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Enthusiast Joined: Jan 10, 2007 Posts: 31 Status: Offline

Hello everybody!

Does anyone of you kow what the maximum range is of the dumb bomb MK84 that is used on the F-16?

Regards.

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Quote:

Max Range: 1nm

http://www.harpoonhq.com/encyclopedia/HTML_Files/aircraft_files/Aircraft_db/1748.htm

Enthusiast Joined: Jan 10, 2007 Posts: 31 Status: Offline

Hello,

So I have o understand it as 1 nautical miles right? But it is only about few meters! How it can be?!

Is this relation correct: The international standard definition is: 1 nautical mile = 1852 meters exactly?

Another thing, in which situation we consiter mk84's max range is 1 nm? For example if I drop it as a projectile with an angle aout 15 deg with a peed about 300 m/sec from an altitude 800 m can its range reach about 6 km with is ~= 3 nm instead of 1 nm?

The bomb release mode would have a huge effect on the range. Dive-Toss (DT) would give you the maximum range, but speed and altitude at the time of release are a huge factor when determining range.

Smitty

Enthusiast Joined: Jan 10, 2007 Posts: 31 Status: Offline

Hi TJSmitty!

So is it be ok if I get in he calculations that the range is about 6 km when dropping from 800 m with 300 m/sec? Although in the tables it had been said that the maximum range is 1 nm?

KAIS wrote:

Hello everybody!

Does anyone of you kow what the maximum range is of the dumb bomb MK84 that is used on the F-16?

Regards.

The farther away you lob a dumb bomb, the more inacurate it will be. No one in their right mind would lob a dumb bomb in war today at its maximum range: far too much risk of collateral damage. I can tell you, however, that the stated maximum range of JDAM is 15 nautical miles under ideal circumstances, perhaps more if dropped by an F-22 from 60,000 ft. doing near Mach 2.

Salute!

Bomb range?

Yikes!!

C'mon you yutes, go look at what Pythag-breath showed us in ancient Greece.

So we drop at 6-8,000 feet, or maybe 2500 meters, or three furlongs, or whatever. Bomb range across the ground will be altitude/tan(dive angle) minus a bit for bomb drag. Slant range at release will be altitude/sin(dive angle).

The drag is a player, but not as much as you think. Still, it's a few degrees below where your jet is pointing.

I have personally tossed MK-82's about 4 miles. Wasn't hard, either. And that was not releasing at 500 knots or 45 degrees climb. Dropped them in actual combat at about 2 miles in a loft, using DTOS, from about 1000 feet AGL.

So when we consider the size of the pipper using CCIP or DTOS, and a few mils computer error, we could use maybe 1 mile and a half, or 2000-3000 meters as a good reference. With JDAM the damned thing could be released 10 miles away if you were high enuf and scooting along at the speed of heat. Hence, JDAM was a fighter/attack pilot's dream.

That's my story, and I'm stickin' to it.

Elite 1K Joined: Apr 26, 2004 Posts: 1215 Status: Offline

Gums, my son the genius said you are correct, but I knew that. Thanks.

Purple

Salute!

Well, haze-breath is getting higher on my list of rational posters here. But no kissing up!

Need to tell sons and grandsons and grandaughters and all the really young'ens the value of geometry and algebra and such.

The actual math required to calculate the bomb "trail" for a specific release condition is not trivial. It involves calculus!!!!! "Trail" is the distance behind you that the bomb impacts the earth (hopefully the target). In a vacuum, the bomb would hit directly below you. But the damned things are unpowered and have drag. So they hit behind you. Really low release altitudes you might see 10, 20 or 50 feet of 'trail'. At higher altitudes like the B-52's used, you might see a thousand feet.

The drag of the bomb is proportionate to the speed thru those pesky molecules of air. Imagine that? Faster means more drag. Uh oh. The bomb slows. So then the drag is less. Sheesh, this is getting complicated. At some point, the gravity forces and the aero forces equal each other and from then on the sucker just drops at a constant airspeed, but still descending at 32.2 ft per second squared (almost ... remember that drag factor). To get a bomb to hit perfectly vertical it has to be dropped from the ionosphere, heh heh.

