Author Topic: Is my math wrong ? (Read 20202 times)

jim gilmore · « **on:** April 11, 2008, 11:06:25 AM »

Ok i'm going to throw some numbers here so please bear with me.
a real p-51 has wing span 37.4 ft. times 12 = 444.8"
a 52" model divided by 444.8 yields .1169
(just doing 4 places for the moment)
p51 to empty wt was 6280 lbs. so multiply that by .1169 we get 734.
So a true scale model should weigh 734 lbs.
I know that's not right but am I doing something wrong scaling wise?

John Miller · « **Reply #1 on:** April 11, 2008, 11:19:11 AM »

You may want to consider figuring the wing loading. Also, when scaling, based on wing area, don't forget to use the sqaure root of the number.

Trostle · « **Reply #2 on:** April 11, 2008, 02:09:59 PM »

Interesting question.

John Miller started down the path to a right answer by suggesting that you needed to compare wing areas which a square function. Then, when comparing volume, there is a cube function. Let's take a somewhat more simple example. For instance, take a model that is 1/10 full size (or 10%, or 1.2" = 1'). The wing area of the model will be 1/100 of the full size airplane (or 1.0%). The volume of the model will be 1/1000 of the full size airplane (or 0.1%). So, if we have a 1/10 model of a full size 6,200 pound Mustang (a model with roughly a 43 inch wing span) and it is the same density as the full size Mustang, then the model will have 1% of the area and would weigh 6.2 pounds. This would be on the heavy side, even for a scale model of that size. The thing that makes full size airplanes work with the comparitively heavier (often much heavier) wing loadings is the Reynolds Number effect that when airfoils get bigger, they can carry a heavier wing loading.

As an interesting aside, the "density" of many modern jet fighters is about the same as a lump of coal. Now, many think that flying our CL models is equivalent to flying rocks on the end of a string, but at least we have a more favorable wing loading for most of our models than if they were carved from a lump of coal.

Food for thought.

Keith Trostle

Paul Smith · « **Reply #3 on:** April 11, 2008, 03:57:33 PM »

True, true, and true.

Linear dimensions are scaled linear.

Areas are scaled to the second power (squared)

Volume is scaled to the cube (third power).

So where does that leave AIRSPEED?

A "lead sled" fighter needs to go almost 200 MPH to get off the ground and 500 to do a loop.

A stunt model takes off at about 20 and stunts between 50 and 60 MPH.

Dynamic similarity is not so easy. That's why the men in Seattle make the BIG BUCKS.

jim gilmore · « **Reply #4 on:** April 11, 2008, 04:15:40 PM »

Ty all I knew I was missing something.
Now it makes sense.

phil c · « **Reply #5 on:** April 11, 2008, 06:05:42 PM »

The scale guys are arguing this all the time, since the rules have something in about realistic or appropriate speeds for the flying. Very hard to do, since the size, speed, area, volume, and the air all have different scaling factors. Small(<6 ft or so) models have to be very light to get anywhere near scale looking flight. CL models need to be even lighter because of the constrained flight area.

don Burke · « **Reply #6 on:** April 11, 2008, 09:11:50 PM »

Keith,
Nothing to do with models, but:
The modern jets are the perfect example of "Give it enough thrust and you can fly anything, and you don't need wings"

Trostle · « **Reply #7 on:** April 12, 2008, 01:48:43 AM »

Quote from: don Burke on April 11, 2008, 09:11:50 PM

Keith,
Nothing to do with models, but:
The modern jets are the perfect example of "Give it enough thrust and you can fly anything, and you don't need wings"

Don,

Back in the 70's, I saw a sign above an engineer's desk at McDonnell in St. Louis that showed an enlargement of a dictionary type entry for "Phantom". After showing the pronunciation for the word and showing that it was a noun, the definition went something like "The triumph of thrust over aerodynamics."

