push the slide rearward. The slide pulls the barrel rearward.

The bullet's forward motion does not cause the barrel's rearward motion. Again, the barrel's rearward motion is caused by the slide - to which the barrel is locked - pulling the barrel rearward.

And what causes the slide to be accelerated rearward? The pressure in the chamber acting against the breech face through the cartridge base.

I believe Mr. Keenan is misapplying Newton's 3d law in reasoning that the bullet traveling down the barrel somehow "creates" a force that moves the barrel rearward. Yes, the barrel - via the rifling - opposes the travel of the bullet because of friction between the rifling and the bullet. So there is a force in the barrel in the direction opposite the bullet travel. However, the rifling only offers so much resistance ("force") in oppositin to the bullet, and this resistance, as Mr. Keenan indicates, is negligible compared to the force with which the bullet is being forced down the barrel by gas pressure. The gas pressure easily overcomes the friction resistance offered by the rifling to the bullet. If the only force offered up by the barrel in the direction opposite the bullet travel is the friction of the rifling (again negligible) - a force easily overcome by the gas pressure force behind the bullet - then what force precisely exists (in terms of the bullet and barrel alone) to cause the barrel to move backwards? Things don't move in a direction without the application of some net force in that direction (See Newton's 1st law). In Mr. Keenan's bullet-barrel version, sorry to say, there is no net force - absent the slide - acting on the barrel to move it rearward. Therefore, the force causing the barrel's rearward movement must come from somewhere external to the bullet-barrel relationship - and it does. It comes from the slide pulling the barrel rearward, driven by the gas pressure in the chamber acting in the direction opposite the bullet travel.

Think of it this way: If I try to push my couch across the floor with say 5 newtons of force, the couch pushes back against me with 5 newtons of force. Same with 10 newtons, 15 newtons, etc... until I overcome the inertia and friction holding the couch in place. The couch can never push back against me with more force than I push against it, and it certainly can't move in the direction opposite I am pushing it. It can't, it's impossible. To do otherwise violates Newton's 3d law. Yet, in Mr. Keenan's version, this is precisely what he is postulating. In the "bullet moves barrel" theory, the barrel not only resists the bullet with more force than the bullet is pushing down the barrel, but this force is sufficient to move not only the barrel rearward, but also the slide (against the recoil spring and it's own mass) violently rearward. Just isn't a proper model as it's not consistent with Newton's laws.

But what about Mr. Keenan's experiment with fixing the bullet in the barrel? Fine experiment, perfectly legit result, but incorrect conclusion about what result shows. By fixing the bullet in the barrel, the bullet and barrel essentially became one unit. What force acted on the bullet acted on the barrel. Upon firing a round, gas pressure pushes the bullet to the left with force "N," and the barrel, being attached to the bullet, wants to go the same direction. But wait a minute, the barrel is also attached (locked) at the other end to the slide/breech face by the locking lugs - essentially making the slide and barrel one unit. Upon firing a round, gas pressure pushes the cartridge case base and the breech and, hence, slide to the right also with force "N" and the slide wants to take the barrel with it. So, let's see, the bullet is trying to pull the barrel to the left, while the slide is simultaneously trying to pull the barrel to the right with equal force. Equal forces acting in opposite directions cancel each other out and no net force equals no change in movement for the slide, the bullet, the barrel which are all interlocked to each other - the whole contraption stays put. What Mr. Keenan created was a closed, fixed system in which the gas pressure exerted equal force in all directions and since everything was locked up tight to everything else in a Rube Goldberg fashion, the pressure strained against it's miniature steel walled cage until it died and dispersed itself, probably in the form of heat. Think of a miniature bomb disposal unit - big round steel container police put bombs into then blow them up. Whole thing just sits there, goes nowhere despite the huge momentary bomb pressure inside. Same principle with Mr. Keenan's experiment.

As for the contention that "pressure does not directly cause the gun to function in a recoil operated pistol as it does in a blowback pistol." Poppycock. The gas pressure created by combusting nitrocellulose is the only energy source/force by which modern semi-automatic firearms (pistol or rifle) operate. It is the only energy source available to get things moving. The only difference between a blowback, recoil, delayed blowback, gas delayed blowback, or gas operated rotating bolt firearm is the manner in which they harness the gas pressure and turn it into mechanical/kinetic energy to move the slide, unlock the slide, hold the slide closed momentarily, drive the piston to move the bolt carrier which rotates the rotating bolt, etc....

Other than these minor points, everything else Mr. Keenan said appears to be spot on.