The F-117 dudes get their LGB's to hit vertical by designating beyond the tgt and then moving the laser spot back, short of the tgt, then finally exactly on the tgt. This technique "shapes" the trajectory of the bomb. The JDAM guidance algorithms do something like that in order to get a vertical "end game". GPS is still not super for vertical positioning, so remove "vertical" from the problem, huh?

Bottomline: All the math crapola at school may help you one day, whether it's working up the guidance module for a bomb or figuring out the length of a 2x4 that spans the corner of your new deck.

Gums sends ...

F-16.net Editor Joined: Sep 23, 2003 Posts: 2862

Always great to see a post from our esteemed warrior/engineer Gums.

( and I would never infer that you sat in on classes with Pythagoras or knew him personally. Laughing

)

Enthusiast Joined: May 31, 2005 Posts: 71 Status: Offline

Small correction to Gums' excellent description: Drag force is actually proportional to the square of the velocity. F = (pACV^2)/2. C is the dimensionless drag coefficient, A is the effective cross-sectional area, V is the velocity of the object (bomb) through the fluid (air), and p is the density of the fluid. C is probably the only constant in that equation since the velocity will vary as Gums noted, the density of air will change in the trajectory of the bomb, and the cross-section presented to the fluid would change as the bomb changes attitude. Of all those, however, velocity is still the major factor because of the exponent. (Note that this is an ideal equation that assumes the fluid striking the cross-section is halted. That doesn't happen here since the air will flow around the bomb as it falls.)

As Gums also noted, there will come a point where the force of gravity is exactly opposed by the drag force and the bomb will be falling at a constant (aka terminal) velocity. I think he meant to say that gravity will continue to act on the bomb at 32 ft/sec^2, not that the bomb will continue to fall at that rate -- that would mean the bomb is accelerating even though it's reached its terminal velocity.

The answer is simple...no geometry required.

With AAR the "range" of a MK-84 on an F16 is limitless!

Gus

Salute!

What Gus said!!

And Loomis is correct - when Pythagorus taught me, he didn't know about the gravity effects or terminal velocity. And I have to tell ya, when I met Newton, that sucker had ALL the answers until this other dude a few hundred years later came up with realtivity!!

One thing about terminal velocity that Loomis brings to the classroom is the fact that in the vacuum example, the bomb would probably hit sooner, as it would keep accelerating by that 32.2 ft per sec per sec. So it would descend more rapidly than in the atmosphere.

What always amazed me was that we mortal pilots could actually hit within 10 meters of a tgt using only seat of the pants and the fixed reticle - TLBAR. That's the basic A-37's 30 degree release from 2,000 - 2500 feet or so. My average during one three-month period when I was a yute was about 35 feet. My most infamous mission was four bombs, starting waaaaaay behind the power curve. First one was 100 feet, second was 20 feet. Next two were shacks. So I had a 30 foot average and won the bet that day. The other guy had dropped three shacks, then "shanked" the last one with a 160 footer, heh heh. Never did live that down, either.

later,

Enthusiast Joined: Jan 10, 2007 Posts: 31 Status: Offline

Hi everybody!
I want to comment about what Loomis said. I agree with him in most of things but he said that the only constant parameters is the drag coefficient, but as I know, drag coefficients depends on reynolds number, hence it depends on the jet velocity, so it also a variable not a constant. This is what I think but maybe I am wrong.

Also he said that A is the cross section that changes due to bomb attitude. But I think that after dropping the bomb, it will get down so I can assume that the cross section that faces the fluid (which in this case is air) is a semi rectanble, its longer side is the length of the bomb and its shorter side is the diameter of the bomb. Can I assume that, what do you think?

Regards

I would agree with the fact of the weapon having an unlimited range when strapped to the F-16. I also believe they can be placed anywhere the user wishes.

To put it bluntly: If you're in range, you'll wish you weren't! LMAO

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Maximum range of the MK84