On another airgun forum, I've been following a discussion that started with the idea that there is a limit to the velocity that can be achieved by  pre-compressed air in an airgun.  One of the ideas of long standing was that the velocity could not exceed the RMS of the molecular velocity which is about 1640 ft/s at room temperature.  However, by experiment, a light for caliber pellet has achieved a recorded velocity of 2162 ft/s using 4000 psi compressed air.  This former idea has been shattered.  

Now on that particular thread,  there has been a lot of conjecture and mathematical/spread sheet models to predict or explain what is going on, but it appears that no one really knows.  Fairly good measurements are needed which brings about my ideas.

What I want to do is make a similar set-up to this device that has achieved these rather high velocities, but put strain gauges on the barrel in at least five places to record the transient pressures.  

I'm familiar with the Wheatstone bridge, and balancing the bridge concepts.  What I want to do is to have the bridges balanced at room temperature and atmospheric conditions.  Then calibrate each of the gauges at discrete pressures; I'll "cork" the barrel and charge the barrel with known values of compressed air.  The output voltages would be recorded for each pressure and gauge preferably by the DAQ device.

Upon firing, I need to be able to record the time and voltage of each gauge.  Ideally, the time and each gauge voltage would be done simultaneously.  However, if the data rate is fast enough, a separate time for each gauge in sequence could be adjusted for.   The start and stop triggers would be when the first gauge (channel 1) reads non-zero and stops when it reads zero again.  

The output to the computer could be as simple as an ASCII  text file.  That could then be put into a spreadsheet for analysis.   I've thought about using a program like Audacity, but sound cards generally require an alternating source and filter out DC sources.  I'm not sure that the upper limit of 20K Hz of most sound cards is fast enough, and the affordable ones only have two channels.

So, the question is, what are the inexpensive options.  I am far from being made of money, especially at this time.  But, time is something I have a bit extra on hand.  Building something on a bread board is also not beyond my abilities.  That kind of thing is something that I enjoy doing.  What is beyond me, is figuring out how to make such a circuit, and which ICs are appropriate to incorporate and where.

Renting equipment is also not completely out of realm either, though I have no idea where one could rent a multi-channel storage oscilloscope for a day or two.

The bandwidth of your strain gauge will system will limit the rate of change your system can detect.


Very fast sensitive accurate measurements are very difficult to pull off.

This part is the problem, I have no idea what assumptions you guys are making to come up with those number or how you are applying things. That seems like a rather… odd … conclusion for this.

I mean, you could apply newton’s laws to the ideal gas law with the assumptions of kinetic theory, to determine the average force on the container wall… Which is what I think yall were playing around with for what I quoted above. But there is no need to do that.  Not when you can measure pressure directly.

Then to actually figure out velocity of the projectile, you integrate the pressure change over the expansion change in the barrel volume. Bullet mass/size is known, so now you have total force and how it changed as the bullet moved down the barrel.  It's basically the exact same things as standard powdered velocity calculations, except the area under the curve looks different as it isn't an impulse. Solving for velocity is then easy:

Based on what you said about 4k psi, 2162 ft/s. If I assume the barrel is .177, length 20 in, 6.7 grain. I get a muzzle velocity of ~2140 ft/s.  When taking into account shot/shot variance, or slight set up changes (slight difference in grain, barrel, pressure, measured velocity) – my calculated result is the same as what the measured result of the setup is.

There isn't a need for expensive test equipment, when the defining equations for something like this are well known.

It is not like the confined gasses are undergoing a lot of chemical reactions like smokeless powder products.

The ideal gas law does start to deviate at high enough pressures.

Here is a link to the thread in question.  Sonic Choking?

The thing about measuring pressure is that it is only done before the pellet moves.  There is no data (that anyone can find) for the pressure drop in the barrel with respect to either time or distance.  The various assumptions and mathematical equations have been not too satisfactory in predicting muzzle velocity.  

The three basic schools of thought are:  F = mA with assumptions on pressure changes and including the mass of the air column behind the pellet; Energy of the gas imparting energy to the pellet; and molecular speed of the air column.  

Various gas equations have been tried but none quite predict actual results without fudge factors.  