(Let me be the first to say, the above is based on my reading of Mr. Keenan's post. It's quite possible he meant to say something different and/or I misinterpreted his wording on some points. If so, I'm sure he will address my mistakes and I provide an a priori mea culpa).
 
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Here is the article:
Jerry Kuhnausen's article here

Jim Keenan -
In his illustrations, and in the accompanying text, he states that the barrel/slide unit does not begin to move until the bullet is out of the barrel. That is simply not true; if it were, the pistol would not work.
Click to expand...
It seems he must be wrong. The pressure would drop off drasticly as soon as the bullet left the barrel. It would be like firing a blank at that point.

1911Tuner -
Have you figured out if the forward drag by the bullet friction is greater than the rearward drag by the case friction?

Higgins -
I agree completely with your explanation for slide recoil. I do not like using the 3rd law because while it is certainly true, it does not explain the source of the force.

I think this explains it best, an analogy:

You got 40 guys standing around outside. Someone really big takes a gigantic open box and turns it upside down over these guys. They are still standing on the ground and are surrounded by the four walls of the box. They all want to get out and run to walls and start pushing. 10 guys against each wall. They all push equally hard and the walls start to bulge. The box stays in position.

Suddenly, one of the walls breaks loose. the 10 guys on that wall push it outward fairly quickly. The 10 guys on the opposite wall now begin pushing whats left of box along the ground. Not quite as fast as the other guys pushed the wall, because they are pushing three times as much weight.

The 3rd law states that the box will move if the wall does. It does not say why. Looking inside the box shows why it moves.
 
They are still standing on the ground
Click to expand...
Your analogy only works if you make this assumption. What's analogous to the ground in a gun? Nothing--the gases are fully enclosed by the cartridge and the bullet.

For it to be truly analogous, they would all have to be fully enclosed in the box like the combustion gases are. So, they're all standing on the bottom of the box and pushing against the sides. NOW when one wall gives way, what happens to the box and why?
 
The barrel does not push the slide rearward.
Click to expand...
There is SOME rearward force transferred to the barrel by the friction between the cartridge case and the chamber wall. I agree that the majority of the rearward force is against the breech, not against the barrel.

Keenan's quote is correct: "pressure does not directly cause the gun to function". Sure, pressure causes the gun to function--it gets the bullet moving. But the recoil force is a result of the bullet (& other ejecta) motion, not directly a result of pressure. What you're saying is like saying that pumping up an airgun is the direct cause of recoil. Sure, there's no recoil without the pumping, but the pumping doesn't cause the recoil, the pellet motion and ejection of the compressed air causes the recoil motion (tiny as it is.)
 
JohnKSa -
For it to be truly analogous, they would all have to be fully enclosed in the box like the combustion gases are. So, they're all standing on the bottom of the box and pushing against the sides. NOW when one wall gives way, what happens to the box and why?
Click to expand...
I think that doesn't work at all, that way. The box would never move.

To analogize a gas, I would have them running around, bumping into each other and the walls.
Keenan's quote is correct: "pressure does not directly cause the gun to function". Sure, pressure causes the gun to function--it gets the bullet moving. But the recoil force is a result of the bullet (& other ejecta) motion, not directly a result of pressure.
Click to expand...
Why couldn't the slide movement be considered as the action, resulting from pressure and the bullet movement as recoil?
 
The box would never move.
Click to expand...
It WILL! The motion of the wall as it gives way, and the motion of the guys pushing against it and now released to move will cause the box to recoil.
Why couldn't the slide movement be considered as the action, resulting from pressure
Click to expand...
Well, in a locked breech gun, the slide CAN'T move based on pressure. The bullet must move and generate recoil in the slide/barrel UNIT before the slide can move.

In a straight blowback gun, I guess you could make the statement--but it's like saying that the 30 guys pushing on the three sides of the box that didn't give out are the reason that the fourth wall broke.
 
It WILL! The motion of the wall as it gives way, and the motion of the guys pushing against it and now released to move will cause the box to recoil.
Click to expand...
I think this is crucial. I think this is the whole cause of all of the preceding theory arguments. I need to look at this to make sure I understand what your saying.
 
Are you saying that their feet pushing back against the floor is the recoil reaction?
 
Well, in a locked breech gun, the slide CAN'T move based on pressure. The bullet must move and generate recoil in the slide/barrel UNIT before the slide can move.
Click to expand...
It seems like it should work. It's easier to imagine with an extremely heavy bullet.
 
