dagabu Posted August 31, 2012 Posted August 31, 2012 Who do I send $ to to get one of these test rigs ASAP!?!?! You can send a PM to Peret for more information but I think a few more weeks to iron out the last of the bugs and then have Peret take a rig to Do It for another trial period is in order first. -dag
WonderBoy Posted September 4, 2012 Posted September 4, 2012 Okay, so after reviewing footage, I believe ours were run numbers 16, 17, and 18. We shot 4 motors, but run 19 does not seem to match up. Could the 19th slot be continually overwritten once the memory is full? I remember seeing at least 1 "demo" after we shot our motors on the rig; run 19 looks to me like it could be recording a hand pressing on it. Here are the motors, all are 1lb whistle type motors: http://www.youtube.com/watch?v=34xpzbelg_s&feature=plcp WB
Peret Posted September 5, 2012 Author Posted September 5, 2012 Okay, so after reviewing footage, I believe ours were run numbers 16, 17, and 18. We shot 4 motors, but run 19 does not seem to match up. Could the 19th slot be continually overwritten once the memory is full? I remember seeing at least 1 "demo" after we shot our motors on the rig; run 19 looks to me like it could be recording a hand pressing on it. The rig has a non-resettable run counter that increments after every test. Since the counter is at 19, only 19 tests were made altogether - the 19th being the one Dag did to make sure the rig would record (so you are right, that was a hand press). This recorded over the first run, run number 1. With the version of software used at PGI, no memory would ever be overwritten. When the 18 memories were full the results would have been lost. However, the run counter would still increment, so if for example you did 22 tests without clearing memory, 18 would be saved, the last four would be lost and the run counter would be at 22. That being the case, either your runs were 15, 16, 17 and 18, or for one of the tests the rig wasn't in record mode.
WonderBoy Posted September 5, 2012 Posted September 5, 2012 Hmmmm is there any sort of auto off timer after pushing the button? If you remember, we were waiting for the multi-color light before lighting the motor. The original video shows me pushing the button on all but the first, on the first I pushed the button before the camera started rolling. So I am fairly certain it should have been in run mode. I keep wanting to think that we started on 16 because it looks right. The first two motors we lit "chuffed" as they entered the delay section, giving that jagged appearance, whereas the third and fourth motors had smooth transitions. Nater's motors also had smooth transitions into the delay section and were shot right before these four. Also, this whole time I have been reading "burn time" as "thrust time", so I was not including the delay section when timing my motors. Soooo I am probably completely wrong. I think it is time for a beer now WB
dagabu Posted September 5, 2012 Posted September 5, 2012 All in all, it looks like the test rig is ready for prime time and I am ready to start some real production after Do-It is completed and Peret has proof of concept and all the debugging done. I cant wait to get a completed rig back here so I can test motors next month. I will be doing end burners (15 or more seconds of constant burn), core burners and 3# whistles that I expect to really push the test rig. There will also be some 5/8" strobes tested as well so I can prove/test the concept of the thrust phase if any with strobe pulses. -dag
nater Posted September 5, 2012 Posted September 5, 2012 Also, this whole time I have been reading "burn time" as "thrust time", so I was not including the delay section when timing my motors. So there is enough thrust during the delay phase for it to record something? When comparing the curves to my videos, I was only looking at what I thought was the thrust phase as well. All in all, it looks like the test rig is ready for prime time and I am ready to start some real production after Do-It is completed and Peret has proof of concept and all the debugging done. I cant wait to get a completed rig back here so I can test motors next month.Great news! I can't wait either.
Peret Posted September 5, 2012 Author Posted September 5, 2012 The sensitivity of the rig with a 100kg load cell is better than 2 grams (1/16 ounce), so it can certainly record thrust during the delay phase, but it is likely to be insignificant compared to the thrust phase and be confused by the loss of fuel weight.
nater Posted September 5, 2012 Posted September 5, 2012 The sensitivity of the rig with a 100kg load cell is better than 2 grams (1/16 ounce), so it can certainly record thrust during the delay phase, but it is likely to be insignificant compared to the thrust phase and be confused by the loss of fuel weight. So the graphs would be showing the thrust phase only?
