how long did it take the bombs to hit the ground from 25 000 ft
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mikewint
Captain
Depends on multiple factors. In the simplest sense Distance traveled under uniform acceleration = (.5)(acceleration due to gravity)(time)^2 power. Using your 25,000ft height and the average gravitational acceleration of 32ft/s^2 (varies with height and latitude). So 39.5s BUT all falling objects IN AIR reach a terminal velocity wherein the downward force tending to accelerate = the upward force of air resistance. Air resistance is a very tricky and impossible to calculate exactly because air density varies with height and temperature of the air, plus we have to consider the cross-sectional area of the bomb and its surface texture and fins.
In the simple example above air resistance was ignored which also would mean that the bomb would continue to accelerate over the entire distance hitting the ground at 862MPH!! Since the speed of sound is roughly 750MPH the bomb hits at MACH 1.2, obviously not true since you can hear bombs as they fall.
So the best estimate I can give you is a terminal velocity of between 550MPH and 650MPH or between 46.5s and 39.3s
ALSO the horizontal velocity of the plane has NOTHING to do with the time of vertical fall. The two motions are INDEPENDENT of each other. As in crossing a river by heading directly across. The river's current determines where you reach the opposite bank
In the simple example above air resistance was ignored which also would mean that the bomb would continue to accelerate over the entire distance hitting the ground at 862MPH!! Since the speed of sound is roughly 750MPH the bomb hits at MACH 1.2, obviously not true since you can hear bombs as they fall.
So the best estimate I can give you is a terminal velocity of between 550MPH and 650MPH or between 46.5s and 39.3s
ALSO the horizontal velocity of the plane has NOTHING to do with the time of vertical fall. The two motions are INDEPENDENT of each other. As in crossing a river by heading directly across. The river's current determines where you reach the opposite bank
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parsifal
Colonel
In a vacuum, the accelaration of any object is 9.8m/s. however in atmospheric conditions, the object will reach terminal velocity, and this will vary according to the density of the atmosphere. At high altitudes, where the atmosphere is thin, there will be a higher terminal velocity, at lower altitudes, the terminal velocity is lower. if the atmosphere was so thick it was a liquid or even solid, the terminal velocity would be very low, or close to zero.
Terminal velocity is where the downward force of gravity equals the air resistance/force of drag and buoyancy, velocity becomes asymptotic to a constant value with net acceleration .
Typically, a skydiver has a terminal velocity of 195 kmh, but this can vary a lot. The world record rate of descent is about 800KmH. A standard rifle bullet has a vertical rate of descent of about 300 KmH or 90m/s
one would expect an iron bomb, with a similar shape to a bullet, would have a similar terminal velocity in the vertical. One would expect the distance travelled by the ordinance to be more than the vertical distance to the ground, since it retains some horizontal movement as well. But its difficult to determine how much of that forward movement is retained, and for how long. If we assume that for a drop at 20000ft the bomb actually travels 25000 ft, or 8000m, at a terminal velocity of 320 kmh, it should take about 90-95 secs for the bomb to make impact.
Terminal velocity is where the downward force of gravity equals the air resistance/force of drag and buoyancy, velocity becomes asymptotic to a constant value with net acceleration .
Typically, a skydiver has a terminal velocity of 195 kmh, but this can vary a lot. The world record rate of descent is about 800KmH. A standard rifle bullet has a vertical rate of descent of about 300 KmH or 90m/s
one would expect an iron bomb, with a similar shape to a bullet, would have a similar terminal velocity in the vertical. One would expect the distance travelled by the ordinance to be more than the vertical distance to the ground, since it retains some horizontal movement as well. But its difficult to determine how much of that forward movement is retained, and for how long. If we assume that for a drop at 20000ft the bomb actually travels 25000 ft, or 8000m, at a terminal velocity of 320 kmh, it should take about 90-95 secs for the bomb to make impact.
Shortround6
Major General
Bomb and bullet may have a similar shape, what they don't have is similar sectional density (weight per unit of frontal area), which is why artillery shells travel soooo much further than rifle bullets even with near the same shape and fired at near the same speeds.
parsifal
Colonel
I get that, but the terminal velocity is still the issue. What is the terminal velocity of a standard 500 lb GP bomb for example. Once we have that, we will be able to get stuck into this problem.
fubar57
General
From(shudder) Wikipedia Bombsight - Wikipedia, the free encyclopedia
The M65 will be dropped from a Boeing B-17 flying at 200 mph at an altitude of 20,000 feet in a 25 mph wind. Given these conditions, the M65 would travel approximately 6,500 feet forward before impact,[8] for a trail of about 1000 feet from the vacuum range,[9] and impact with a velocity of 1150 fps at an angle of about 77 degrees from horizontal.[10] A 25 mph wind would be expected to move the bomb about 300 feet during that time.[11] The time to fall is about 37 seconds.
