Aces High Bulletin Board
General Forums => The O' Club => Topic started by: Ciaphas on May 01, 2019, 07:25:21 PM
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I have an object that weighs: 6300lbs
This object falls from 12,000 ft at 1,714 ft per mile and travels 7 miles (36,960ft). How fast is it traveling?
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at 1,714 ft per second
How fast is it traveling?
Seems you answered your own question.
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updated original post as follows:
1714 ft per mile
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updated original post as follows:
1714 ft per mile
but there's 5280 feet per mile according to google. And why are you still using the Imperial measurement system - are you guys still a british colony?
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Over the course of a mile (5280 ft) you fall 1714 ft.
If we were still a British colony, we would be using the metric system.
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I have an object that weighs: 6300lbs
This object falls from 12,000 ft at 1,714 ft per mile and travels 7 miles (36,960ft). How fast is it traveling?
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So the object ends up 24,960 feet below ground?
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1714 ft x 7 = 11,998 ft
12,000 ft / 7 = 1714.28571 ft per mile
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Something is clearly wrong with the equation.
1 mile = 5280 ft
an object that travels 1714 feet travels about .325 miles and cannot travel 7 miles
Feet per mile is not a measurement for speed or velocity. you need a distance factor and a time factor - e.g. Feet per second, miles per hour, etc.
Or are you trying to figure out how the Millennium Falcon made the Kessel Run in less than 12 parsecs? :devil
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it's all there. It's in the wording.
An object falls from 12,000 ft at a rate of 1714 ft per mile travelled and travels total distance of 7 miles, how fast, in mph is the object traveling?
It would appear to me that:
Mass = 6300 lbs
Velocity = 1714 ft per mile
Height = 12,000 ft
Distance is 7 miles (36,960 ft)
It can be calculated without time but it requires starting velocity, ending velocity, acceleration and gravity.
To solve this we have to assume that 1714 ft per mile is the starting velocity. we must now find the rate of acceleration based on gravity and object weight and in turn this will give us it's final velocity over distance.
v^2=u^2+2as
V = final velocity
U = initial velocity
A = acceleration
S = distance
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feet per mile measures exactly one thing - how many feet are in a mile, which is 5280 feet. That is a measurement of distance, not velocity.
either your units are mislabeled or you are missing a factor of time.
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This is why I need this solved.
We had a T-6 Texan go down about 30 miles north of our base. We are tasked with locating the two ejections seats but they are only willing to give us basic information.
ie GVW 6300 (full load), altitude of ejection 12k and a distance traveled from aircrew to impact (7 miles).
all help is appreciated.
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You’re saying they lost 1700 feet vertically for every mile covered horizontally, no?
Just remember it is going to be an arc. Lateral distance will decrease with time. You’re gonna’ need to break out your Trig and Calculus books.
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that's what the information provided eludes to.
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15
Coogan
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terminal velocity?
semp
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This is why I need this solved.
We had a T-6 Texan go down about 30 miles north of our base. We are tasked with locating the two ejections seats but they are only willing to give us basic information.
ie GVW 6300 (full load), altitude of ejection 12k and a distance traveled from aircrew to impact (7 miles).
all help is appreciated.
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When do the pilots separate from the seats and does their chute open immediately?
And does the seat have a chute of it's own? If so, when does it open?
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[edit]Sorry I was being flippant....
If you had to estimate the average square plate surface for drag to be? 6^2ft?
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roughly 2.5 seconds for crew separation, the seat does have a drogue chute that stabilizes and allows the seat to allow crew seat separation. The seat without crew is roughly 200 pounds.
this all happens within 2.5 - 3 seconds after seat initiation.
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Sorry bud. I've had a couple of beers, but "1714 ft per mile is the starting velocity" doesn't make any sense to me. Can you double check that?
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roughly 2.5 seconds for crew separation, the seat does have a drogue chute that stabilizes and allows the seat to allow crew seat separation. The seat without crew is roughly 200 pounds.
this all happens within 2.5 - 3 seconds after seat initiation.
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but the seat freefalls after this?
Also, there still seems to be too many variables in play. Seems to me that the attitude of the Texan when the ejection occurred will play a huge factor in determining the probable search area. Even if we can figure out this "feet per mile" problem, I do not see there being enough info to go by.
Might be easier to figure out where the plane was when the crew ejected relative to the crash site and figure out the trajectory of the crew from that point and go with the assumption that the seats would go farther than the crew.
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but the seat freefalls after this?
Also, there still seems to be too many variables in play. Seems to me that the attitude of the Texan when the ejection occurred will play a huge factor in determining the probable search area. Even if we can figure out this "feet per mile" problem, I do not see there being enough info to go by.
Might be easier to figure out where the plane was when the crew ejected relative to the crash site and figure out the trajectory of the crew from that point and go with the assumption that the seats would go farther than the crew.
I agree with you, I've been on many crash scenes and the ejection seats are typically easy to find. The weather is the factor here, the aircrew were no doubt tossed around in their chutes by the wind, disorientation is going to play a negative roll in this one. Unfortunately the seats do not interface with the FDR or any other system in the cockpit other than 1 electrical connector for the seat actuator.