Well, the bullet can be extremely heavy or the acceleration can be very high. Either contributes equally to a large force. In the firearm case it's the acceleration that's large.
 
Isn't it simply two masses free to move in relation to each other? The only real difference is the disproportionate size. Why would one have preference as the one "acting"?
 
re: General Response

Jim...I don't know which offers the most resistance, but I would have to guess that the bullet does....mainly due to its having to swage down to get past the rifling.
______________________


Higgins...Great post. We're on the same page, with one exception. I don't think that it's all in the slide. The barrel eventually winds up pushing on the slide too...it just occurs a split second after the slide starts its rearward pull. Nothing is everything...Everything is something. I DO, however, believe that the slide is the primary mechanism that provides say....90%
of the action. The barrel's influence ends quickly...as soon as the link unlocks it from the slide.

I completely agree with the "Pushing against a Wall" theory. The slide and barrel being locked together, and being pushed/pulled in opposite directions with equal force can't move and will remain static until something happens to make the force unequal. That unbalancing of the force vector occurs when the bullet moves...but it occurs gradually rather than suddenly. The balanced vector isn't "broken" as per Kuhnhausen...but rather makes a transition. <------Total theory here.

I'm going to propose another experiment for Jim with a cross-post on both threads in a bit. Things are pretty chaotic around here early mornings.

Stand by...

Tuner
 
Another Experiment

Here 'tis. There is a duplicate post on the other thread. Mr Keenan...Ya game?
______________________

Okay....Let's recap Jim Keenan's experiment in which he threaded the muzzle end of a barrel and inserted a rod to block bullet movement. A
set screw was threaded into the muzzle, positively blocking any chance of
bullet movement.

We learned two things from this.

One is that the 1911 pistol is far stronger than many of us imagined.

Two is that if the bullet doesn't move, the slide won't move. This
SEEMS to fly in Kuhnhausen's face on his "Balanced Thrust Vector"
theory of recoil, in which he appears to state that the bullet exits and
THEN the slide recoils...which is wrong. If the bullet is gone BEFORE the slide moves...the slide won't move. There is no resistance(the bullet) to the pressure to act against the slide and cause it to move. No launching pad, if you will.

I submit that because the barrel was pushed forward...and because the
slide and barrel were mechanically connected...the slide COULDN'T
move in the opposite direction. The barrel was keeping it static via the locking lugs. it would be the same as two cars linked together by a chain
and pulling in opposite directions. Assuming equal vehicle weight, power,
gear ratios/torque multiplication, and equal traction...neither one CAN move...until something breaks to make the balanced vector UN-equal.

I propose another experiment. I'll remove the locking lugs from a USGI
Colt barrel and mail it to Jim. Hopefully. he still has the rod and screw,
so that all he needs to do is thread the barrel and go. I realize that the gun will then be blowback operated, but the point is this:

If the mechanical connection between the barrel and slide is broken,
the pressure will be free to pursue the path of least resistance, and the slide will move...probably violently. I suggest a strong recoil spring and a shock buffer. I'll include a buffer with the barrel.

Standin' by...
 
That unbalancing of the force vector occurs when the bullet moves...but it occurs gradually rather than suddenly.
Click to expand...

That I think would have to be the case, otherwise we would have to assume the powder charge would develop max pressure at the instant of firing, which is not true.

If the barrel has a part in starting the slide rearward, wouldn't there be telltale wear marks on the slide where it is impacted by the barrel?
 
Impact Marks

Howdy Delmar. You asked:

If the barrel has a part in starting the slide rearward, wouldn't there be telltale wear marks on the slide where it is impacted by the barrel?
____________________

Not unless there's a lot of fore/aft play in the barrel when it's in-battery, or
the locking lugs aren't bearing an equal load. Slap-seated lugs...or impact
equalized lugs often show damage to the barrel and slide lugs with a
stepped appearance. This can also occur whenever there's a linkdown timing problem with the barrel, but that particular damage usually produces a rolled or rounded off locking lug at the top...Flanging of the lug is also usually seen. Flanging is a burr or sharp ridge on the top of the lug that can be felt with a fingernail raking the lug from back to front.

The stepped damage on the lugs likely occurs when the bullet slams into the rifling and yanks the barrel forward against the slide lugs closer to the pressure spike...which crashes them harder due to having a longer running start. When the play is less, the barrel and slide lugs butt together earlier, before the speed and force have built up to a degree that the steel is peened on impact. Demonstrate this by whacking two hammer heads together from an inch apart...and then from a foot apart.