Geo
The M65 will be dropped from a Boeing B-17 flying at 200 mph at an altitude of 20,000 feet in a 25 mph wind. Given these conditions, the M65 would travel approximately 6,500 feet forward before impact,[8] for a trail of about 1000 feet from the vacuum range,[9] and impact with a velocity of 1150 fps at an angle of about 77 degrees from horizontal.[10] A 25 mph wind would be expected to move the bomb about 300 feet during that time.[11] The time to fall is about 37 seconds.
Geo
mikewint
Captain
Parsifal, The actual path distance traveled by the bomb is immaterial to this question as the motion is in two planes and the two motions are independant of each other. Thus the vertical velocity is exactly the same whether the object is dropped from a moving aircraft or a stationary cliff. A bullet frired horizontally and a bullet dropped from the barrel at the same instant reach the ground at the same time. Your time of 90s is vastly too long.
Air resistance varies as the cube of the cross-section thus a bullet and a bomb will not have a similiar terminal velocity.
Arty shells travel further than bullets because they are fired above 45 degrees into thinner atmosphere thus reducing air resistance to the horizontal velocity while the added upward vertical velocity increases time of flight (range). Arty shells fired north or south are also affected by corollis forces producing curved paths
Air resistance varies as the cube of the cross-section thus a bullet and a bomb will not have a similiar terminal velocity.
Arty shells travel further than bullets because they are fired above 45 degrees into thinner atmosphere thus reducing air resistance to the horizontal velocity while the added upward vertical velocity increases time of flight (range). Arty shells fired north or south are also affected by corollis forces producing curved paths
GregP
Major
In a vacuum, the bombs would fall straight down from the bomb bay and stay right under the bomb bay as the aircraft moves forward. It would accelerate at 9.8 m/s^2 but the horizontal velocity would remain unchanged. Let's say the B-17 is bombing from 25,000 feet at 180 mph and somehow is doing it in a vacuum.
180 mph is 80.46 m/s and 25,000 feet is 7,620 m. The bomb continues to move in the x (horizontal) direction at 80.46 m/s and accelerates from an initial value of zero in the y (vertical) direction to a final value of 386.46 m/s downward and impacts the ground in 39.435 seconds. 346.46 m/s is 864.5 mph downward and 180 mph forward. The combined speed of the bomb would be 883 mph in a vacuum.
In the real world, the plane cannot fly in a vacuum; it flies in air. The real secret to the Norden bombsight, or any other bombsight, is that it computes the effects of air resistance and shows the point of impact of the bomb continuously in the sight. When the target shows up in the crosshairs, press the release. The bomb's terminal velocity in real air would vary with altitude, but gets slower as the bomb gets closer to the ground because the air gets denser as you approach the ground.
In the figure below, the solid line is how the bomb would fall in a vacuum. It will stay right under the bomb bay as it descends and gains vertically velocity. In real life, it follows the approximate path of the dotted line. The amount it falls shorter than vacuum is determined by the altitude of release, the cross-sectional area of the bomb, the drag of the bomb, and any winds that the bomb falls through on the way down.
I doubt of the real bomb will get to 860+ mph and would guestimate the real speed to be in the 600 – 650 mph range depending on bomb type. I'd not be surprised to be off either high or low. I have books on artillery fire that explain a lot, but they are all several hundred pages of math.
Since the vertical velocity decreases by about 1/3, I'd expect the fall time to be in the 59 second range from 25,000 feet if I try a terminal velocity of, say, 600 mph. I'd therefore expect anywhere from 50 seconds to about 1 minute 10 seconds as a first-order estimate.
180 mph is 80.46 m/s and 25,000 feet is 7,620 m. The bomb continues to move in the x (horizontal) direction at 80.46 m/s and accelerates from an initial value of zero in the y (vertical) direction to a final value of 386.46 m/s downward and impacts the ground in 39.435 seconds. 346.46 m/s is 864.5 mph downward and 180 mph forward. The combined speed of the bomb would be 883 mph in a vacuum.
