55 replies to this topic

[Doom]

I know you say that you disagree that they are inversely related but you also just said they were.
"Static friction can increase the maximum pressure, which can shorten ideal barrel length"
i.e., More static friction = Less barrel length.

No, more static friction CAN EQUAL, not does equal, less barrel length. It is possible, and likely, but not necessarily true. You understand the restrictions. Just make them clear by avoiding absolute statements like "more static friction = lower ideal barrel length".

Significantly?
I understand this in terms of theory but for this application I really don't think it's necessary to take temp change into consideration.
You really think the temperature change of the gas in the chamber is really going to effect anything to a degree where it should even be considered?

Even if it does change to an amount that should factor in then fine, use PV=NRT. You still end up with the same effect. The pressure increase in with a tightening ring before it overcomes static friction, regardless of if the temp changes a little.

The temperature changes significantly as can be shown in both theory and experiment. The temperature must change because the thermal energy of the gas is converted into dart kinetic energy and there is no heat transfer to make up the difference. A simple calculation using an equation I linked to shows that the temperature in the gas of a Nerf gun drops to about -105 F if it starts at 80 F. That's a drop of about 33% in thermal energy (and the percent drop is the same regardless of what temp. it starts at for calorically perfect gases). That's significant. It causes the pressure to drop faster than it would under isothermal conditions, so neglecting the temperature drop results in inaccurate muzzle velocity estimates.

And you don't seem to understand my point, which was that Boyle's law is not an appropriate model, not that your understanding is entirely wrong. I never said using adiabatic process relationships show that increasing static friction can't help, for example. (I'm not sure if you are confused about what I said.)

Also, the ideal gas law can not be used alone to derive these relationships because it says nothing about heat transfer. If you read the link I posted you would have seen that the adiabatic process relationships detailed are derived from the ideal gas law and the first law of thermodynamics.

Edited by Doom, 25 December 2011 - 08:43 PM.

[Zorns Lemma]

Since you don't seem to carefully read or perhaps fully comprehend what people have told you, I'll try to sum it up:

1. This is a creative modification of Vulcan belts, but it had also been done before. For example, applying brass to the chain achieved the exact same effect. More importantly, the effect is NOT CERTAIN and aside from numerical computation and approximation, must be empirically measured. Even at non extreme cases, increasing static friction may not decrease ideal barrel length.

2. While high school classroom physics gives a good idea for gore these systems work, they are entirely inadequate to model with any certainty. Nerf pneumatics is a world of energy loss (friction/drag, heat exchange, sound, etc) so many of your classroom staples such as impulse, recoil, energy, and gas equations are inadequate to make predictions within statistical "certainty"

Edit: For example, simple physics tells us to eliminate all volume between the dart and propulsion system ("eliminate dead space"). However, numerical approximation AND empirical testing show that some volume increases muzzle velocity.

Edited by Zorn's Lemma, 25 December 2011 - 09:43 PM.

[BOSS9]

Empirical testing shows that some volume increases muzzle velocity.

Really? Cool. Do you have more information on this, like what kind of testing has been done? This interests me.
This shit is soo off topic now. You can PM me info if it wouldn't fit in the thread. If you feel like it, or care.

Edited by BOSS9, 25 December 2011 - 10:33 PM.

[Darthrambo]

The temperature changes significantly as can be shown in both theory and experiment. The temperature must change because the thermal energy of the gas is converted into dart kinetic energy and there is no heat transfer to make up the difference. A simple calculation using an equation I linked to shows that the temperature in the gas of a Nerf gun drops to about -105 F if it starts at 80 F.

Is this true? You are saying that the air inside a nerf gun drops to -105 F?
If this is the case it's pretty shocking to me. It seems pretty impossible for a nerf gun to drop air temp that much just by compressing it. (I'm not saying that it's impossible, just that I don't get how it would be). If the change in temp is that much then it's way more than I was assuming which would account for why I was under the impression that temp change in air was negligible.

What equation are you using to come to this?

As for the rest of the stuff it sounds like we are in violent agreement about what we are saying.

If I'm not mistaken you were just pointing out a variable that I left out that does come into play, and pointing out that the relationship of static friction vs barrel length is not linear.

If this is the case we are totally on the same page

I am ignoring certain variables like change in temp because we can't really measure that for nerf and because the general statement of what I'm saying still holds true.

In the end we are just talking about theoretical physics. While I do really enjoy that, my original intent was to just make a general statement about the relation between static pressure and barrel length that could be applied to nerf.