This is why I want to measure the pressure with respect to time at discrete points in the barrel.  This would at least give something to model the actual pressure behind the pellet.  

By all means, if you can add constructive stuff to that thread, please do.

This is where OP has me confused. But yes, it is much more complex with powders, different impulse lengths, intensities, and heats. All of which makes using ideal gas law, for powders, a harder task. Which is why you usually see more of a generalized "rule of thumb."

But in this case, it's a known pressure and volume. If you wanted to use ideal, you could. But there is no need when it can be directly measured. It really is as simple as integrating pressure over barrel length. You've got all your forces and all your masses. Done.

Thanks, imma look through there to see what all assumptions are being made.

I think I found the problem. No one had been integrating the pressure change over the barrel length. It wasn't even mentioned until yesterday in that thread.

The majority of the calculations that I saw were all linear.

Using lyod's values from his shot on the 29 apr - I again got a calculated value (~2300 ft/sec) damn close to what the empirically measured value was. Pretty sure my difference is only due to initial chamber volume difference and I'm not taking any frictional forces into account.  

Pretty sure that is only for slight pressure changes, internal to the media itself. ei - In the pressurized chamber, the wave front of slight changes is material limited. However, large pressure changes or anything external; and that doesn't apply.  -eta- I double checked on this, my explanation isn't 100% technically accurate. A large pressure change (Ideal gas law) will increase the temperature, thus altering maximum wave front propagation. So, the wave front speed is different by virtue of pressure heating the gas.

So, two buildings down from me they use compressed air to propel projectiles ~5000 ft/s to test projectile designs. I pulled that thread, as I work with people that have worked there and on various projectile systems. What they have is similar to a light-gas gun system. Essentially just a spring piston, but without the hydrogen as they aren't shooting as fast as what is typically seen in a gas gun setup.  
 
For the airgun folks, having a high pressure chamber in tandem with the dump valve directly in front of a step down pressure coupler should allow for some pretty fast speeds; sans going to a spring piston system.

After all, powder projectiles are gas driven. It's simply a different generating source.   In addition to the outgassing, it is super heated.




Having said all that... calculating maximum velocity would be done exactly the same way I did it before. Wall pressure integrated over barrel volume (as that accounts for the above) to get total force applied.


I will however note, my engineering degrees are in EE.



Right or wrong, I love threads that actually talk math. It's a good thing this wasnt posted in GD, half the replies would have been "satan limits projectile velocity to test your faith."

I updated what I posted above, with a better explanation. Calculation would be done the same way.


For what the OP was wanting to do, a single strain gauge or pressure transducer would do the trick. You want to model the pressure over time curve (only need a single measurement location, as all other locations can simple be calculated from that)- as that will give you more accurate calculations. It should be different with the variously sized dump valves.


ETA - having popped into some other threads and looking at what all they are doing. I've got two main comments. 1: To simply get closer calculations in general, they need to start integrating. 2: For moving past mach 1, the guns need to use a better pressure release system than a simple dump valve and use higher pressure in the charge cylinder.  That might not be possible if wanting to keep the gun highly portable.  I've mainly been thinking about this as a lab type setting.

Yeah, I guess I didn't realize those dudes were trying to do this by essentially lightly modifying existing guns...

I very much was of a lab mindset. I used to work in one of the National Research labs, and like I said above; I can throw a rock and hit dudes going super fast on air. But a spring piston with a burst valve bears little resemblance to what they are working on.  


So... good luck .

Part of the whole discussion is that the pressure decrease may not be a simple integration with respect to the remaining pressure inside the reservoir.  This is the whole thing about sonic choking.  The idea is that the pressure, for a very large reservoir would be relatively constant up to some point in the barrel just behind the pellet. After that point the pressure behind the pellet then drops while the constant pressure column continues at some constant speed.  

Now the reason for the multiple strain gauges is to confirm or disprove various hypotheses.

Unless one can show that the integration is relatively straight forward, and that it does a good job of modeling real world results there is going to be doubt as to the validity of the mathematical model.