Peret Posted September 5, 2012 Author Posted September 5, 2012 So the graphs would be showing the thrust phase only?The full graph shows everything. The zoom graphs, the ones Dagabu posted, just show the thrust phase - well more precisely, they show that portion of the curve where the thrust is more than 5% of the peak thrust, to the nearest whole second either side. This is the part that's integrated and used to calculate the impulse. The integral is the area under the curve, so 1/10th second at 25 pounds is worth 10 seconds at 4 ounces - the delay phase isn't going to add any significant amount to the impulse. The analysis program can show either the full 11 second trace or just the impulse. I was in a hurry or I'd have made it more sophisticated, but I figure the sophisticated stuff is better done by opening the file in Excel. Going a bit further into the math, the impulse has units of momentum. If you divide it by the weight of the rocket plus header, you have the speed after the thrust phase, from which you can work out how high it will go. If you look at run 17 for example, the impulse is 382 lb.ft/sec, so a 2 pound header will be coasting upward at close to 200 ft/sec after the half-second thrust phase. It will be retarded by gravity at 32 ft/sec^2, so by Newton's laws of motion, v = u + ft (u initial velocity, v final velocity, f acceleration, t time) neglecting air resistance u = 200, f = -32, and solving for v = 0 (apogee) 0 = 200 - 32t t = 200/32 = approx 6 seconds After 6 seconds the rocket will be stationary in the air at the highest point. s = 1/2 f t^2 (s distance) s = 32*6*6/2 = 576 feet When I have a moment I'll incorporate that math into the analysis program.
dagabu Posted September 5, 2012 Posted September 5, 2012 (edited) The full graph shows everything. The zoom graphs, the ones Dagabu posted, just show the thrust phase - well more precisely, they show that portion of the curve where the thrust is more than 5% of the peak thrust, to the nearest whole second either side. This is the part that's integrated and used to calculate the impulse. The integral is the area under the curve, so 1/10th second at 25 pounds is worth 10 seconds at 4 ounces - the delay phase isn't going to add any significant amount to the impulse. The analysis program can show either the full 11 second trace or just the impulse. I was in a hurry or I'd have made it more sophisticated, but I figure the sophisticated stuff is better done by opening the file in Excel. Going a bit further into the math, the impulse has units of momentum. If you divide it by the weight of the rocket plus header, you have the speed after the thrust phase, from which you can work out how high it will go. If you look at run 17 for example, the impulse is 382 lb.ft/sec, so a 2 pound header will be coasting upward at close to 200 ft/sec after the half-second thrust phase. It will be retarded by gravity at 32 ft/sec^2, so by Newton's laws of motion, v = u + ft (u initial velocity, v final velocity, f acceleration, t time) neglecting air resistance u = 200, f = -32, and solving for v = 0 (apogee) 0 = 200 - 32t t = 200/32 = approx 6 seconds After 6 seconds the rocket will be stationary in the air at the highest point. s = 1/2 f t^2 (s distance) s = 32*6*6/2 = 576 feet When I have a moment I'll incorporate that math into the analysis program. Bravo! The apogee is probably the single most important piece of information I can use from the curves and the reason why this has been such an exiting project to be part of. Imagine if you will, making a girendola and weighing the frame plus heading, add the weight of the motors and do a test curve of a motor. Theoretically, you could have a dola rise slowly, VERY slowly and then transition to a faster fuel and then stall, rise, stall rise and so on. Much like Tony S dolas but without all of the hundreds of test motors. -dag Edited September 5, 2012 by dagabu
Peret Posted September 12, 2012 Author Posted September 12, 2012 I've been doing some rocket science, looking for a better way to present the data. Here it is presented in height versus time format for runs 15 to 18 that were depicted as force curves a couple of pages back. These curves give the apogee height and time to apogee, so in conjuction with the impulse graph that shows the burn time you can use it to find the delay needed. To do this requires some assumptions, which are as follows: Each motor has a 5 inch ball headerThe aerodynamic drag coefficient is 0.7The gross weight is 1 kilogram The aerodynamic drag coefficient is somewhat guesswork and my guess can only be confirmed by timing the flight with an actual rocket, but pencil and paper calculations indicate the heights and flight times are "sensible", that is, probably within 10%. I will add a "Rocket Science" button to the PC program to allow the weight, size, shape parameters to be edited and present the output in various ways.
dagabu Posted September 13, 2012 Posted September 13, 2012 Sweet-sassy-molassy! That is HUGE Peret. Thanks man. -dag
Peret Posted October 3, 2012 Author Posted October 3, 2012 Very well, though I only had one participant, Mike Garrett. The trace was identical to those he took from similar motors on a rig of the traditional type and is posted on Passfire. Judging from the very high proportion of rockets that rose 20 feet and power-dived, or in some cases never lifted out of the launch tube, a lot of other people would have done well to avail themselves of the testing opportunity.
