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Stefan Theory


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#1 DTReaper

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Posted 01 October 2005 - 09:31 PM

Ok here is my theory my theory is that the more mass you have the greater the distance the stefan will go. but the best weight is dependant on the power of the gun, and wind resistance on the dart decreases its range so I came up with this thoretical equasion (sp?)

distance=(massXpower of gun)/wind resistance

That is distance equals mass times power of the gun divided by wind resistance.
But mass can only equal a certain percent of the power of the gun (currently undetermined).
Also the less mass th faster it accelorates (sp?) and the higher the top speed but it gets less distace it gets. the heavier it is the slower it moves but it gets better range and hits harder,
So thats my theory please add your thoughts opinians and questions and I'll get back to you as soon as I can thank you.

-DTR

Edited by DTReaper, 01 October 2005 - 09:33 PM.

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#2 PissBacon

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Posted 01 October 2005 - 10:18 PM

There are more complicated physics at work here than you realize. You should understand that the only reason a dart can only go a certain distance is from gravity and wind resistance (more accurately aerodynamic drag) (Newton's first law). This means really you have a basic force (force = mass x acceleration, Newton's 2nd) which is being hindered by the previously mentioned forces.

To make an accurate formula you would need to incorporate gravity's negative acceleration, quantify friction the dart suffers, and figure out the force the dart is projected with (muzzle velocity x dart weight). This is to make a simple formula too, a more accurate one has to include other factors like windspeed and imperfect darts.

There may be some formulas I am unaware of that can approximately model what I'm discussing, but I haven't taken any physics classes yet so I wouldn't know.

EDIT:
http://nerfhaven.com...?showtopic=3223 (esp. cranky monky's post)
http://www.glenbrook...DKin/U1L6a.html
After thinking this out further, I realized you can calculate how long it will take the dart to fall if fired level from, say, 5 feet off the ground. Then use the muzzle velocity to calculate how far it would travel in that timeframe. That would be a rough estimate, but it would ingore resistence on that muzzle velocity as it travels through the air

Edited by PissBacon, 02 October 2005 - 12:33 AM.

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#3 tenesee23

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Posted 02 October 2005 - 11:22 AM

distance=(massXpower of gun)/wind resistance

Why would you be dividing wind resistance. Would you not be subtracting it? Say for example, we could measure wind resistance in units of Bob. If there was 4 bob, it would bring a ussually 100 feet shooting gun down to 25 feet. I understand where wind resistance does take effect, but not that great of one.
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#4 DTReaper

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Posted 02 October 2005 - 12:45 PM

The air resistance may be minus Im not sure but my second theory is correct 9the one concerning weight and distance.That is why I use tack stefans in indoor wars, because they leave welts at close range and sting alot.

-DTR
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#5 WratH

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Posted 02 October 2005 - 01:01 PM

Who cares? Just shoot somebody.
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#6 DTReaper

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Posted 02 October 2005 - 01:04 PM

Well Im just pointing out that the tack stefan has a home in indoor wars for me at the very least. However the heavier darts are needed out doors for the increased range unless you get in a massive amount of cc fighting in outdoors game.

-DTR
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#7 AirApache

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Posted 02 October 2005 - 01:35 PM

I think...you are looking WAY too much into this. Unless you're doing it for a Physics project or something, there is no need to be thinking about this.

Sure, you can make the most physically perfected dart that gets amazing distance with every shot with perfect accuracy, but what's the point? Where's the fun? Might as well go shoot real guns if you're looking for that.

I'm not bashing you for this topic or anything, because it has a certain degree of interesting-ness, but just in general...
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#8 ompa

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Posted 02 October 2005 - 01:52 PM

Air actually did a math speech on barrel/dart physics for math class. Needless to say, it was rather funny.

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#9 LordoftheRing434

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Posted 02 October 2005 - 10:27 PM

Excellent, if I need a physics project this year, I'm definitely going to look into this. I think the amount of wind resistance would be subtracted as well. Of course it's going to take some distance out of play, but it's not going to reduce it by a fraction. If distance=(mass x velocity)/wind resistance, you are going to get radically affected outcomes with just a bit of wind. Rather, gravity is going to make more of an affect than wind. Gravity will remain a constant force on the dart, and wind resistance is going to be a changing variable. These two should be added together and fit into the formula in a different spot. Perhaps D=(m x v) - (g + w). This might even take a step into calculus, but I don't have that til next semester, so I'm not even going to think of it that way. I'd take more time into thinking this out, but it's still the weekend, so I'm going to ignore everything with numbers and equations for now.

~Rings

Edited by LordoftheRing434, 02 October 2005 - 10:30 PM.

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#10 chiefthe

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Posted 03 October 2005 - 08:26 AM

The equations are pretty straightfoward until you add drag in.

For the non-drag situation the point mass travels as in a parabola, and mass doesn't even enter into the equations:

x=(v_0 cos (theta))t

y=(v_0 sin (theta))t -0.5(g)(t)^2

where:
x= distance traveled along the ground
v_0= initial velocity
theta= initial firing angle
t= time the projectile is in the air
g= acceleration due to gravity

You can iterate through this, or solve for the t intercepts in the y equation to see how long it will take the mass to land.

For the drag situation the point mass travels approximately in a parabola, but the math gets a little stickier, and you have to iterate and work in two dimensions. Any first year college physics text will have an example of this iteration.

In general, you can get a decent estimate of distance with a low profile item (like a Nerf dart) doing the calc without drag.