To clarify what is going on here, there are two different kinds of airgun power-plants.  The first is a chamber filled with atmospheric air that is suddenly compressed by a spring.  This is the usual break barrel, side lever, or under lever air gun that most folks are familiar with.  The second type comes in many flavors.  However, these rely on a pre compressed volume of gas ( usually air or CO₂) that is released by a valve of some sort into the barrel just behind the pellet.  

This sonic choking idea relates to the latter type.  A dump valve with the pressure right behind the pellet with little to none of an air passageway is the purest form of the pre charged pneumatic.

For choked flow, necking down to match bore size from a larger volume = no longer an issue. Or... if using a choke plate/necking smaller than bore; you can simply increase pressure to overcome choke conditions. It's going smaller than the upstream volume, then suddenly expanding that causes the issue. Just dont do that. Neck down to the same bore. Or if the "pinch" is for some reason required (maybe the valve is there), then increase pressure.

Now... not knowing the exact setup in an air rifle... But necking down to a smaller diameter than the barrel will cause problems or using an orifice place can cause problems. However, they can both be overcome with higher pressure... You take an efficiency hit though.  Not knowing how these guns are set up... to gain the most efficiency and speed they should be necked down to bore size to avoid any of the choking issues. At that point, the only place you would see choking, would be at the end of the barrel. But at that point the projectile is already gone and it doesnt matter any more.

Is there a particular reason that airgun folks are using orifice plates or are necking down to below bore size? Also, can the charge chambers not hold higher pressure? Wondering if there is some particular reason the highest pressure tried is only ~4k psi?

The pressure limits are based a lot on how much a human can inflate the main reservoir with a hand pump (in a reasonable amount of time), or where one can get the charging air cylinders filled.  The SCBA cylinder (one of the common ones) can only be filled to 4500 psi. Finding places to fill to this pressure is not available in many parts of the country.  I'm in one of those.  The second option is to use a SCUBA tank which are typically filled to only 3000 psi.  The highest pressure commercially available cylinder is 6000 psi nitrogen. The rental / leasing fees typically make this a rather expensive option, unless one shoots many thousand of rounds per month.

If one fills the rifle to the pressure of the charging cylinder, then he must get the cylinder filled much more often.  This requires a compressor at home, or a trip to the dive shop.  In my case, since I use a SCUBA tank at 3000 psi, I fill my guns to 2200, and 2450 psi.  This allows several fills before I have to make a 70 mile round trip to the dive shop.  

Now as far as efficiency, there is the number of shots per fill in the cylinder that many worry about.  For that, it is found that restricting the transfer port in a standard PCP with a "second order knock open valve"  allows many more shots with no significant change in velocity---up to a point.

The typical PCP has a high pressure storage reservoir sealed to the barrel by a poppet valve.   A heavy hammer strikes the valve to momentarily open the valve.  This releases a relatively consistent puff of high pressure air.  In a well tuned gun, as the pressure reduces in the reservoir, the valve opens just a little bit more, which keeps the velocity pretty consistent for the dropping pressure. The peak velocity occurs in the middle of the shot string which makes a bell curve when plotted on a velocity/shot count graph.  By choking the transfer port, the bell curve is flattened and becomes wider.

Most PCP (pre-charged pneumatic) designs position the valve in the high pressure tube that is mounted below the barrel.  The puff of air, then, makes a U-turn to proceed down the barrel.  A few position the valve directly behind the barrel, and these have the high pressure cylinder as part of the stock.  One improvement on the higher end guns is the addition of a regulator.  This allows for a second "medium" pressure chamber in front of the valve which gives a much more consistent shot string.  

The whole high velocity thing is about one individual who proposed some time back that there the maximum velocity that a PCP gun can achieve is the rms speed of the air molecules.  That original ceiling has been broken quite resoundingly.  Now the point is to find out two things.  First, is there a limit to the velocity?  And second, if so, what is it?  

From a practical standpoint, shooting an airgun at more than about 900 ft/s is really not worth the effort.  With the diabolo pellet velocities above that will make the pellet trajectory go haywire.  Also, wind drift is minimized somewhere between 800 and 850 ft/s.  This goes for cast bullets as well which are being used in big bore air rifles.   (Now if a cast rifle bullet can be pushed out at 2400 ft/s, then this would be quite interesting.  It may be achievable with hydrogen, but most folks are not going to want to carry a compressed cylinder of hydrogen in their car for a trip to the range.)