Peret Posted October 9, 2012 Author Posted October 9, 2012 I was hoping we'd be shipping this week, but Dag's accident has put a crimp in that plan. 1
dagabu Posted October 9, 2012 Posted October 9, 2012 Sorry everyone, I have surgery tomorrow and will have a good idea of when I will be able to get the completed units out the door on Friday. -dag
tangent Posted December 8, 2012 Posted December 8, 2012 The full graph shows everything. The zoom graphs, the ones Dagabu posted, just show the thrust phase - well more precisely, they show that portion of the curve where the thrust is more than 5% of the peak thrust, to the nearest whole second either side. This is the part that's integrated and used to calculate the impulse. The integral is the area under the curve, so 1/10th second at 25 pounds is worth 10 seconds at 4 ounces - the delay phase isn't going to add any significant amount to the impulse. The analysis program can show either the full 11 second trace or just the impulse. I was in a hurry or I'd have made it more sophisticated, but I figure the sophisticated stuff is better done by opening the file in Excel. Going a bit further into the math, the impulse has units of momentum. If you divide it by the weight of the rocket plus header, you have the speed after the thrust phase, from which you can work out how high it will go. If you look at run 17 for example, the impulse is 382 lb.ft/sec, so a 2 pound header will be coasting upward at close to 200 ft/sec after the half-second thrust phase. It will be retarded by gravity at 32 ft/sec^2, so by Newton's laws of motion, v = u + ft (u initial velocity, v final velocity, f acceleration, t time) neglecting air resistance u = 200, f = -32, and solving for v = 0 (apogee) 0 = 200 - 32t t = 200/32 = approx 6 seconds After 6 seconds the rocket will be stationary in the air at the highest point. s = 1/2 f t^2 (s distance) s = 32*6*6/2 = 576 feet When I have a moment I'll incorporate that math into the analysis program. Hi, noobie here. I haven't seen an update to this thread in about 2 months. If this project is still a go, I'd be interested in one. Please add me to the list. @dag - hope your wing is on the mend! On the math above, is there any way to get similar numbers to figure out apogee for a shell launched from a mortar? thanks, -t
Peret Posted December 8, 2012 Author Posted December 8, 2012 (edited) On the math above, is there any way to get similar numbers to figure out apogee for a shell launched from a mortar?Tangent, I spent quite some time working on that problem a couple of years ago. The problem with shells is knowing how fast they are moving when they leave the mortar - if you know that, the rest follows. For a rocket you can calculate everything if you know the propelling force at any given time, which the rig gives you. I don't know of any way to measure the shell speed. One way to figure it out is to fire a dummy shell, like a baseball, and time it with a stopwatch from launch until it hits the ground again. You have to make some correction for air resistance, which is very difficult, but theoretically you could calculate the speed (and thus the height) working backwards from the time. Ned Gorski uses the approximation that it takes twice as long to fall from apogee as it takes to get there, then uses the Newton formula s = ut + 1/2 ft^2, where f is the acceleration due to gravity, ut is zero, and t is 2/3 the total flight time. Then s is the distance it fell, ie the apogee height. This only works for a small range, since the air resistance is proportional to some power of speed between 2 and 3 and increases sharply with speed. For a slow launch the up/down ratio will be closer to half and half, and for a very fast launch it will be more than 1 to 2. Another way is to fire a live shell, video it, and count frames from the time you see the flash until the time the sound arrives. Sound travels at about 1100 feet per second, so if the film is running at 30 frames/sec the height is about 36 feet per frame. This only gives you the burst height though, not the true apogee. The test stand project is still going, and about to ship, but the first order is sold out. We'll be making more. Edited December 8, 2012 by Peret
tangent Posted December 8, 2012 Posted December 8, 2012 TY! You kind of lost me mid-para, (the math) but I will revisit when more awake. Glad to know the project is still a go! Thanks, -t
tangent Posted December 8, 2012 Posted December 8, 2012 On the test stand project - what are you charging per unit and what kind of timeline are you talking about? thanks, -t
dagabu Posted December 8, 2012 Posted December 8, 2012 The units are in progress as we speak, I should have several ready for shipping very soon but will not give dates until Peret and I have been able to lock some things down. I see no treason why we would not be shipping by the new year. -dag
MrB Posted December 9, 2012 Posted December 9, 2012 Perhaps a stupid question, but how long can a recording session actually be, before it runs out of memory?Also, the selected loadcell has a 2g accuracy? Does the software record start, and end weight, or just weight-loss after the run? (I'm guessing weight loss, since the unit is most likely zeroing it self when the system is armed, yes?) And whats the weight of this thing. I'm thinking shipping to Sweden might be a bitch...B!