Now, you are asking why do my heavier darts seem to fly further if mass doesn't enter into the equation? It has to do with maintaining the center of mass forward of the center of pressure to eliminate the net torque on the body and keep the low drag nose facing forward. By adding mass you are bringing the center of gravity closer to the tip and further forward of the center of pressure. When the mass is too far back, the dart could spin out, or otherwise not keep the low drag nose pointed forward.

These pages do a much better job explaining it than I can:

http://en.wikipedia....ter_of_pressure

http://en.wikipedia....al_significance

http://www.grc.nasa....lane/rktcp.html


chiefthe
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#11 DTReaper

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Posted 03 October 2005 - 01:18 PM

If mass has no effect then how is it that a tack stefan will go only 50' while as another heavier dart fired from the same gun will go a considerable amount furter.
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#12 cxwq

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Posted 03 October 2005 - 01:37 PM

Now, you are asking why do my heavier darts seem to fly further if mass doesn't enter into the equation? It has to do with maintaining the center of mass forward of the center of pressure to eliminate the net torque on the body and keep the low drag nose facing forward. By adding mass you are bringing the center of gravity closer to the tip and further forward of the center of pressure. When the mass is too far back, the dart could spin out, or otherwise not keep the low drag nose pointed forward.

^
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#13 DTReaper

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Posted 03 October 2005 - 01:38 PM

Sorry about that I got bored reading his post after the first two paragraphs.

-DTR
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#14 cxwq

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Posted 03 October 2005 - 01:40 PM

If you don't want to learn about the real physics of nerf then why did you start the goddamn thread in the first place? Oh yeah, just to make us read your pointless and uninformed drivel. Thanks.
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#15 LastManAlive

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Posted 03 October 2005 - 04:03 PM

distance=(massXpower of gun)/wind resistance

Why would you be dividing wind resistance. Would you not be subtracting it? Say for example, we could measure wind resistance in units of Bob. If there was 4 bob, it would bring a ussually 100 feet shooting gun down to 25 feet. I understand where wind resistance does take effect, but not that great of one.


I didn't even use the ("3-finger-grip technique"). If anything I (abused) it. And if anythign else, I am (a bad bad girl). You don't just pull (a broom handle) out of your ass. That's why I suggested (sucking you off) in woods or indoors. Not much (exhibitionism) with walls or trees (hiding) it (right?) Was that (my fist in) my ass?
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#16 LordoftheRing434

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Posted 03 October 2005 - 07:18 PM

Sorry about that I got bored reading his post after the first two paragraphs.

-DTR

Yeah, Cx said it right. Show interest in your own thread, otherwise it's pointless.

I, however, am interested in what chiefthe had to say. It's the same thing we're going over in my physics class right now and for some reason, I understood that post better than my teacher is explaining it. Thanks! :D chiefthe, just out of curiousity, how would you go about finding the v_f (final velocity)? The formulas we have learned involve calculations where the acceleration is known, but it seems that's not the case here.

LMA, what the fuck is that?

~Rings
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#17 chiefthe

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Posted 03 October 2005 - 08:15 PM

Ah! But you accelleration is known--it is actually a decelleration due to gravity. That said:

The time it takes something to fall from a given height (h) when its initial angle is 0 degrees can be derived from the y equation (you can derive it yourself for other angles, zero is just the one I don't have to look up):

y=(v_0 sin (theta))t -0.5(g)(t)^2+h

y=0-0.5(g)(t^2)+h

set y=0 to see when it hits the floor:

t_floor=sqrt(h/(0.5g))

Using a little calculus magic, the velocity equations are:

v_x=v_0

(remember, no drag, so there is no opposing force slowing the x-velocity, so it has the same velocity it started out with---this equation is only for the zero degree case, for the general case: v_x=(v_0 cos(theta)) )

v_y=-(g)(t)

(negitive because the vector is pointing down---this equation is only for the zero degree case, for the general case: v_y=(v_0 sin(theta))-(g)(t) )

So, going back to vector algebra:

v=sqrt(v_x^2 + v_y^2)

If you want to solve for v_f, then plug t_floor from above into the v_y equation, then use the resulting v_x and v_y to put into the v equation above.

LOTR434, I hope I'm not doing your homework... ;)

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#18 NinjZ

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Posted 03 October 2005 - 08:39 PM

Cheifthe ftw!

Thanks for taking the time to give us those equations.
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#19 LordoftheRing434

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Posted 03 October 2005 - 08:44 PM

No worries, this is still beyond what we're doing. It's only the first month, and we've just touched a bit of vector analysis and acceleration/deceleration so far. I get the main idea at hand, but I'll need to read through it a few more times. I'm surprised I understood as much as I did. Wow, thanks for answering my questions. Maybe it's a good thing I'm looking towards a major in Physical Therapy... :D

Edited by LordoftheRing434, 03 October 2005 - 08:50 PM.

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#20 LastManAlive

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Posted 04 October 2005 - 04:28 PM

  LMA, what the fuck is that?

~Rings

Admin (mocking) me (because) my feelings (are irrelevant).
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#21 Pineapple

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Posted 04 October 2005 - 04:57 PM

Chiefthe spoke. She represents the very core of the people who are making your toys to play with, and is giving the bottom line of the so-called "theory" behind the darts, manufactured or homemade.

Reaper has been quiet since her second barrage so I guess that settles that.

Some of those equations are giving me flashbacks :(. That's why I'll never be an engineer.

Thank God for you folks.


-Piney-
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<!--quoteo(post=209846:date=Feb 5 2009, 06:27 PM:name=boom)--><div class='quotetop'>QUOTE(boom @ Feb 5 2009, 06:27 PM) View Post</div><div class='quotemain'><!--quotec-->
It's to bad you live in hawaii I bet there are not many wars there.Wait what am I saying<b> you live in hawaii you lucky bastard.</b>
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