Really it sounds like we are pretty much not disagreeing with each other about anything once we get to the point where we actually understand where the other person is coming from.

Anyhow I really do enjoy bouncing ideas off of someone else who like the physics of nerf. Thanks for the discussion man

[Doom]

Is this true? You are saying that the air inside a nerf gun drops to -105 F?
If this is the case it's pretty shocking to me. It seems pretty impossible for a nerf gun to drop air temp that much just by compressing it. (I'm not saying that it's impossible, just that I don't get how it would be). If the change in temp is that much then it's way more than I was assuming which would account for why I was under the impression that temp change in air was negligible.

What equation are you using to come to this?

Adiabatic cooling is well understood. Your refrigerator would not work if this process did not exist.

If you followed the discussion to my link you would have seen I was using one of the adiabatic process relationships I linked to on Wikipedia. Specifically, T_2 / T_1 = (P_2 / P_1) ^ ((gamma - 1) / gamma). gamma for air is 1.4. Solving for T_2 leads to T_2 = 300 K * ((14.7 psia / (50 psig + 14.7 psia)) ^ ((1.4 - 1) / 1.4)) = 196 K = -107 F.

This process does not actually describe the temperature drop in a Nerf gun very well. In fact, the final temperature here is usually too high because the barrel gases do the work and the gas chamber gases don't --- this assumes that both do. Also, this applies only for air guns --- springers are more complicated. If you want to get more accurate estimates of the temperature drop for more general situations, write a computer simulation.

As for the rest of the stuff it sounds like we are in violent agreement about what we are saying.

In the end we are just talking about theoretical physics. While I do really enjoy that, my original intent was to just make a general statement about the relation between static pressure and barrel length that could be applied to nerf.

Really it sounds like we are pretty much not disagreeing with each other about anything once we get to the point where we actually understand where the other person is coming from.

I'm stating the facts, not arguing. You seem to want to make everything into an argument as Langley pointed out. We do disagree on a few points, and you are stubborn.

Really? Cool. Do you have more information on this, like what kind of testing has been done? This interests me.

Dead space can improve performance. I have shown this with computer simulations. Don't get too excited, though, as the amount of dead space that does improve performance is very small (like 5% of the barrel volume, if I recall correctly), the performance gains over zero dead space are minimal (a few percent gain in energy efficiency), and dead space is hard to eliminate in most designs. Assuming zero dead space is best is a good approximation of reality.

Edited by Doom, 26 December 2011 - 05:23 PM.

[Zorns Lemma]

pointing out that the relationship of static friction vs barrel length is not linear.

It is not that the relationship is nonlinear, it is that sometimes there is none at all.

[Darthrambo]

It is not that the relationship is nonlinear, it is that sometimes there is none at all.

Just saying something like this in a statement without any evidence or explanation means nothing. Unless you explain your how you came to your conclusion I have no reason to consider what you are saying.

[Zorns Lemma]

Just saying something like this in a statement without any evidence or explanation means nothing. Unless you explain your how you came to your conclusion I have no reason to consider what you are saying.

No, more static friction CAN EQUAL, not does equal, less barrel length. It is possible, and likely, but not necessarily true.

I am just trying to help you by thoroughly reading Doom's posts and highlighting information you seem to ignore or discard. Arbitrarily increasing static friction under the assumption that you can then reduce barrel length and still achieve optimum acceleration is flawed and the results should be empirically tested to be statistically significant before we go out and declare results.

[Darthrambo]

I understand what you are saying. What I was saying is that the variables that I am omitting are too hard to measure and too small to really matter for the purpose of foam darts. For something that needed complete precision, sure things like the variable of temp matter.

If the air temp in a chamber of a nerf gun really drop from 80 F to -105 F like doom claims this would not be the case but I find this extremely hard to believe. I'm not saying it's not true, I'm saying I need to seem some evidence before I believe that a nerf plunger tube can decrease the temp of the air inside by 185 F.

What I'm saying is I don't see any reason to pay attention to statements that are made without any evidence to back them up.

Edited by Darthrambo, 27 December 2011 - 04:01 PM.

[Doom]

If the air temp in a chamber of a nerf gun really drop from 80 F to -105 F like doom claims this would not be the case but I find this extremely hard to believe. I'm not saying it's not true, I'm saying I need to seem some evidence before I believe that a nerf plunger tube can decrease the temp of the air inside by 185 F.

Adiabatic cooling is real and ubiquitous. Refrigerators, air conditioners, and engines all use this process. The process is easily understood theoretically with adiabatic process relationships as I've shown.