Peret Posted December 9, 2012 Author Posted December 9, 2012 Theoretically the recording could run for about three and a half minutes, or seven minutes at slow sample rate. For the application it was designed for, fireworks, the standard recording period is 1 second pre-trigger and 10 seconds post-trigger. This covers pretty much everything I've ever seen, but the software allows this period to be doubled for really long burning motors (20 seconds). It could be reprogrammed for other applications if I was sufficiently motivated. Specs on affordable load cells are a bit vague. They do come with a calibration certificate but I don't put a lot of trust in it, since a lab would charge 10 times the cost of the cell to test it properly. The spec sheet says 2%. I don't know whether that means absolute accuracy, linearity, or some combination. You can calibrate the rig yourself to any degree of absolute accuracy you choose, so it would only be linearity that bothered you. Bear in mind that load cells are mechanical objects and all kinds of mechanical events affect the reading - thermal expansion, hysteresis, internal friction etc. These mostly affect results at the bottom end, so you won't get accurate and repeatable results if you use a 100kg load cell to weigh 10 grams but it will be pretty good in the kilogram range. If you want more low-end accuracy you could use a smaller load cell, but the thrust is so disproportionally huge compared to the static weight that you have to accept some sort of trade-off, and it is after all thrust you want to measure. The internal resolution of the system is 21 bits, about 1 in 2 million, but that's about 50 times better than the load cell is capable of reading. The system zeros itself when it's turned on. When it's armed it makes a note of the start weight and records that, together with the end weight, with the data file. All the taring and corrections are done by the analysis software to calculate the weight of fuel consumed. You can measure this off the rig if necessary for greater accuracy, by weighing the motor before and after burn on a more sensitive scale. The made-up system is pretty heavy, over 5kg. The electronics part weighs next to nothing and costs little to ship, if you were willing to make your own metalwork.
MrB Posted December 9, 2012 Posted December 9, 2012 Thank you for the detailed answer. I'm going to look in to the "making my own" stand, but quite honestly, i think i would like the whole thing, and surface shipping doesn't have to be that expensive. Never cheap, but then, getting the parts together, and making it myself isn't going to be free either, so it might be worth it.B!
tangent Posted December 10, 2012 Posted December 10, 2012 Tangent, I spent quite some time working on that problem a couple of years ago. The problem with shells is knowing how fast they are moving when they leave the mortar - if you know that, the rest follows. For a rocket you can calculate everything if you know the propelling force at any given time, which the rig gives you. I don't know of any way to measure the shell speed. One way to figure it out is to fire a dummy shell, like a baseball, and time it with a stopwatch from launch until it hits the ground again. You have to make some correction for air resistance, which is very difficult, but theoretically you could calculate the speed (and thus the height) working backwards from the time. Ned Gorski uses the approximation that it takes twice as long to fall from apogee as it takes to get there, then uses the Newton formula s = ut + 1/2 ft^2, where f is the acceleration due to gravity, ut is zero, and t is 2/3 the total flight time. Then s is the distance it fell, ie the apogee height. This only works for a small range, since the air resistance is proportional to some power of speed between 2 and 3 and increases sharply with speed. For a slow launch the up/down ratio will be closer to half and half, and for a very fast launch it will be more than 1 to 2. Another way is to fire a live shell, video it, and count frames from the time you see the flash until the time the sound arrives. Sound travels at about 1100 feet per second, so if the film is running at 30 frames/sec the height is about 36 feet per frame. This only gives you the burst height though, not the true apogee. The test stand project is still going, and about to ship, but the first order is sold out. We'll be making more. When kicking this arround, I had a number of thoughts. First off, Skylighter has a couple of charts: The amount of black powder lift needed for single break shells.http://www.skylighter.com/fireworks/help/shell_building_charts_and_tables.asp#lift and Shell Burst Heights.http://www.skylighter.com/fireworks/help/fireworks_display_charts_and_tables.asp#burst However, these would lony be ballpark figures and would change depending on the lift charge, the weight of the shell and shell diameter vs ID of launch tube, etc. To determine velocity, we used to sandwitch a frame with 2 pieces of AL foil seperated by a sheet of paper, and an identical one a foot away and measure the difference in time then the circuites were completed. This could also be done by breaking a light beam (laser or as in alarm systems) effecting a sensor that is normally on. Other approaches from rocketry are embetting some electronics in a shell with eithor GPS or an acellorometer(sp?). One method from, I believe, an older edition of the Handbook of Medle Rocketry involeves mounting an aiming stick attached to a protractor on a tripod, or similar and placing it a known distance from the launch point and recording the angle at apogee. Determining the size/height of things in images is part of the field of photogrammetry. An interesting paper on this subject can be found here:https://www.cia.gov/library/center-for-the-study-of-intelligence/kent-csi/vol5no2/html/v05i2a02p_0001.htm Does that suggest any good approaches to you? As to shipping things to Sweeden, it seems like most of the weight is in a single component that should be able to be obtained locally. thanks for the feedback! -t
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