Interesting to note weights of the P-51D, one of the top fighters at the end of WW II (1945) and the F-4E, our front line fighter from the 60's to mid 70's:

P-51D: empty 7,125 lb, normal loaded 10,100 lb, max 12,100 lb (wing area 233 sq ft, 1,695 HP Merlin)

F-4E: empty 30,238 lb, max 61,795 lb (wing area (530 sq ft, two J-79 engines, each rated at 17,900 lb thrust max afterburning.)

The F-16 and the F-15 in particular represent a significant improvement in thrust to weight over the F-4 while the F-22 is off the charts.

(Numbers are general, not included here to start any kind of debate on absolute accuracy or currency)

Keith

Al Rabe · « **Reply #8 on:** April 14, 2008, 07:23:16 PM »

With regard to Keith's numbers for the F-51, the V-1650 is displacement, not horsepower. The V-1650-3, -7 and -9A are all rated at 1490 horsepower. As for weights, there are no specified weights published for the airplane. The P-51 D & K reference manual give the (dry) weight as approximately 7,000 lbs. The T.O. 1F-51D-1 (pilot's handbook) lists the normal take-off weight with full internal fuel and armament as approximately 9600 lbs. The maximum take-off weight with full internal fuel, two externally mounted 1000 lb bombs and six 5 inch rockets is approximately 12,500 lbs.

Al

Trostle · « **Reply #9 on:** April 15, 2008, 03:05:06 AM »

Quote from: Al Rabe on April 14, 2008, 07:23:16 PM

With regard to Keith's numbers for the F-51, the V-1650 is displacement, not horsepower. The V-1650-3, -7 and -9A are all rated at 1490 horsepower. As for weights, there are no specified weights published for the airplane. The P-51 D & K reference manual give the (dry) weight as approximately 7,000 lbs. The T.O. 1F-51D-1 (pilot's handbook) lists the normal take-off weight with full internal fuel and armament as approximately 9600 lbs. The maximum take-off weight with full internal fuel, two externally mounted 1000 lb bombs and six 5 inch rockets is approximately 12,500 lbs.

Al

Gee thanks Mr. Rabe. I am sure that there are any number of different specifications that have been published for the P-51 that will list different numbers. I probably have at least 10 references in my library that give some numbers on the P-51D. Though the book I used may not necessarily be the most authoritative (certainly not like a factory manual), I used William Green's "Volume 4, Fighters, Warplanes of the Second World War, 1961" that listed the power plant for the P-51D-25-NA as "One 1,695 hp. Packard V-1650-7 Merlin twelve-cylinder Vee liquid-cooled engine". (My post made no mention of the 1,650 horsepower as you suggested.) It is possible that the reference I quoted is incorrect as another William Green book "The Great Book of Fighters 2001" lists the V-1650-7 Merlin used on the P-51D with 1,450 hp. I have other references, most of which are probably more authoritative than the 1961 Green book that show the Packard Merlins used on the P-51D rated at different ratings from 1,490 hp to 1,590 I suppose that I should have researched this important matter more thoroughly before posting published information. (It is interesting that another book, "Aircraft of the Fighting Powers, Vol 7" lists the power Plant for the P-51H as the "Packard-built Rolls-Royce Merlin V-1650-11" with "over 2,000 hp at rated altitude using water injection". The weights quoted come from the earlier Green book and are basically the same in the later Green book.

I think it is really good that one with your vast experience with full scale P-51 operations can take the time to show us that you probably have operating manuals so that the information that was included in my post as general information can be more precisely defined. Thanks.

Keith

Al Rabe · « **Reply #10 on:** April 15, 2008, 09:54:13 AM »

Keith,

Your welcome. And the wing area happens to be 240.06 sq. ft.

Al

Larry Renger · « **Reply #11 on:** June 06, 2008, 04:15:58 PM »

I don't know how many times I need to write this, but Ron St. Jean covered it in the dark ages in MAN and I updated it within the decade.