Stepped, impact damage to the slide lugs is worse when the total lug engagement is less than half of lug depth. Demonstrate by whacking the hammer heads together when they're offset to one another instead of
straight on with heads perfectly aligned.

When there is nearly zero play between slide and barrel lugs, they strain
against one another without the running start/momentum buildup...and
build tension between them gradually. Demonstrate by touching the
hammers together and starting pressure from that point instead of building speed and momentum before they hit.

Interestingly, it was the stepped damage in some pistols that really got me thinking about the balanced thrust vector. I'm now at the point of believing
that the slide initiates the rearward movement of the barrel...and the barrel
begins to offer some help at some point through the rearward thrust against the case through the breechface. This probably is a very small
amount of push, due to the pressure starting to fall off after the spike
and allowing the lighter barrel to be pulled rearward by the heavier slide
which has its momentum established...and because the barrel travels
for such a short distance before starting to unlock. The slide is the major player because of its greater mass and its resulting larger amount of kinetic energy.
 
What if instead of locking the bullet to the barrel, as in Mr K's test, lock the bullet to the frame and placing the frame in a secure holding device. This would hold the bullet from moving but would not lock the barrel/slide to the bullet.
 
I completely agree with the "Pushing against a Wall" theory. The slide and barrel being locked together, and being pushed/pulled in opposite directions with equal force can't move and will remain static until something happens to make the force unequal. That unbalancing of the force vector occurs when the bullet moves...but it occurs gradually rather than suddenly. The balanced vector isn't "broken" as per Kuhnhausen...but rather makes a transition. <------Total theory here.
Click to expand...

Precisely the case. If we could see inside the case, and watch a slow-motion video of ignition while also having the ability to determine pressure, resulting force, etc, etc, things would look like this:

The primer ignites, and in turn the powder begins to burn. Pressure within the case builds rapidly. For a tiny moment in time, the bullet remains locked to the case via the static friction between the two, but the force on the base of the bullet resulting from the building pressure quickly overcomes this. During this moment when nothing moves, the net force on the whole system is zero. The force exerted radially on the case walls is balanced by a reaction force exerted by the chamber on the case. The force exerted on the base of the bullet is balanced by the static frictional force of the crimp. The force exerted on the case head is balanced by a reaction force exerted by the breech face.

Now the static frictional force is overcome, and the bullet starts to move. At this point, it's easier to analyze the situation using the conservation of momentum principle. Before anything moved, the total momentum was zero; momentum is equal to mass times velocity. To simplify the problem, imagine there is no recoil spring; otherwise, this adds an external force to the system and complicates things a bit. Without the recoil spring, final momentum of the system at any point- whether the bullet is still in the barrel or way downrange, we're still considering it as part of the system. Neglect air resistance, as well; it only muddys the problem unnecessarily. Condier the moment the bullet breaks contact with the barrel; at this point the bullet has momentum, since it has both mass and velocity. In order to conserve momentum, the slide/barrel (still locked at this point) will have equal momentum, but in the opposite direction. This implies that they have to move, because they have to have velocity.

If the recoil spring is replaced, it's going to exert an impulse on the slide/barrel (later just the slide after unlocking, but we're not going that far in the future), an impulse equal to force times the duration of time the force was exerted. In practice, determining this impulse would require integration, but it's enough to know the principle for now. The law of conservation of momentum actually states that the final momentum of a system is equal to the initial momentum plus the sum of applied impulses. So in reality, with the spring exerting an impulse, the slide/barrel assembly will be traveling rearward a little slower than what we'd get from the first method I described.

As far as the gradual unbalancing of the vector- put simply, the force on the base of the bullet, and therefore also on the case head, is going to vary with time, as the bullet travels down the barrel. On the one hand, powder continues to burn, increasing pressure, which would imply that force also increases, but at the same time the bullet moves, effectively expanding the area containing the gasses, so pressure is not going to increase that much. The force on the bullet at any given time is equal to the pressure at that moment multiplied by the bullet's base area, and since area isn't going to change, force is directly proportional to pressure. Not only is this force varying, but it's being applied continuously as the bullet travels down the barrel, as opposed to a single force applied at a moment in time only, like hitting a baseball into the pitcher's skull.

If I'm explaining things that are obvious to you, don't take it personally. Frankly, I just like to talk about the way things work, and how everything interacts. It's no wonder I'm an engineering major.

Side question, off-topic: Out of curiousity, what part of NC are you in?
 
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