The pressure changes are large and significant.

As soon as the projectile has swept out its own length the volume has doubled.
Pressure (absent a burning propellant) is half.

Barrel pressure (even with a propellant) is dropping like a rock as the volume behind the bullet increases.
It is a closed system at that point.

The velocity of a projectile is the integral of the pressure behind it over barrel length.

That assumes it is an instantaneous impulse, with starting volume only equal to that of the projectile itself. Depending on that initial chamber volume, the projectile moving only its length could result in large or negligible pressure change. I don't know that starting volume though. Once back in the office I'll check to see what I used for an initial chamber volume.



Yes, which is why I said the folks predicting speed on the airgun forum need to integrate pressure change over the length of the bore. Some of them are using initial pressure and just calling it good, assuming it's linear. No accounting for pressure change or impulse shape.



Correct. I think we're both saying the same thing.

I'm not completely sure that this is true.  Consider a very large high pressure chamber, say a bank of SCBA 88 cubic foot cylinders (at 4500 psi) with the valves fully open.  (The size of the SCBA cylinders is measured by the number of cubic feet of atmospheric air that it holds.)  These multiple cylinders in turn feed (with large pipes) the somewhat smaller reservoir that contains the valve.  The valve is designed to quickly fully open and  then close just as the pellet leaves the barrel.  This gives the high pressure chamber an essentially infinite volume, and the maximum possible force on the pellet for its whole length of travel down the barrel.

Now consider a very long barrel, say 12 ft long and .25 caliber shooting a pellet of 25 grains or 1.11 x 10 ^-4 slugs.  There is no integration required behind the pellet here since the pressure will be essentially constant (This is where the principle assumption fails in my opinion).  Integration may be warranted  for the slight mass reduction of the air in front of the pellet.  Friction would also be essentially constant.

That gives us a force on the base of the pellet of about 221 lbs.  Assume a friction force of six pounds (probably on the high side), that leaves us with 215 pounds of force acting on the base of the pellet for the whole 12 ft. length.

F = mA or A = F/m leaves us with a constant acceleration of nearly 1,937, 000 ft/s².  Assuming constant acceleration, that gives a muzzle velocity of  about 11,620,000 ft/s.  That I do not believe.

Your math is incorrect.

In this example, its very straight forward as the pressure is constant. Velocity is simply (((2*pressure*bore_crossesction*144)/mass)^.5)/12, with velocity known, energy (in ft-lbs) is (mass*velocity^2)/2, momentum is just mass* velocity.

Vp = ~1990 ft/s
E = 221 ft-lbs
F = 0.22 lb-sec

It looks like you are conflating momentum with energy, then dividing energy by mass again.

If the pressure does not vary significantly you will just up integrating a constant.
Still works just fine.

Pressure over the time it is applied.

I was not using energy at all.

What I did was take  F = mA, where F is the force applied, m is the mass, and A is the acceleration.  

F = (P x a) - f , where P is the pressure pushing on the pellet, and a is cross sectional area,  and f is the friction.

The mass of the pellet is 25 grains.  Converting that to slugs:  7000 grains in a pound, and 32.174 pounds in a slug, gives us 1.11 x 10 ^-4 slugs.

Solving for A gives us: A = F / m = (P x a - f) / m

A = (( 4500 pounds / in² x .049 in²) - 6 pounds ) / 1.11 x 10 ^-4 pounds / ft/s²  

A ~ 1,932,000 ft/s²

What you did, then, was take the energy of force acting through a displacement, and using E = 1/2 m V², and solved for V.  You also converted length to inches instead of keeping with feet.

What I get when I do that without converting to inches is as follows:

E = F x d = 1/2 x m x V²,

where E is energy, F is the sum of the forces acting on the pellet, d is the distance or length of the barrel, m is mass of the pellet in slugs.