Weight = K * wing area * wingspan

That formula works so well over huge scale ranges that it is frightening.

Plug in the numbers for the original model to determine "K" for that model type, then solve for the weight with the new model's span and area. IT IS THAT SIMPLE!!!

Robert McHam · « **Reply #12 on:** June 06, 2008, 08:22:46 PM »

Scaling can lead to some interesting number conclusions to be sure. And clever marketing as well.
Anyone remember the ads (I think it was Hot Wheels) that claimed scale speeds of over 200 mph?

As for the flying models. if one were to make a one tenth mustang he may also consider the horsepower as well. If the Merlin supplied 1,490 horses then the one tenth scale model would need 149 hp!

Suppose also that the engine rpms were one tenth? Or the energy that is contained in the fuel itself were a tenth? How about the aluminum skins being a tenth in thickness? There is more but you get the idea.

Interesting numbers indeed.

Some things must remain constant and be adusted for accordingly. Air density for example...

Too many numbers can just make the whole thing crazy! Just another reason that the bumblebee cannot possibly fly when you run the numbers and yet it does quite well.
This is simply a myth and is explained in this page:
http://en.wikipedia.org/wiki/Bumblebee see Bumblebee myths.

Robert

Ralph Wenzel (d) · « **Reply #13 on:** June 07, 2008, 03:55:13 AM »

Now, let's demand that that full-scale P-51 try to approach our mythical CLPA 5' turn radius . . .

That one should drive the number-crunchers absolutely bonkers

Trostle · « **Reply #14 on:** June 22, 2008, 02:22:52 PM »

Quote from: Robert McHam on June 06, 2008, 08:22:46 PM

(clip)

As for the flying models. if one were to make a one tenth mustang he may also consider the horsepower as well. If the Merlin supplied 1,490 horses then the one tenth scale model would need 149 hp!

(clip)

There is more but you get the idea.

Interesting numbers indeed.

(clip)

Robert

If you are going to "play" with numbers, you should avoid intentional distortion of facts and at least keep area and volume considerations in mind. The total displacement of that 1/10 engine is reduced by 1,000. Just as was explained in the weight discussions earlier in this thread, to compare the power required to fly a 1/10 Mustang, the comparable engine power rating would be 1/10 x 1/10 x 1/10 or 1/1000 of the full scale version or 1.49 horsepower. Certainly, this is not beyond what would be reasonable to power a 1/10 scale model of the Mustang.

Keith

(edit for clarification)

Kim Mortimore · « **Reply #15 on:** June 22, 2008, 08:35:28 PM »

Quote from: Larry Renger on June 06, 2008, 04:15:58 PM

I don't know how many times I need to write this, but Ron St. Jean covered it in the dark ages in MAN and I updated it within the decade.

Weight = K * wing area * wingspan

That formula works so well over huge scale ranges that it is frightening.

Plug in the numbers for the original model to determine "K" for that model type, then solve for the weight with the new model's span and area. IT IS THAT SIMPLE!!!

It seems to me there is a simpler method than that, using a single parameter only (wingspan, or length for example). Wouldn't the scaled up (or down) weight be a function of the ratio of the chosen parameter cubed? For example, a 48" span stunter weighing 40 ozs enlarged to 60" would be:

48 x 48 x 48 = 110,592

60 x 60 x 60 = 216,000

216,000/110,592 = 1.95 x 40 oz = 78 oz

Bearing in mind that this is theoretical exercise with probably only ballpark relevance to actual builds, since we can seldom predict actual finished weights exactly in any case.

Kim Mortimore

jim gilmore · « **Reply #16 on:** June 23, 2008, 07:22:21 PM »

i'm almost sorry I asked. My head hurts thinking about this.

