Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Pneumatic air test force 2
- Thread starter skaboy607
- Start date
LittleInch
Petroleum
Force = pressure x area
Volume of air is not really relevant once it is more than a couple of pipe diameters.
The problem is how will the end cap or blank ( two different things) actually fail? It is a bit far fetched to work on the basis is simultaneous bolt failure, but I guess you could so after that it is Force = mass x acceleration.
However this is a "pop". How far the end would actually go is then dependant on the trajectory / orientation of the end ( gravity gets in the way here) and some assumed air resistance, but if horizontal I would go for the point it hits the ground ignoring that.
I don't know for something like this, but my first case guess is between 5 to 10m??
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Volume of air is not really relevant once it is more than a couple of pipe diameters.
The problem is how will the end cap or blank ( two different things) actually fail? It is a bit far fetched to work on the basis is simultaneous bolt failure, but I guess you could so after that it is Force = mass x acceleration.
However this is a "pop". How far the end would actually go is then dependant on the trajectory / orientation of the end ( gravity gets in the way here) and some assumed air resistance, but if horizontal I would go for the point it hits the ground ignoring that.
I don't know for something like this, but my first case guess is between 5 to 10m??
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
- Thread starter
- #3
Hi,
Thanks for the reply.
Does the energy release from the pipe/system containing the compressed air not depend on volume though? i.e. a smaller volume of air will not have as big an energy release as a large volume. This is only the case for air though and not for water as it is incompressible?
Would the force not be different depending on the medium? I can't see 0.5 bar of air having the same force at failure as 0.5 bar of air?
Or am I mnisunderstanding?
Thanks for the reply.
Does the energy release from the pipe/system containing the compressed air not depend on volume though? i.e. a smaller volume of air will not have as big an energy release as a large volume. This is only the case for air though and not for water as it is incompressible?
Would the force not be different depending on the medium? I can't see 0.5 bar of air having the same force at failure as 0.5 bar of air?
Or am I mnisunderstanding?
-
1
- #4
If you refer to the document "Standard Pneumatic Test Procedure Requirement for Piping System AB 522" by ABSA, the equation to calculate the stored energy is given as follows. The code specifies a limit of 1677 kJ for safety reasons.
The stored energy may be calculated using the following formula providing that nitrogen
or air is used as the test medium:
E=2.5 x Pat x V [1 (Pa/Pat)0.286 ]
Where:
E = stored energy in kJ
Pa = absolute atmospheric pressure, 101 kPa
Pat = absolute test pressure in kPa
V = total volume under test pressure in m3
LittleInch
Petroleum
skaboy607,
You're now asking a different question. Total energy stored in a system is different from the force applied to a single item being ejected from the piping. Search this forum for pneumatic testing and look at the document by zdas04 for a different view.
Pipe and pipelines are different, so not sure why you're doing this??
Don't be silly. Force = Pressure X area. No exceptions and no changes. Force from 0.5 bar pressure of anything on the same surface area will result in the same force.
I think what you're thinking about is that, in reality, air escaping from a pipe which has a blank flange detach will continue to supply some force as the blank lifts off to a certain (fairly small) distance from the pipe. Water escaping will fall off in velocity much faster and in reality, this extra force for slightly longer as the flange lifts off actually doesn't make much difference.
However to a bystander, the force of the gas then exiting the rupture at high velocity will make it seem worse.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
You're now asking a different question. Total energy stored in a system is different from the force applied to a single item being ejected from the piping. Search this forum for pneumatic testing and look at the document by zdas04 for a different view.
Pipe and pipelines are different, so not sure why you're doing this??
Don't be silly. Force = Pressure X area. No exceptions and no changes. Force from 0.5 bar pressure of anything on the same surface area will result in the same force.
I think what you're thinking about is that, in reality, air escaping from a pipe which has a blank flange detach will continue to supply some force as the blank lifts off to a certain (fairly small) distance from the pipe. Water escaping will fall off in velocity much faster and in reality, this extra force for slightly longer as the flange lifts off actually doesn't make much difference.