I'm hoping that after the aircrew debrief, we will know more. All attempts to reach the crash site by recovery personnel were called off due to weather. We will have to try again in the morning. I just pray that if someone stumbles across them, that they have enough sense to leave well enough alone and contact the authorities.
Thanks for the help guys
:salute
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Seems like you’re sincere in your question and have more details then you’ve presented.
If any part of the numbers you posted are accurate, I think someone’s messing with you.
Go down to the machine shop and bring me a left handed monkey wrench and be quick. And don’t come back without it!
Either that, or someone wants to teach the value of research. Hopefully that’s the case but I doubt it. In the sprit of having some fun with it, …
According to Wikipedia, the listed gross weight of your T-6 Texan II is 6,300 lbs. Well that’s helpful, it matches one of your numbers. As least you might know which T-6 that clown was using. The top speed, not even the cruise speed, for that plane is 364 mph, or ~533 ft/sec. if it were a legitimate exercise. You can find the model of ejection seat be googling the plane and then google the seat to see if a terminal V or other details are given for the seat.
V in your post is velocity: V is distance traveled over unit time, usually seconds in this kind of problem, but can be hours, minutes or seconds. Thus 1714 ft per mile is meaningless as given. The seat falls from 2.3 miles alt and it might travel 1714 ft (third of a mile) horizontally?
It can become a simple spreadsheet exercise targeting a range of distances depending on a cruse of 320 or max at 364 mph, unless you know the actual speed. Plug in the numbers for acceleration of gravity Ag, and terminal velocity. And you just added another vector for wind drift. :rolleyes:
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Seems like you’re sincere in your question and have more details then you’ve presented.
If any part of the numbers you posted are accurate, I think someone’s messing with you.
Go down to the machine shop and bring me a left handed monkey wrench and be quick. And don’t come back without it!
Either that, or someone wants to teach the value of research. Hopefully that’s the case but I doubt it. In the sprit of having some fun with it, …
According to Wikipedia, the listed gross weight of your T-6 Texan II is 6,300 lbs. Well that’s helpful, it matches one of your numbers. As least you might know which T-6 that clown was using. The top speed, not even the cruise speed, for that plane is 364 mph, or ~533 ft/sec. if it were a legitimate exercise. You can find the model of ejection seat be googling the plane and then google the seat to see if a terminal V or other details are given for the seat.
I work on the ejection seat for this aircraft and the ESUP MK16 for the T-38 (2011 - present) and previously the ACES II Ejection seat (1998-2011)
V in your post is velocity: V is distance traveled over unit time, usually seconds in this kind of problem, but can be hours, minutes or seconds. Thus 1714 ft per mile is meaningless as given. The seat falls from 2.3 miles alt and it might travel 1714 ft (third of a mile) horizontally?
I know that for an aircraft in a steady rate of decent from 12,000 ft (actual: 11,049) covering 7 miles to impact (impact site = roughly 951 ft, rounded down to 0 to make it a bit for the sake of simplicity) would require a decent of 1714 ft per mile (actual 11578.42 ft per mile)
It can become a simple spreadsheet exercise targeting a range of distances depending on a cruse of 320 or max at 364 mph, unless you know the actual speed. Plug in the numbers for acceleration of gravity Ag, and terminal velocity. And you just added another vector for wind drift. :rolleyes:
Since the speed nor time from punching out until impact are known at this time, I am trying to figure out an educated guess as to the speed the aircraft was traveling when the sortie went south. This will allow me to figure a time taken for the aircraft to impact the ground from punching. This will also allow me to figure out wind sheer and drag forced on the seats upon exiting the cockpit due to blunt wind shear and wind speed affecting the drogue chute and help me better map out probable impact sites for the seats, of course not exact but within a reasonable area.
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Too much I don't know about the problem but that helps. If you have a spreadsheet program, Excel, you can plug in these numbers and formula as a starting point.
This can give the initial velocity V. It calculates the distance D in feet traveled in one hour at any given speed and converts it to feet per second.
320 mph (cruise speed)
5,280 ft/mile
1,689,600 D ft/hour (320 x 5280)
3,600 second/hr (60sec x 60 min)
469.3333333 Vsec(ft) (feet per second) (D distance in feet per hour / second in an hour)
Further, more useful then the 1714 number is the time to impact for the aircraft. However, that is independent of the ejection seat. It's expected the plane accelerate as it descended. But using the initial Vsec(ft) and a straight line decent as you suggested, (unlikely?) we can estimate the time to impact thus.
7 miles
5,280 ft/mile
36,960 horizontal feet traveled
469.3333333 Vsec(ft) (from above)
78.75 seconds to impact (36,960 ft / Vsec(ft) )
Comparing the 79 seconds to the actual time can help define the curve for the plane.
:cheers:
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terminal velocity?
semp
Eventually but not initially. Depends on which way they ejected as they would travel in some sort of arc before reaching TV.