Here's an introductory-level video showing adiabatic cooling. Don't believe me? Here's another introductory-level video showing adiabatic cooling. Still don't believe me? Here's another demo with a plunger (which acts the same as a dart). On some larger spud guns you can even see the condensation without high speed cameras. (And to be clear, it is not possible to measure the temperature of barrel gas in real time. Condensation is one way to see that the temperature has dropped for fast processes.)

Both springers and pneumatics use adiabatic heating and cooling. I've pointed out before that apparent pressure losses due to leaks could be due to heat transfer because the air in a tank gets very hot when you pump. So unless you shoot soon after pumping, you will lose some pressure even if you have perfect seals.

Edited by Doom, 26 December 2011 - 09:22 PM.

[Darthrambo]

Adiabatic cooling is real and ubiquitous. Refrigerators, air conditioners, and engines all use this process. The process is easily understood theoretically with adiabatic process relationships as I've shown.

Here's an introductory-level video showing adiabatic cooling. Don't believe me? Here's another introductory-level video showing adiabatic cooling. Still don't believe me? Here's another demo with a plunger (which acts the same as a dart). On some larger spud guns you can even see the condensation without high speed cameras. (And to be clear, it is not possible to measure the temperature of barrel gas in real time. Condensation is one way to see that the temperature has dropped for fast processes.)

Both springers and pneumatics use adiabatic heating and cooling. I've pointed out before that apparent pressure losses due to leaks could be due to heat transfer because the air in a tank gets very hot when you pump. So unless you shoot soon after pumping, you will lose some pressure even if you have perfect seals.

I never doubted that it happens. I want to make that very clear

What I'm doubting is that it happens to a degree that needs to really be taken into consideration for nerf springers.
I really can't believe that a springer can drop air temp in the chamber from 80F to -105F without some evidence.

For example look at the video you linked with the syringe as the plunger tube. The drop in temp is really small and needed to be rescaled so really see effect.

I'm not doubting adiabatic cooling happens. What I am doubting is that it happens to an extent that needs to be considered specifically to the relationship of barrel length to static friction in springers.

[Doom]

What I'm doubting is that it happens to a degree that needs to really be taken into consideration for nerf springers.
I really can't believe that a springer can drop air temp in the chamber from 80F to -105F without some evidence.

I've provided strong theoretical and experimental evidence. What more do you want? I can't provide more evidence. How about you justify your statements for once? You request evidence from others but don't provide any yourself.

How are the adiabatic process relationships unclear? The derivation of these relationships is detailed on the page. Where does this go wrong if the temperature does not drop as much as the relationships suggest?

Also, to be clear, both heating and cooling happen in springers in a fraction of a second. The end temperature is probably about atmospheric for most springers. No, that does not imply that the temperature in the middle is constant or that temperature does not matter. Air guns, on the other hand, often cool down isochorically (i.e., under constant volume and mass --- P/T = const. for ideal gases) before they are shot.

For example look at the video you linked with the syringe as the plunger tube. The drop in temp is really small and needed to be rescaled so really see effect.

The expansion ratio of a Nerf gun is larger, so the temperature change is larger. The same video mentioned that compressed air cans (which expand greatly) often freeze water, so this is perfectly consistent with that video.

Honestly, if you're going to be so stubborn, I'll quit while I'm ahead. I try to use these forums to explain the mechanics of Nerf guns, but I keep running into kids who take a high school physics class and then think they're experts. I should know better by now.

Edited by Doom, 27 December 2011 - 04:51 PM.

[Daniel Beaver]

Honestly, if you're going to be so stubborn, I'll quit while I'm ahead. I try to use these forums to explain the mechanics of Nerf guns, but I keep running into kids who take a high school physics class and then think they're experts. I should know better by now.

What you're seeing here is similar to what happens when climate scientists try to argue about the finer points of their long-term models on internet forums. They are trying to advance the state of human understanding, but all the fucking amateurs come out of the woodwork and try to hijack the conversation. It's honestly embarrassing to even read them sometimes. It is good to ask questions to experts, but it is presumptuous to challenge them unless you are very sure of your argument. And while that may sound harsh and elitist, it is important to understand and respect the limits of your own knowledge. Without the presumption of competence in discourse, we cannot be sure of the truth of it's results. So while I am much more competent than the super-majority of the forums when it comes to the physics and engineering of nerf, there are still member whom I defer to in some cases (Doom being one of them).