However to a bystander, the force of the gas then exiting the rupture at high velocity will make it seem worse.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
The big fallacy in all of the stored energy calculations is delivery. A 100 km 2 m pipeline at 10 MPa has a LOT of stored energy, but when that cap blows off the vast majority of the stored energy is isolated from the failure by a shock wave and is absolutely irrelevant to the failure dynamics. In reality, the energy released is limited to a couple of joints of pipe.
LittleInch has it absolutely correct, f=P*A and f=m*a so acceleration = (P*A)/m. If your blank does fail all at once then you can get an initial acceleration normal to the axis of the pipe from which you can determine a max velocity and a distance to impact with the ground.
The sudden and complete failure is less unlikely than you might think. While this seems far fetched, we find the lids of tanks blow off all at once. When I've investigated these over-pressured atmospheric tank failures I've found the failure tends to occur on the tank side of the top weld in the heat affected zone (HAZ). The failure must be ductile (brittle failure tends to result in cracks and shrapnel instead of symmetry). You see a ductile yield pattern on remaining metal that is pretty uniform. When the metal necks down to the point it cannot contain the force it seems to launch the lid before minor rips can relieve the pressure.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
LittleInch has it absolutely correct, f=P*A and f=m*a so acceleration = (P*A)/m. If your blank does fail all at once then you can get an initial acceleration normal to the axis of the pipe from which you can determine a max velocity and a distance to impact with the ground.
The sudden and complete failure is less unlikely than you might think. While this seems far fetched, we find the lids of tanks blow off all at once. When I've investigated these over-pressured atmospheric tank failures I've found the failure tends to occur on the tank side of the top weld in the heat affected zone (HAZ). The failure must be ductile (brittle failure tends to result in cracks and shrapnel instead of symmetry). You see a ductile yield pattern on remaining metal that is pretty uniform. When the metal necks down to the point it cannot contain the force it seems to launch the lid before minor rips can relieve the pressure.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
- Thread starter
- #8
Hi,
Thanks for all the replies.
Perhaps I am a little simple in my thinking. Because I understood the stored energy for air to be greater than water I associated this with an increase in force. I understand now that force is the same but that we are talking about a different variable.
I am trying to work out for my company what would be deemed as a safe testing pressure for pneumatics. I am not in the pipeline industry but building services.
When using Victaulic couplings for example, there is only two bolts that hold the coupling together and the blank sits in a groove. A small release in the tension of one of the two bolts could cause the blank to fly off if the system is under pressure.
Thank you for the replies.
Thanks for all the replies.
Perhaps I am a little simple in my thinking. Because I understood the stored energy for air to be greater than water I associated this with an increase in force. I understand now that force is the same but that we are talking about a different variable.
I am trying to work out for my company what would be deemed as a safe testing pressure for pneumatics. I am not in the pipeline industry but building services.
When using Victaulic couplings for example, there is only two bolts that hold the coupling together and the blank sits in a groove. A small release in the tension of one of the two bolts could cause the blank to fly off if the system is under pressure.
Thank you for the replies.
You can do an experiment at home. Your water system is at about 40 psig. You probably have an air compressor (for airing up tires if for nothing else) that can be set at 40 psig. Look at the hose diameter on the air compressor and adapt your garden hose to the same size.
Now take a ping pong ball and hold it over the end of the garden hose and turn it on. The ball flies out of sight. Do the same thing with the air hose. The ball flies out of sight.
Now try it again with a golf ball. With water the golf ball goes up a couple of feet and stays there. Same outcome with air. Pressure is pressure, force is force. As long as a pressure acts on a surface the force is the product of pressure times area. In near field effects, your Victaulic fitting would reach about the same velocity before it falls out of the flow stream in either an air test or a water test. Once the projectile falls out of the flow stream, in either case the remaining pressure will cause the fluids to flow harmlessly (except for the risk of the liquid soaking into carpets etc) until the pressure is depleted.