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Right, once the seat has been exposed to wind sheer and the effects of this wind sheer have been felt by the drogue chute, the seat will fall like a stone with some degree of horizontal velocity, also gaining speed within its own arc. I'm going out on a limb but I presume that the seat will be found roughly 6.75 - 7.25 miles from the impact site with about a 500 yard margin for error. I should know when I get to work tonight.
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one of the unique aspect of this ejection is the effects of the weather. Most ejections are caused through mechanical/electrical fault and the aircrew is seldom affected by the weather in the manner that these two pilots were.
Typically pilots will locate the seats if they can and it becomes a non issue but high winds and parachutes = no bueno. We also have three terrain types to search: fields, heavily wooded areas and a lake.
like I said before, this is a very non-typical situation. It would be nice to have an ELT on the seat but MartinBaker decided that we didn't need one.
Thanks for all of the help!
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I wondered if you had a seat locator beacon. Guess not. :(
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What did the pilots have for breakfast?
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You dont have enough information to solve the problem.
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we have three pieces of information that can help us solve this.
Knowns:
height: 12,000
Distance: 7 miles (convert to ft)
estimates:
time to impact: ~79 sec
This can solve for speed.
speed= 318.987 mph
This is in a controlled decent and is not dependent on wind and the lack of pilots controlling the aircraft.
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I would think that to work up some probability on where you would find the seats you want to know how the seat behaves at a given ejections speed, in other words how how far it would travel without regard to wind before its descent is more or less vertical and then modify that for the speed of the aircraft when the ejection happened. Is there any source of test data for this? Or failing that is there a database for similar accidents that would have useful information you could extrapolate from?
I would think failing any more information you could start with two points on a line that is an extension of the heading of the aircraft when ejection occurred. One point would be the furthest distance an object could have traveled assuming no drag if you dropped it from the aircraft at the same time as ejection the second point would be to use the same starting conditions but use a number for drag based upon someones informed opinion. If you erred on the side of too much drag the seat should be somewhere between the points with some allowance for wind effect.
You could also just start with the one point on the line using the assumption of no drag as that would be the furthest the seat would likely be able to .travel.
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What about if they started on a conveyor belt?
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Turns out, the pilots have no idea where they punched. You would think that RAPCON or the Tower would have that information or an approximate location. So there are three terrain types that are possible sites for the ejection: Field, dense tree and the dreaded lake... .
9 hrs of searching and nothing to show for it.
a tid bit: the texan created a 5 ft deep crater on impact.
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If you know the plane's heading in the last minutes, and the impact point, the seat should be within, say 80 seconds flight time back along the flight path, adjusted for wind. Where's the lake? :devil
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less than a 1/4 mile from impact. It's been pretty remarkable the amount of disorientation that has been shown by aircrew.
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Turns out, the pilots have no idea where they punched. You would think that RAPCON or the Tower would have that information or an approximate location. So there are three terrain types that are possible sites for the ejection: Field, dense tree and the dreaded lake... .
9 hrs of searching and nothing to show for it.
a tid bit: the texan created a 5 ft deep crater on impact.
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RAPCON radar track should have been saved. If so, it can be reviewed to determine this point of ejection. More than likely, the accident investigation board has already done this. Also, winds aloft data for that day and time should be available to help determine drift components. Keep in mind that ejecting is a very high stress event and knowing the exact point over the ground is probably way down at the bottom of the pilot’s priority list, especially if it was an out of control event, in the wx, etc.
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Turns out, the pilots have no idea where they punched. You would think that RAPCON or the Tower would have that information or an approximate location. So there are three terrain types that are possible sites for the ejection: Field, dense tree and the dreaded lake... .
9 hrs of searching and nothing to show for it.
a tid bit: the texan created a 5 ft deep crater on impact.
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When I was an instructor in San Antonio I saw where a T-6 crashed south of Stinson Field. It was funny how the Air Force had a big canvas wall erected along the road but we buzzed right over it taking photos while doing touch-and-goes. Lol.
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haha, they will be debriefing the pilots after the safety team gets here tomorrow
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Feet per mile is not a linear equation. The mile is the X and the 1714 ft is the Y. If it has not been solved yet, give me a few minutes.
Given your 79s estimation, the speed the object travels, on a lateral plane, is 318.987 mph. The problem is the fact that 318 mph is a mean number and it likely only traveled that speed once. The speed of the object is much slower at miles 0-2 than it is from 4-7. Also, this problem was impossible without time.
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This conversation is irrelevant at this point.
The aircraft either flew through or got pulled in to a tornado, this is as per the aircrew.
Those seats could be anywhere at this point.
Not sure how something like this happens as I'm not a pilot. One bit of information worthy of note is the T-6 is not weather radar equipped, neither is the T-38. This may have played a major role in to this event.
I do appreciate all of the help guys!2
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The aircraft either flew through or got pulled in to a tornado, this is as per the aircrew.
Those seats could be anywhere at this point.
Ouch! Thankfully the crew is alright.
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one of the pilots face had ice on it when he hit the ground. crazy stuff for sure
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Reminds me of Lt. Col. William Rankin. “The Man Who Fell Up.”
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These two cats are lucky to be alive for sure