V² = 2 E / m = 2 x 215 pounds x 12 ft / 1.11 x 10 ^-4 slugs

V = (2 E / m = 2 x 215 pounds x 12 ft / 1.11 x 10 ^-4 slugs)^.5 = 6818 ft/s  ETA:  This is exactly what is derived in the chart in the next post: V = (2PAL / m) ^1/2

I still have a hard time believing this.  

Oh, and I get the ~1900 ft/s when I use a 12 inch barrel instead of a 12 ft one.

ETA:  I see  what my big mistake was in the first case.  I was trying to use distance traveled instead of time.  Just like mentioned above.  When I solved for time:

d / A = t²  => t = .002489

Then solving again for velocity by putting in the time:  V= 1/2 x A x t, I now get a reasonable 2400 ft/s.  Note:  A few microseconds makes a bit of a difference in final velocity.

ArimoDave,  

Your answer is many, many, many orders of magnitude from being correct.

Please review the following (No need for integrating Phat, as it is constant. It's simply P.):

Remember, make sure that all of your units are in the correct form for the equation you are using. You cant be mixing and matching. Also order of operations when solving.



image source: http://closefocusresearch.com/assets/images/ballistic_calculation_derivation.gif

I thought I was being careful about units.  The force is pressure in pounds per square inch, while the area is in square inches.  These cancel leaving pounds.  At this point, I can dispense with using inches.

Now energy is in terms of foot pounds, and the distance in feet.  So I kept with those.  Mass in the USCS is slugs, so I converted the pellet weight in grains (while on the earth) to slugs.  M =  pellet wt in grains / 7000 grains per pound / 32.174 pounds per slug.

I realize that since the pressure is constant---the area too--- and we don't need to integrate it. (By the way, my preferred method of integration from point A to B is to plot the curve; cut out a rectangle; weigh the piece of paper and determine its density per area; cut out the curve and weigh what is below the curve.  I then calculate the area based on that.)

I'm still trying to find my mistakes, since it has been a very long time since I have done much of this stuff.

However, that chart is useful, and I thank you for posting it.

You were working everything with a base unit of in (hint: psi), but when calculating velocity you switched to ft. Calculate everything through first, then convert at the end.

So to find velocity you should have used 144 (12 ft), then at the very end convert back to ft.

Oh, wait one minute. I started to do this, then remembered that mass is in slugs which is pound-feet/s².

Dividing that by 12 with that value for mass is not the right thing to do.  Remember that the inch units cancelled when calculating the force on the pellet.

Now let me try this using feet as my base unit with respect to pressure and area.   The pressure will be 4500 psi x 144 square inches per square foot = 64800 pounds per square foot.  Likewise, the area will be (.125 inches / 12 inches per foot)² x pi = 3.4088 x 10^-4 square feet.  Multiplying these together gets me the same ~221 pounds of force.  Subtracting that 6 pounds for friction still results in the 215 pounds I've been using.  

Vp = (2*4500*.05*144)/0.00011)^.5  then divide by 12 to get back to ft. - With velocity, you can get whatever energy measurement you want to you. Now that we have velocity, we can get whatever measurement we want. Joules, ft-lbs, newtons, momentum, etc.

When converting grain to slug, you are already in the correct units. By using slug, and psi and 144 everything is in the correct unit. Then, once finished - convert back to feet.

What about barrel friction of the pellet in the bore?  What about head loss as the gas flows down the barrel?  What about the pressure change as the gases expand down the barrel?  That pressure is not constant unless you have an infinite reservoir and friction-less gases flowing into the barrel.  As the pressure drop so does the temperature and that increases your pressure drop.  It's going to be modeled best with an adiabatic model.  Then as you approach the speed of sound (for the gas conditions in the bore) you will have to deal with shock wave propagation in the barrel that can choke flow.  Most of its easy to model until you get to the shock wave stuff.  The pressure at the base of the pellet will not be the pressure in the reservoir except at the very beginning.  -Rambling.

You have touched on most of the things about why I wanted to measure the pressure in the barrel at various places.  It is the angst in the thread and place in question.

outofstep,

You just cannot divide the mass by 12 unless you convert it first to pound- inches per second squared.  

My results become the same when you convert it first.