The source of all the fear and superstition surrounding pneumatic tests is a paper written on NASA letterhead (the paper is no longer available from NASA) by a summer intern (who I sincerely hope failed out of his engineering program and is now a floor manager at Wal-mart). This neophyte calculated the energy stored in an entire pipeline and assumed that every bit of this energy was instantly available at any particular point along the line. What utter nonsense. I hope the senior NASA engineer that approved the paper has been demoted to washing test tubes. For the energy in any control volume along that pipe to participate in an explosion, that energy must be present at the site of the failure. The energy in a fluid stream is transported by mass flow. Moving mass to the failure site takes time. Even at sonic velocity of 1300 ft/s, a cubic inch of mass located 5 miles away would take 20 seconds to reach the failure--at 20 seconds any projectiles have been at rest for 18 seconds or more and have been out of the flow stream for over 19.5 seconds. After any projectile has moved something like 6 times the diameter of the opening it will be removed from available force and will have reached maximum velocity. From that point on you have a decelerating projectile and a blowdown stream.
Force is force. The difference between a pneumatic static test and a hydrostatic test is the amount of mass that must be removed from the system to reach atmospheric pressure. For gas PV=nRT, so the amount of mass that must be removed can be huge. For liquid the amount of mass is a function the Bulk Modulus (a very small number) so you only have to remove a very small fraction of the total mass to depressurize.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
Now take a ping pong ball and hold it over the end of the garden hose and turn it on. The ball flies out of sight. Do the same thing with the air hose. The ball flies out of sight.
Now try it again with a golf ball. With water the golf ball goes up a couple of feet and stays there. Same outcome with air. Pressure is pressure, force is force. As long as a pressure acts on a surface the force is the product of pressure times area. In near field effects, your Victaulic fitting would reach about the same velocity before it falls out of the flow stream in either an air test or a water test. Once the projectile falls out of the flow stream, in either case the remaining pressure will cause the fluids to flow harmlessly (except for the risk of the liquid soaking into carpets etc) until the pressure is depleted.
The source of all the fear and superstition surrounding pneumatic tests is a paper written on NASA letterhead (the paper is no longer available from NASA) by a summer intern (who I sincerely hope failed out of his engineering program and is now a floor manager at Wal-mart). This neophyte calculated the energy stored in an entire pipeline and assumed that every bit of this energy was instantly available at any particular point along the line. What utter nonsense. I hope the senior NASA engineer that approved the paper has been demoted to washing test tubes. For the energy in any control volume along that pipe to participate in an explosion, that energy must be present at the site of the failure. The energy in a fluid stream is transported by mass flow. Moving mass to the failure site takes time. Even at sonic velocity of 1300 ft/s, a cubic inch of mass located 5 miles away would take 20 seconds to reach the failure--at 20 seconds any projectiles have been at rest for 18 seconds or more and have been out of the flow stream for over 19.5 seconds. After any projectile has moved something like 6 times the diameter of the opening it will be removed from available force and will have reached maximum velocity. From that point on you have a decelerating projectile and a blowdown stream.
Force is force. The difference between a pneumatic static test and a hydrostatic test is the amount of mass that must be removed from the system to reach atmospheric pressure. For gas PV=nRT, so the amount of mass that must be removed can be huge. For liquid the amount of mass is a function the Bulk Modulus (a very small number) so you only have to remove a very small fraction of the total mass to depressurize.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
- Thread starter
- #10
Ok, thank you David that is very helpful.
I understand that the Victaulic blank after a certain point of time will be out of the flow stream.
If I estimate the energy in a system to approx. 250 kJ, is there a way of approximating how far a mass of x will travel?
That way, we are able to establish a safe exclusion zone for a given air pressure test.
I understand that the Victaulic blank after a certain point of time will be out of the flow stream.
If I estimate the energy in a system to approx. 250 kJ, is there a way of approximating how far a mass of x will travel?
That way, we are able to establish a safe exclusion zone for a given air pressure test.