EddyR · « **Reply #17 on:** June 24, 2008, 08:59:16 AM »

I think another thing about scale flying is how fast we fly compaired to the real aircraft. We have two of the US scale team members in our club and in talking to them I found out scale models fly very fast compaired to the real aircraft. Even at slow speed they are a long way out of the scale speed.
Ed Ruane

L0U CRANE · « **Reply #18 on:** June 30, 2008, 06:14:40 PM »

...Just to tickle the "realistic appearance in flight" a bit more...

Full size airplanes, flying at - say - cruise speed, move forward a certain numbers (or fractions) of fuselage length per second. Flight airspeed would look "accurate" if a model moved at a similar rate of progress through the air, and also, perhaps, at a relatively comparable, fairly steady, distance from the observer.

Example: a 24.5' long Piper J-3 lands at 38 MPH. That's about 56 fps, or about 2.25 fuselage lengths per second.

A model J-3, 3' long, "flying?" at 2.25 fuselage lengths per second is blazing along at almost SEVEN feet per second...

... and cruises at 73 MPH, or about 106 fps, or 4.33 fuselage lengths per second. The 3' long model has to streak around at about 13 fps...

A bumblebee may overcome the laws of aerodynamics by brute force and lack of proper college level training, but I think scale modelers have to settle for a general impression of what looks as leisurely as a cruising or landing "real" J-3 in terms of actual airspeeds that allow it to happen.

A Mustang should look "fast" and a J-3 Cub should look "slow" - in a relative, undefined sliding scale that gives us a believable "impression" of how the model's prototype "should" look in flight.

Perhaps other cues have a role, here. A J-3 on landing approach is buoyant, floating, a bit bouncy and sensitive to disturbance from turbulence and cross-wind. A Mustang would be expected to bore on in with a minimum of such flutteriness... A scaled twin-engine WW2 medium bomber should not wallow around like an RC trainer flown by a hamhanded RC newby... Things like that...

IMHO...

Pinecone · « **Reply #19 on:** February 02, 2009, 01:47:14 PM »

Quote

The modern jets are the perfect example of "Give it enough thrust and you can fly anything, and you don't need wings"

Not really. Take the Northrup T-38A Talon. 25 foot wingspan, about 50 feet long.

You rotate at 135 knots, and lift off at around 170. If you immediately fly a single engine patter (lost an engine right after takeoff) you would flying down final at 175 knots.

Climb at 300 knots, because the normal 250 below 10,000 feet is scary.

Normal loop is flown at a 450 knot entry speed and it about 9,000 feet in diameter.

And the glide ratio is greater than a Cessna 172.

Cessna 172 - 8.8 to 1

T-38 - 9.6 to 1

Of course the C-72 is at like 50 - 60 knots, the T-38 is at 240 knots.

jim gilmore · « **Reply #20 on:** February 02, 2009, 01:57:59 PM »

Isn't glide ration a kind of misnomer ?
a rock has a glide ration relative to how hard you thjrow it and how soon it hits the ground ?
Does the t-38 remain at 240 during that 8.8to 1

rustler · « **Reply #21 on:** February 02, 2009, 02:40:00 PM »

Slightly off topic - back in 1980(?) Steve Smith produced the first successful flying wing F2C model and won the World Champs with it. It led the way for every F2C model since. I had the honour(?) of drawing the plan for it. A tame aerodynamics expert looked it over and pronounced it a triumph of horsepower over design.

Dick Fowler · « **Reply #22 on:** February 02, 2009, 02:59:18 PM »

Quote from: jim gilmore on February 02, 2009, 01:57:59 PM

Isn't glide ration a kind of misnomer ?
a rock has a glide ration relative to how hard you thjrow it and how soon it hits the ground ?
Does the t-38 remain at 240 during that 8.8to 1

Not really. Every aircraft has and optimum glide speed (usually found in the POH). So you arec expected to set up the aircraft to achieve this value. Once done, a quick look at the altimeter will give you a good idea of how long you have before impact! So yes it is ho[pe that you can maintain the 240 Knot airspeed. Lift/Drag is the big deal here.