What you propose is like using cm for everything except leaving the mass in Kg.  The Kg is N-m per second squared.  If you fail to convert the mass to N-cm per second squared, you will be off by a bunch.  

Again, if you complete what I did using feet as the base unit instead of inches even for pressure and area the results will be the same as mine.  Also, if you convert every thing to newtons and meters, run the calculations,  then convert back to ft/s, it will be the same as mine.  

Converting the slug into pound-inches per second squared means multiplying it by 12 first.  That puts a 12 in the denominator, which cancels with the 144 to leave 12.  Then, we are right back where we started.  

V = (2 P A 12L / 12 m) ^ 1/2

You are not dividing just the mass by 12, you are dividing the final velocity by 12  to get back to ft - because the answer is in inches based on the units you plugged in.

Again, order of operations and making sure you are using the correct units for the equation.

 
You going to need to sample you strain gauges faster than 20ks/s  if you have 5 sensor spread along a 20 inch barrel (~5 inches apart) and expect a ~2000fps muzzle velocity then you have a touch over .2 msec between the last two sensors.  At 20ks/s you only get four reading between the fourth and fifth sensors.  Again just quick and dirty calculations.  You would really need to get to 100ks/s or more to get decent data and faster in this case is almost always better.




Measurement computing has some cost effective USB driven DAQ systems.  Fore $199 you can get 500ks/s aggregate sample rate.  If you have five sensor, thus five channels you could get 100ks/s/ch.




http://www.mccdaq.com/usb-data-acquisition/USB-204.aspx




You would still need to get some strain gauge amplifiers

ETA Maybe these?


http://www.industrologic.com/sgaudesc.htm

Again, order of operations and making sure you are using the correct units for the equation.

That is the whole thing.

When you divide the result you are dividing the mass part which was not converted the way the equation was written. If the equation looked like this:

V = ((2 P A L / 12) / m ) ^ 0.5

then I would agree.  

Another way to write the equation you are referring to is this:

V = (((2 P A L )/ 12) x ((1 / m) / 12))) ^ 0.5

In this latter distributed denominator form, you can see that you are dividing the mass without correcting its units beforehand.

500 kS/s for a 1-2 mSec event that only needs ~ 10 samples to more than adequately get the slope...


OP, basicly ANY microcontroller from the past 2 decades can sample at way more resolution than you will ever need for this project.

mcb,

I found a a couple of chips that I think I can do a bit better than that. Both handle simultaneous sampling on up to six channels: one at 600 kS/s the other at 1500 kS/s.  I will need a bunch of other stuff like the appropriate op-amps,  resistors, capacitors, and a suitable bread board and adapters to build it.

The two chips in question are:  LTC 1408-12 and LTC 2351-12.


Oh, and the barrel is not going to be a measly 20 inches, its going to be 45 inches.  If I had a longer piece on hand it would be even longer.

... if using slugs and inches to get to a final solution in ft/s

it is (((2*PSI*cross_section*barrel_length)/mass)^.5)/12

cross section is in in^2, length is in inches, mass is in slugs.  If you dont want to use slugs then you convert everything else to match.

1 slug = 1 (lb-s^2)/ft  - you cant just arbitrarily divide it by 12.


If it helps, 1 slug = 143 N. Work everything through in the proper SI, then convert at the end. That might be easier for you. Or just use what I posted above, it is correct.  

One last time with units in place of letter values:

First, however, I need to correct what I have written was the units of mass.  F=mA  =>  m=F/A   or m = (lbs) / (ft/s² )

Now we take our equation and substitute units instead:

V² = (2 P A L) / m  = ((Unitless)  (Lbs/in²) (in²) (ft)) / ((lbs) / (ft/s²)) = ft²/s²

Taking the square root then leaves us with the  units ft/s as required.

With your conversion at the end results with this before the conversion:

V² = (2 P A L) / m  = ((Unitless)  (Lbs/in²) (in²) (in)) / ((lbs) / (ft/s²)) = ((in) (ft))/s²

Taking the square root leaves you with ((in^1/2) (ft^1/2)) / (s)  which is not a good velocity.  There are mixed units in this case.