LittleInch
Petroleum
Ha,
I knew Dave would answer this way better than me...
Don't mistake "energy" for pressure. You could have the same "energy" in a high pressure low volume piece of piping as a low pressure much larger system. However the force on your blank end will be very different, therefore no, you can't work out distance of a projectile from energy when the force is only available to the projectile for a few milliseconds.
If needed covering vulnerable ends with sandbags or building a "catch wall" around them might be required or to reduce your exclusion zone.
One thing we haven't talked about is what your material is. If it's something like PVC or a similar hard brittle material, you could get fragmentation much easier than for steel of some ductile material like PE. I escaped near death as a student when my PVC 2" diam tube I was pressure testing fragmented at about 35barg. Small sharp pieces of PVC flew all over the test lab at high velocity and how we didn't injure someone is part miracle, part one of my nine lives... Small amount of energy, but potentially catastrophic results.
Take care and assume the worst might happen so don't stand in front of the blind...
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
I knew Dave would answer this way better than me...
Don't mistake "energy" for pressure. You could have the same "energy" in a high pressure low volume piece of piping as a low pressure much larger system. However the force on your blank end will be very different, therefore no, you can't work out distance of a projectile from energy when the force is only available to the projectile for a few milliseconds.
If needed covering vulnerable ends with sandbags or building a "catch wall" around them might be required or to reduce your exclusion zone.
One thing we haven't talked about is what your material is. If it's something like PVC or a similar hard brittle material, you could get fragmentation much easier than for steel of some ductile material like PE. I escaped near death as a student when my PVC 2" diam tube I was pressure testing fragmented at about 35barg. Small sharp pieces of PVC flew all over the test lab at high velocity and how we didn't injure someone is part miracle, part one of my nine lives... Small amount of energy, but potentially catastrophic results.
Take care and assume the worst might happen so don't stand in front of the blind...
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
-
1
- #12
skaboy607,
You get to use your basic equations of motion from Dynamics that you never thought you'd see again. First let us assume that the fitting launches horizontally from a meter above the floor. In the "y" direction, I know that gravity wants the projectile to fall:
S=h=1m=(g/2)*t2 ∴ tfall = (2*h/g)1/2
If you assume that the force is applied for the first 6 pipe diameters of travel (not a perfect assumption since the force adheres to an inverse square relationship and will bleed off rather than being on/off, but it is pretty close), so (combining the definition of pressure and the 1st law):
f=P*A=m*acap ∴ acap=(P*A)/m
S=6D=(1/2)acap*tacc2 ∴ tacc=[(2*6D/acap)]1/2
Now you know how long the force is applied to the projectile. The velocity at the end of acceleration is:
v=acap*tacc
Now the distance traveled is:
Stravel=v*(tfall-tacc)+6D
If you work all this out for a 0.5 lbm projectile from a 2-inch line at a 60 psig test you'll see a velocity at the end of the acceleration period (which is 0.01 s) of 105 ft/s and a distance traveled of 47 ft. A 2 m line at 10 MPA would launch a 100 kg closure almost 900 m before it fell 1 m.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
You get to use your basic equations of motion from Dynamics that you never thought you'd see again. First let us assume that the fitting launches horizontally from a meter above the floor. In the "y" direction, I know that gravity wants the projectile to fall:
S=h=1m=(g/2)*t2 ∴ tfall = (2*h/g)1/2
If you assume that the force is applied for the first 6 pipe diameters of travel (not a perfect assumption since the force adheres to an inverse square relationship and will bleed off rather than being on/off, but it is pretty close), so (combining the definition of pressure and the 1st law):
f=P*A=m*acap ∴ acap=(P*A)/m
S=6D=(1/2)acap*tacc2 ∴ tacc=[(2*6D/acap)]1/2
Now you know how long the force is applied to the projectile. The velocity at the end of acceleration is:
v=acap*tacc
Now the distance traveled is:
Stravel=v*(tfall-tacc)+6D
If you work all this out for a 0.5 lbm projectile from a 2-inch line at a 60 psig test you'll see a velocity at the end of the acceleration period (which is 0.01 s) of 105 ft/s and a distance traveled of 47 ft. A 2 m line at 10 MPA would launch a 100 kg closure almost 900 m before it fell 1 m.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
- Thread starter
- #13
Thank you Dave. This is excellent. Had forgotten all about kinematics but it is coming back to me now! Finding this subject very interesting.