In the real world, the plane cannot fly in a vacuum; it flies in air. The real secret to the Norden bombsight, or any other bombsight, is that it computes the effects of air resistance and shows the point of impact of the bomb continuously in the sight. When the target shows up in the crosshairs, press the release. The bomb's terminal velocity in real air would vary with altitude, but gets slower as the bomb gets closer to the ground because the air gets denser as you approach the ground.
In the figure below, the solid line is how the bomb would fall in a vacuum. It will stay right under the bomb bay as it descends and gains vertically velocity. In real life, it follows the approximate path of the dotted line. The amount it falls shorter than vacuum is determined by the altitude of release, the cross-sectional area of the bomb, the drag of the bomb, and any winds that the bomb falls through on the way down.
I doubt of the real bomb will get to 860+ mph and would guestimate the real speed to be in the 600 – 650 mph range depending on bomb type. I'd not be surprised to be off either high or low. I have books on artillery fire that explain a lot, but they are all several hundred pages of math.
Since the vertical velocity decreases by about 1/3, I'd expect the fall time to be in the 59 second range from 25,000 feet if I try a terminal velocity of, say, 600 mph. I'd therefore expect anywhere from 50 seconds to about 1 minute 10 seconds as a first-order estimate.
CobberKane
Banned
- 706
- Apr 4, 2012
A lot of people really have trouble getting their head around this. I've spoken to experienced shooters who point blank refuse to accept the above example. Gotta admire their self belief, backing their opinions against Newton in this area!
Regarding the terminal velocity of bombs - weren't the tallboy and grand slam supersonic? if so, there must be quite a range according to the purposes of various bombs, from incendiary to armour piercing
GregP
Major
Not sure if the tallboys were supersonic or not, could be so and have heard that in the past with no references to support it, but there certainly WAS a wide range of drag and shapes in bombs, making a single bombsight a complex item. Looking at the shape of a tallboy, it might easily be possible that it was supersonic given the weight and a reasonably high drop altitude. Whether they were supersonic or not, their mass would certainly impart tremendous momentum, p (p = mv), making their penetrating power impressive. The mass was 10.1 times (an order of magnitude higher) that of a 2,000 pound GP bomb which was, in itself, an impressive instrument of destruction.
Many ships were sunk with 500-pound bombs.
Many ships were sunk with 500-pound bombs.
The maths is here for anyone who is interested. Greyman's 25.5 sec from 10,000 feet is pretty much the same as the exemplar used in this exercise.
http://msemac.redwoods.edu/~darnold/math55/deproj/sp09/wilsonpepper/cpandcwfinal.pdf
Extensive tests were done using dummies of the nuclear weapons to be dropped on Japan. Not exactly GP bombs but the variations are less than you might expect. The results are tabulated here.
Cheers
Steve
http://msemac.redwoods.edu/~darnold/math55/deproj/sp09/wilsonpepper/cpandcwfinal.pdf
Extensive tests were done using dummies of the nuclear weapons to be dropped on Japan. Not exactly GP bombs but the variations are less than you might expect. The results are tabulated here.
Cheers
Steve
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There's an altitude vs. time-of-fall graph in this article:
http://legendsintheirowntime.com/Content/1943/Fl_4310_bombsight.pdf
Numbers seem to tie in to Greyman's OK.
Have various other bits and pieces, will try to upload tomorrow, beddy-byes now, kid wants to go running in the morning...
http://legendsintheirowntime.com/Content/1943/Fl_4310_bombsight.pdf
Numbers seem to tie in to Greyman's OK.
Have various other bits and pieces, will try to upload tomorrow, beddy-byes now, kid wants to go running in the morning...
mikewint
Captain
Greg, you need to consider only the bomb's vertical velocity which is zero at time of release. It is also unnecessary to convert to metric measures. Average g over the US is 32ft/s^2 (as I stated earlier it varies with height and latitude). At bomb release Vvert = 0. Once released the bomb will accelerate negatively at approximately 32ft/s for every second of fall opposed by air resistance (impossible to know exactly). The bomb will accelerate (at a decreasing rate until a = 0) to it's terminal velocity (also not constant over the entire fall as air resistance is constantly changing). Once at terminal velocity it will fall at a constant rate until impact.