By the way,

I thinking about it a bit more, my high velocity figure must be closer to what may be achieved if there were no losses rather than the lower ~2000 ft/s.  With my .25 caliber Armada, I can get 1000 ft/s with a .25 grain pellet (with a heavy enough hammer spring) at 2500 psi.  It has only an 18" barrel.  

Still, I think there are significant losses in the air flow, and that is what I want to measure.   I'd like to know by how much, when, and where the pressure drops behind the pellet.

... dude

So, go check out the wiki page on Slug (mass) . You will see how it relates to F = m*a. Remember, acceleration is the 1st derivative of speed.

You can't take the square root of the unit description- that is nonsensical. For example;  sqrt(25 ft/s^2) absolutely does not equal 5 (ft^.5)/(s)  

When everything is in the correct form, if it helps, pretend the unit description isnt there. Just focus on the math.   But you have to know, what you want the final units in and what the standard units that the formula uses are.

If you take what I wrote about the units of the slug, you will find that it can be rewritten in the same form as on that Wiki page.  I said it is:

m = (lbs) / ( ft/s²)  

Wnen you have a fraction in the denominator, then that part of the bottom fraction can then go to the numerator.  "Invert and multiply"

Ergo, m = (lbs) (s²) / ft.


The units certainly do multiply, divide, square, and so-on.  

Consider one foot being multiplied by another foot.  You get a square foot.  Multiply it again by a foot and you get a cubic foot.  If you multiply it by a scaler (unitless) you just get that multiple of what you started with.  So If you multiply 2 ft by 4 you get 8 linear ft.  If you multiply a foot by an inch, you get an inch wide strip one foot long.  Multiply that by one cm, you get a bar that is, one cm wide, by one inch thick, by one foot long.  

When you divide a distance (ft) by time (s) you get speed.  Divide it again by time (s), and you get acceleration.  The old way to express acceleration was: feet per second per second.  

Taking roots does the same operation to the units as it does the values.  If you take the square root of a square you wind back to something linear.  So, if you take the square root of the inch-foot you wind up with the square root of 12 inches.   This is the missing piece in your equation.  Instead of dividing by 12. you need to divide by 12^1/2 (square root of 12).

There is so much wrong here, I don't even know where to begin...


I'm not trying to be a dick, but where was the last math/science class you left off on? Its very clear now that you haven't had physics or calculus. I'm just trying to get a baseline, so I can explain it better to you.

https://www.khanacademy.org/math/algebra/units-in-modeling/intro-to-dimensional-analysis/v/dimensional-analysis-units-algebraically

Yes you can treat units algebraically, in fact you should.

One final time on the calculation except this time all in SI units.

V² = (2 P A L) / m

P = 31026408 N / m²

A = 3.167 x 10^-5 m²

L = 3.66 m

m = .0016 Kg

Plugging those in, and taking the square root leaves one with 2120 m/s.

Converting that to ft/s using my TI-86 calculator is 6956 ft/s.  

I think we are now done with this part of the exercise.

It is now time to go back to the original intent: strain gauge DAQ.

Dude... he even says in the video it isnt applicable once you get to higher level math/physics/engineering.

What he discussed was the most basic of unit conversions. But you need to understand, all of those conversions MUST described the EXACT same thing.  12 inches is the exact same thing as a 1 ft.  10 km/h is the exact same thing as 9.11 ft/s. You can convert between whatever standard of measurement all the live long day. But ONLY if the BOTH describe the EXACT same thing.

You CAN NOT randomly convert between measurements that describe very different events.  Volumes, forces, rates of changes dont work that way.


If I travel 10 ft in 2 second with constant velocity, I was traveling at 5 ft/s... but my acceleration was ZERO. It was not 2.5 ft/s^2 (dividing by time again).  

If I have something with dimension 3.92ft x 15ft x 17ft it giving me a volume of 1000 ft^3... WTF does 10ft represent (just cube rooting everything)? Nothing. It represents nothing because it doesnt work like that.  The cube root of 1000 ft^3 is 10 ft^3.  However, I can convert that 1000 ft^3 into cm^3 or yards^3 or whatever I want, because it describes the same thing. This is literally the most basics of measurements.