A colleague of mine has also come proposed an alternative by calculating the available energy within a kg of air. Just going through the figures now.
As a side question albeit linked to my calculations. For Boyles Law, does this apply when a compressor is used to increase the pressure within a pipe?
Example
1m length of 300dia pipe, both ends are blanked off-the volume is know and the pressure it atmospheric (p1v1)
Then compressor is used to raise pressure to 1.5 bar. p2 is known but doing the maths means reduction in volume? But compressor would surely increase volume?
A colleague of mine has also come proposed an alternative by calculating the available energy within a kg of air. Just going through the figures now.
As a side question albeit linked to my calculations. For Boyles Law, does this apply when a compressor is used to increase the pressure within a pipe?
Example
1m length of 300dia pipe, both ends are blanked off-the volume is know and the pressure it atmospheric (p1v1)
Then compressor is used to raise pressure to 1.5 bar. p2 is known but doing the maths means reduction in volume? But compressor would surely increase volume?
As LittleInch and I have both tried to convey to you, total energy is irrelevant. Just like sunlight that never enters the Earth's atmosphere is irrelevant to the temperature on the Earth, stored energy that does not get to the failure while particles are available to become projectiles is irrelevant to the projectile calculation. It is really easy to calculate total potential energy, but so what? There is simply no way to use that energy in an explosive decompression.
As to your compressor question it does not make any sense at all. I don't care what the source of additional pressure is, PV = nRT for an ideal gas like air. For your example V1=V2 (the amount of expansion that you would get by going from local atmospheric pressure to 1.5 bara is too small to measure). If the compressor discharge is at ambient temperature then you have increased the mass in the system to n2=1.5*n1 (assuming local atmospheric pressure is 1.0 bara)
See my point? I cannot comprehend what "maths" would require volume to decrease.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
As to your compressor question it does not make any sense at all. I don't care what the source of additional pressure is, PV = nRT for an ideal gas like air. For your example V1=V2 (the amount of expansion that you would get by going from local atmospheric pressure to 1.5 bara is too small to measure). If the compressor discharge is at ambient temperature then you have increased the mass in the system to n2=1.5*n1 (assuming local atmospheric pressure is 1.0 bara)
See my point? I cannot comprehend what "maths" would require volume to decrease.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
- Thread starter
- #15
Thank you for continued responses. Apologies if I am taking a while to understand. I try to understand every detail until it is clear in my head.
The reason I was asking about the "available energy" was because it was used in a colleagues calculation. You have now reiterated the point about available energy so thank you.
With regards to Boyles law and the question of volume I was applying the following logic but perhaps the equation is not applicable. Perhaps it is only for a closed system with no external input.
p1v1=p2v2
v2 is (p1v1)/p2
v1 is atmosphere pressure, v1 is volume within the pipe
p2 is 5 bar more than p1 or 50 kpa or 1.5 time atmosphere - this makes v2 smaller.
For the blanking flange being subject to the force for 6D - is this when an object is within a pipe or at the end of it?
Also, where is the compressibility of the fluid taken into account? The equations only refers to the pressure which as I have been reminded of numerous times is the same whether the fluid is water of air.
Using your ping pong and golf ball analogy, the forces are the same but where is it taken into account in the equations above about your statement below. If force is force and the plate can go the same distance before it falls out of the flow field, then why is air considered worse than water?
"Force is force. The difference between a pneumatic static test and a hydrostatic test is the amount of mass that must be removed from the system to reach atmospheric pressure. For gas PV=nRT, so the amount of mass that must be removed can be huge. For liquid the amount of mass is a function the Bulk Modulus (a very small number) so you only have to remove a very small fraction of the total mass to depressurize."