Without more definitive date we can only make assumptions. Assume the bomb accelerates for 1/2 the distance or 12,500ft then Vvert goes from 0 to 550mph over 12,500ft. The time required to cover this distance is the same as if the bomb had covered this distance at the average velocity of 275mph. Since D = vt; t = 30.99s The bomb then covers the remaining 12,500ft at the terminal velocity of 550mph requiring t = 15.496s. This gives a total time of 46.5s Using the same logic for a more streamlined bomb and a terminal velocity of 650mph we get t = 26.224s and t = 13.112s or a total time of 39.3s
Thus the extremes are 46.5s and 39.3 or most likely the average of the two about 42.9s
As a note, the worlds first programable computer was ENIAC (Electronic Numerical Integrator And Computer) its main function was to calculate arty firing tables. It simply took too long for humans to do all the massive calculations required to predict the landing point of a shell
Without more definitive date we can only make assumptions. Assume the bomb accelerates for 1/2 the distance or 12,500ft then Vvert goes from 0 to 550mph over 12,500ft. The time required to cover this distance is the same as if the bomb had covered this distance at the average velocity of 275mph. Since D = vt; t = 30.99s The bomb then covers the remaining 12,500ft at the terminal velocity of 550mph requiring t = 15.496s. This gives a total time of 46.5s Using the same logic for a more streamlined bomb and a terminal velocity of 650mph we get t = 26.224s and t = 13.112s or a total time of 39.3s
Thus the extremes are 46.5s and 39.3 or most likely the average of the two about 42.9s
As a note, the worlds first programable computer was ENIAC (Electronic Numerical Integrator And Computer) its main function was to calculate arty firing tables. It simply took too long for humans to do all the massive calculations required to predict the landing point of a shell
And G.P.O. engineer Tommy Flowers gets overlooked again
The Mk 1 Colossus ran in November 1943 for the first time. I don't want to start a row about what the term 'programmable' means
Cheers
Steve
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The wiki entry for the bombsight mentions a "Terminal Ballistic Data" file, which I've tried to upload in this post, but the site boots me out, perhaps due to size of the file. It looks like the very highest speeds achieved by US bombs from 35,000 feet are just in excess of 1,200 feet/sec, doesn't seem to vary greatly with weight of bombs.
I have another pdf file with weapons data, which also has terminal velocity curves, maxing out at about 1,050 ft/sec for 500 lbers and 1,075 ft/sec for 1,000 lbers, when dropped from 32k feet.
There's a bombardier's information file at this rather interesting site:
View Items
It has some Navy ATF data, and a small copy of a bomb table referred to when setting data into the Norden, on which I've highlighted relevant time of fall and altitude entries.
I have another pdf file with weapons data, which also has terminal velocity curves, maxing out at about 1,050 ft/sec for 500 lbers and 1,075 ft/sec for 1,000 lbers, when dropped from 32k feet.
There's a bombardier's information file at this rather interesting site:
View Items
It has some Navy ATF data, and a small copy of a bomb table referred to when setting data into the Norden, on which I've highlighted relevant time of fall and altitude entries.
Attachments
For what it's worth, I had a look at my other pdf file (43-odd megs, watermarked to USSBS.com) for Grand Slam and Tallboy data, apparently the GS had a striking velocity from 30k feet of 1,375 feet/sec, which apparently converts to 937.5 mph, so yes, supersonic.
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GregP
Major
Nice paper mhuxt!
The problem is determining the cross-sectional area and all the constants, but it is a pretty decent read.
Thanks!
I was using a text on artillery fire and assuming a zero angle of elevation and the height as distance below the start point where the target is located. It gives the same results.
Hi Mike!
I don't think so ... if you consider only the vertical velocity, you will fail to take into account for the slowing of the horizontal velocity by air resistance and will fall short of you aim point. The TIME OF FALL would be correct if you know the terminal velocity and the equation of it, but the HORIZONTAL TRAVEL DISTANCE would be wrong by a a bit and you'd impact short of target.
The problem is determining the cross-sectional area and all the constants, but it is a pretty decent read.
Thanks!
I was using a text on artillery fire and assuming a zero angle of elevation and the height as distance below the start point where the target is located. It gives the same results.
Hi Mike!
I don't think so ... if you consider only the vertical velocity, you will fail to take into account for the slowing of the horizontal velocity by air resistance and will fall short of you aim point. The TIME OF FALL would be correct if you know the terminal velocity and the equation of it, but the HORIZONTAL TRAVEL DISTANCE would be wrong by a a bit and you'd impact short of target.
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