As to OP, I answered that already. Make sure you sample at Nyquist rate. Anything in the past 2 decades is capable of this. Shouldnt cost you more than 20 bucks for everything.  Any arduino is more than capable of this, strain gauges are dirt cheap.

I can confirm that number for velocity.

Have you taken a look at what kind of sensitivity you need out of your strain gages to accurately detect pressures in the barrel?  The best foil gages/amplifiers have a noise floor of about 1 microstrain (depending on who you ask).  Rule of thumb for a good measurement is an SNR of about 10.  Say then that your minimum resolvable signal is about 10 microstrain.

I don't know how thick your barrel is, but using the thin cylinder hoop strain calculation in "Mechanics of Materials" for a steel wall thickness of 0.08" (total guess based on my Beeman), I.D. of 0.25" with an internal pressure of 4500 psi, I get about 200 hoop microstrain (and stresses well below the UTS of the steel).  Waving my hands around, this gives your minimum consistently resolvable pressure to be about 230 psi.  This number gets bigger as your barrel thickness and/or noise floor increase.

Maximizing precision would require determining the minimum safe thickness for the barrel (as nervous as that makes me ).  It all depends on what you hope to get from your measurements.

ETA:  You can always double the sensitivity by using 2 gages in an additive half-bridge configuration at each axial location on the barrel.  I can't remember, but this may introduce a common-mode noise issue since the gages need to be on opposite sides of the Wheatstone bridge (since they're both undergoing elongation in your configuration).  Somebody else may know.

The Arduino could maybe sample at the right rate (does that ADC have a fast mode?  Can't remember).  Problem with those is that the noise floor, especially when running fast, is absolutely balls, and is compounded by the ADC's 12 10 bit resolution.

I've used some cheap but high-quality IC bridge drivers from TI before, but I can't for the life of me remember what they were.  Ran em off the 5v bus on a hacked-up USB joystick .  Still, even those things have a shitload of internal filtering to get the noise down.  That might screw with the sampling rates he needs.

If you use kg, the velocity number is incorrect. Keep track of what units are and what they are measuring. Using N will get the answer in m/s - then convert to ft/s.

OP is going to hurt himself or someone else.

Even total junk arduinos can sample at ~ 10 kHz. 10 bits for ~ 4500 psi... yeah that, more than enough resolution. That's 4 psi quantization bins.

Or he could snag a raspberry pi or a yard sale Oscope. I would personaly go Oscope - there are some badassed solutions that piggyback off of your Android based phone that give Mhz sampling rates and the full capabilities of a lab level scope.

His velocity number is correct given the numbers he shows in the quoted post.  I'm not quite sure what you're trying to say here- kg is the SI unit of mass. His m/s -> ft/s conversion is correct also.

It's not the psi quantization that's the issue, I'm worried about the quantization of the small voltage that he will see across the Wheatstone bridge.  He's going to be measuring very small strain, which without some moderately thoughtful amplification and use of a proper reference voltage on the ADC will end up causing problems.

Reading through the datasheet again, Atmel quotes a +/- 2 LSB accuracy and 1 LSB integral nonlinearity on their ADC.  My suspicion is that this applies only when using the 5v Vcc supply as reference.  Atmel even recommends a bunch of wacky shit in the datasheet to improve SNR (shutting down the CPU during a conversion, etc).

Converting to SI, N needs to be used, not kg. All units need to be in the correct form.

When using psi and slug the velocity will be in inches, which is why you need to divide by 12 to get to ft/s instead of in/s.  slugs are in lb ft/s^2, it should be converted into newtons as it is in kg m/s^2.  He was at least able to convert the pressures and lengths correctly (you can straight google calculators for that), but then stopped after that and decided getting the slug conversion wasn't important. Slugs goes to N to preserve correct units - you then get an answer in m/s. That answer matches what I original posted it to be.

OP keeps doing stuff that is analogous to "10 minutes * 1.5 blueberries/bucket = 15 chimichangas s^2  - Yall are dumb, clearly 10*1.5 = 15"  ignoring the fact that the equation itself is nonsensical because the units are all fucked up.