The reason I was asking about the "available energy" was because it was used in a colleagues calculation. You have now reiterated the point about available energy so thank you.
With regards to Boyles law and the question of volume I was applying the following logic but perhaps the equation is not applicable. Perhaps it is only for a closed system with no external input.
p1v1=p2v2
v2 is (p1v1)/p2
v1 is atmosphere pressure, v1 is volume within the pipe
p2 is 5 bar more than p1 or 50 kpa or 1.5 time atmosphere - this makes v2 smaller.
For the blanking flange being subject to the force for 6D - is this when an object is within a pipe or at the end of it?
Also, where is the compressibility of the fluid taken into account? The equations only refers to the pressure which as I have been reminded of numerous times is the same whether the fluid is water of air.
Using your ping pong and golf ball analogy, the forces are the same but where is it taken into account in the equations above about your statement below. If force is force and the plate can go the same distance before it falls out of the flow field, then why is air considered worse than water?
"Force is force. The difference between a pneumatic static test and a hydrostatic test is the amount of mass that must be removed from the system to reach atmospheric pressure. For gas PV=nRT, so the amount of mass that must be removed can be huge. For liquid the amount of mass is a function the Bulk Modulus (a very small number) so you only have to remove a very small fraction of the total mass to depressurize."
OK, one more time. P1*v1 only equals P2*v2 IF NO MASS IS ADDED TO THE SYSTEM. You are adding mass to increase the pressure so Boyles Law doesn't apply.
Your statements
Think for yourself for a minute. A piece of piping becomes a projectile when it separates from the piping. The projectile remains in the flow stream until it gets too far from its original position for the stream to impact it or it moves to the side. Many experiments have shown that somewhere around 6 times the diameter of the flow stream is a practical limit for the stream controlling the acceleration of the projectile.
Your question about compressibility makes me think that you are in far far over your head and I can't help. "Compressibility" is really a measure of departure from ideal behavior. At best it is a tertiary factor in the projectile discussion.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
Your statements
are gibberish and I can't figure out what the heck you are talking about, but Boyles Law does not apply when you are adding mass.
Think for yourself for a minute. A piece of piping becomes a projectile when it separates from the piping. The projectile remains in the flow stream until it gets too far from its original position for the stream to impact it or it moves to the side. Many experiments have shown that somewhere around 6 times the diameter of the flow stream is a practical limit for the stream controlling the acceleration of the projectile.
Your question about compressibility makes me think that you are in far far over your head and I can't help. "Compressibility" is really a measure of departure from ideal behavior. At best it is a tertiary factor in the projectile discussion.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
skaboy607. Here is a link to Pressure Testing (Hydrostatic / Pneumatic) Safety Guidelines.
- Thread starter
- #18
Ok understood. They are only gibberish if boyles law doesn't apply. If it did, they would make perfect sense.
Everyone has to be over there head at some point - it is how we learn.
Surely compressibility is the whole reason that air is potentially dangerous (with poor engineering design) and water is not so I cannot see how it is considered tertiary.
Thanks for your help anyway. You have got me almost there.
EDIT - what value of acap do you get for the 2" example above. Trying to prove the conversion. I get:
tfall = 0.45s
acap = 3698m/s2
tacc = 0.0128s
v = 47.48m/s
stravel = 21.35m
Everyone has to be over there head at some point - it is how we learn.
Surely compressibility is the whole reason that air is potentially dangerous (with poor engineering design) and water is not so I cannot see how it is considered tertiary.
Thanks for your help anyway. You have got me almost there.
EDIT - what value of acap do you get for the 2" example above. Trying to prove the conversion. I get:
tfall = 0.45s
acap = 3698m/s2
tacc = 0.0128s
v = 47.48m/s
stravel = 21.35m
- Status
Similar threads
- Question
- Question
- Question