[Zorns Lemma]

I really can't believe that a springer can drop air temp in the chamber from 80F to -105F without some evidence.

The science shows the equations Doom repeatedly links to (have you actually looked at them!?) give an excellent model for what is going on.

I mean, if the temperature doesn't change, where does the energy come from?

[Darthrambo]

The science shows the equations Doom repeatedly links to (have you actually looked at them!?) give an excellent model for what is going on.

I mean, if the temperature doesn't change, where does the energy come from?

First, I already said that I'm not doubting that there is temp change. I even underlined it.

Second, unless I'm totally off base, the energy doesn't come from temperature change, it's lost to temperature change.

I've provided strong theoretical and experimental evidence. What more do you want? I can't provide more evidence. How about you justify your statements for once? You request evidence from others but don't provide any yourself.

How are the adiabatic process relationships unclear? The derivation of these relationships is detailed on the page. Where does this go wrong if the temperature does not drop as much as the relationships suggest?

Also, to be clear, both heating and cooling happen in springers in a fraction of a second. The end temperature is probably about atmospheric for most springers. No, that does not imply that the temperature in the middle is constant or that temperature does not matter. Air guns, on the other hand, often cool down isochorically (i.e., under constant volume and mass --- P/T = const. for ideal gases) before they are shot.



The expansion ratio of a Nerf gun is larger, so the temperature change is larger. The same video mentioned that compressed air cans (which expand greatly) often freeze water, so this is perfectly consistent with that video.

Honestly, if you're going to be so stubborn, I'll quit while I'm ahead. I try to use these forums to explain the mechanics of Nerf guns, but I keep running into kids who take a high school physics class and then think they're experts. I should know better by now.

Sweet jesus you guys really need to calm down. I never claimed to be and expert. I'm not even claiming that I'm right, so how the hell can that be called stubborn? All I said was that I don't see how it's possible for the pressure to change the temp that much and I'm trying to understand it.

So for those equations you'd use Cv =1.01, Cp= .718?

Edited by Darthrambo, 27 December 2011 - 07:52 PM.

[Doom]

Second, unless I'm totally off base, the energy doesn't come from temperature change, it's lost to temperature change.

Energy is conserved outside of nuclear reactions. So there's no "energy loss" here. The energy comes from the gas's internal energy, which is controlled by temperature.

Sweet jesus you guys really need to calm down. I never claimed to be and expert. I'm not even claiming that I'm right, so how the hell can that be called stubborn? All I said was that I don't see how it's possible for the pressure to change the temp that much and I'm trying to understand it.

So for those equations you'd use Cv =1.01, Cp= .718?

You didn't say you were trying to understand this, just that you don't think it's true. And regardless of whether or not you think you are right, I hope you understand how your posts can be frustrating. I encourage you to learn more about thermodynamics and ballistics --- both are worthwhile endeavors.

The ratio of specific heats, not the individual values of them, is all you need for these equations. As I said, gamma (the ratio of specific heats, which also is called k) for air is 1.4 over the temperature range we are interested in: http://www.engineeri...ties-d_973.html

Be careful when converting between Kelvin and Fahrenheit. I usually work solely in metric units to keep everything simple but I will convert when sharing with others.

Edited by Doom, 27 December 2011 - 08:29 PM.

[Zorns Lemma]

Second, unless I'm totally off base, the energy doesn't come from temperature change, it's lost to temperature change.
All I said was that I don't see how it's possible for the pressure to change the temp that much and I'm trying to understand it.

The air is the medium for energy transfer from the propulsion mechanism to the dart. The temperature change is significant because that is how the air gets the energy to increase in volume and do work on the dart, moving it forward.

dU = δQ + δW where Q and W are heat added to the system, and work done on the system, respectively

[Darthrambo]

Energy is conserved outside of nuclear reactions. So there's no "energy loss" here. The energy comes from the gas's internal energy, which is controlled by temperature.

I guess what I meant to say is kinetic energy transformed to heat energy. Either way you aren't saying comes from temp change like that guy was implying with "I mean, if the temperature doesn't change, where does the energy come from?" right?

The air is the medium for energy transfer from the propulsion mechanism to the dart. The temperature change is significant because that is how the air gets the energy to increase in volume and do work on the dart, moving it forward.

dU = δQ + δW where Q and W are heat added to the system, and work done on the system, respectively

Totally get what you were saying now.

You didn't say you were trying to understand this, just that you don't think it's true. And regardless of whether or not you think you are right, I hope you understand how your posts can be frustrating. I encourage you to learn more about thermodynamics and ballistics --- both are worthwhile endeavors.
The ratio of specific heats, not the individual values of them, is all you need for these equations. As I said, gamma (the ratio of specific heats, which also is called k) for air is 1.4 over the temperature range we are interested in: http://www.engineeri...ties-d_973.html

Be careful when converting between Kelvin and Fahrenheit. I usually work solely in metric units to keep everything simple but I will convert when sharing with others.

Yeah, I get the frustration especially since people on the internet seem to assume everyone else is speaking with a contentious tone.
For this Kelvin is the temp used in the equations correct? I'd actually prefer to stay away from Fahrenheit, I just adds a a conversion to do.

So if i understand correctly by using "k" you will be able to find the temp change for a final pressure, if you know that initial pressure and temp, and this is regardless of volume?

Edited by Darthrambo, 27 December 2011 - 08:53 PM.

[Doom]

So if i understand correctly by using "k" you will be able to find the temp change for a final pressure, if you know that initial pressure and temp, and this is regardless of volume?

The volume can be found through the ideal gas equation. You'll need as many equations as unknowns and that's it.

[Hammy]

FFS it's only a Vulcan, get over it.

I cut my vulcan belt into sections of 4 or 6 rounds, then secured those sections to my other blasters using rubber-bands, and use them as dart holders.

I donated my Vulcan to charity


Applictaion is spelt application, but those who are good at equations are usually not good at spelling. And there should be an additional 's' or 'a' in the title along with a 'the' to make all the grammar correct.

Edited by Hammy, 28 December 2011 - 08:41 PM.

[Darthrambo]

What I don't get is how the gamma works in the ideal gas equation.

If you know the change in volume how do you determine how much of a pressure change vs temp change? I'm guessing gamma would dictate that but how does that fit in with the ideal gas equation?

[Doom]

The ratio of specific heats does not appear in the ideal gas equation unless you rewrite it to appear (which replaces temperature with an energy). It's in a completely different equation here. I'm not sure what you are asking, but I can explain how these relationships are derived.

The adiabatic/isentropic process relationships are derived from the following equations:

P = rho * R * T,

P * V = m * R * T, and

P * V ^ gamma = constant.

The first two are just different ways to write the ideal gas law. rho is mass density (rho =: m / V). R is the specific gas constant, not R_bar, which is what you use in chemistry class --- R =: R_bar / M where M is the molar mass of the gas.

Noting that in the adiabatic process the total mass of gas is conserved, i.e., m is a constant, so (P_1 * V_1) / (R * T_1) = (P_2 * V_2) / (R * T_2) = m, you can derive every relationship listed on the Wikipedia page I linked to.

Edited by Doom, 28 December 2011 - 10:53 PM.

[Darthrambo]

Let me try to re-frame my question a little more clearly.


With PV=nRT as it applies to a plunger tube, we only know:

Volume initial and volume final

n, (or mass or)

And the constant, R

Pressure and temp both change when the volume decreases. My question is how do you know how much each changes. How do you relate Charles law and Boyles law?

[Doom]

Pressure and temp both change when the volume decreases. My question is how do you know how much each changes. How do you relate Charles law and Boyles law?

You know how much each something changes either from the geometry or starting and atmospheric pressures. And you find what you don't know using the equations.

For example, using P * V ^ gamma = constant we can find the volume a gas adiabatically expands to from a certain pressure to atmospheric pressure.

So if we have a simple piston assembly that starts off at atmospheric pressure, temperature, and has a set (known) volume and we compress this piston assembly to a new known volume we can find the new pressure to be

P_compressed = P_atmospheric (V_atmospheric / V_compressed) ^ gamma .

And we can use the ideal gas law to find the compressed temperature from here. T_compressed = (P_compressed * V_compressed) / (m * R) .

Charle's law and Boyle's law do not factor into this. Charle's law only applies for ideal gases with constant pressure. Boyle's law only applies for ideal gases with constant temperature. Neither pressure or temperature is constant here.

[Darthrambo]

Charle's law and Boyle's law do not factor into this. Charle's law only applies for ideal gases with constant pressure. Boyle's law only applies for ideal gases with constant temperature. Neither pressure or temperature is constant here.

Yeah, I just didn't know how to determine how much of a change in each would occur when neither was held constant.

So for this you'd use:

Gamma= 1.4
V= Liters
P= ATM
M= kg
T= Kelvin
?

Thanks a ton for taking the time to explain this btw.