Howard Rush · « **Reply #23 on:** February 05, 2009, 10:36:20 PM »

Glide ratio of a rock is zero: no lift / some drag. You could get a little lift by spinning it.

L0U CRANE · « **Reply #24 on:** February 06, 2009, 04:48:07 PM »

Hmm, is the initial rise on a ballistic trajectory due to lift?

A very fast, heavy airplane has a lot of momentum in the direction it was going when the thrust stopped, but that path will be altered by gravity, eventually. Sooner or later... I guess we should ask Captain Sullenberger about that.

Dick Fowler · « **Reply #25 on:** February 06, 2009, 07:26:29 PM »

Lou..."Hmm, is the initial rise on a ballistic trajectory due to lift? "

Rhetorical... right? Never mind. Having followed most of your post over the years... I think that makes my question rhetorical too!

Pinecone · « **Reply #26 on:** February 09, 2009, 10:23:24 PM »

No, it is truely a glide ratio. The wing is flying due to airspeed, but instead of an engine providing thrust, you go down sacrificing altitude (potential energy) for speed (kinetic energy).

So yes, the T-38 is flying forward 9.6 feet for every foot it drops. If you start out with more than 240 KIAS, you could pull the nose up and convert that kinetic energy (speed) to potential energy (altitude) as you intercept the best glide of 240 knots.

And it will have the same 9.6 to 1 ratio at 50,000 feet and at 100 feet.

Same thing with a glider at 50 to 1.

A rock or a bullet loses speed as time goes on. But they are dropped under the influence of gravity the entire time. So at the beginning they are moving forward faster than down, and later they are moving down faster and faster, while losing forward speed due to drag.

minnesotamodeler · « **Reply #27 on:** February 10, 2009, 05:41:58 AM »

Is it true (as I've heard) that if you fire a bullet aimed exactly horizontal, and drop a bullet from the same height at the same time, they will hit the ground simultaneously? Showing, I guess, that gravity isn't affected at all by velocity?

Dick Fowler · « **Reply #28 on:** February 10, 2009, 07:24:56 AM »

Quote from: minnesotamodeler on February 10, 2009, 05:41:58 AM

Is it true (as I've heard) that if you fire a bullet aimed exactly horizontal, and drop a bullet from the same height at the same time, they will hit the ground simultaneously? Showing, I guess, that gravity isn't affected at all by velocity?

Yes that's true. Most times this classic problem is conducted in a vacuum to eliminate drag.

What is really happening here is that the force of gravity begins to act on the fired bullet and the dropped bullet at the same time. Both bullets will accelerate toward the ground at the same rate with no regard for the forward velocity of the fired bullet. The gun didn't impart any force toward or away from the ground so the only force acting in a vertical direction is gravity. Both bullets are acted upon by gravity in the same manner and thus travel the same distance to the ground at the same rate and arrive at the same time.

Probably more than you want to know!

minnesotamodeler · « **Reply #29 on:** February 10, 2009, 09:08:41 AM »

Thanks, good to have confirmation...kinda hard to get one's mind around the concept.

jim gilmore · « **Reply #30 on:** February 10, 2009, 02:20:23 PM »

First off let me say that this thread has lived way longer than nI ever expected when I was only wondering if my method of scaling was correct and why It didn't scale the wiegjt as well as the size.

Second question and maybe I'm mis-understanding the actual definition of glide ratio.
So a bullet, arrow, rocket, or any unguided item has a zero glide ratio even any out of control aircraft that doesn't have a control to set it in flight to a glide position ?

Dick Fowler · « **Reply #31 on:** February 10, 2009, 02:56:21 PM »

Quote from: jim gilmore on February 10, 2009, 02:20:23 PM

First off let me say that this thread has lived way longer than nI ever expected when I was only wondering if my method of scaling was correct and why It didn't scale the wiegjt as well as the size.

Second question and maybe I'm mis-understanding the actual definition of glide ratio.
So a bullet, arrow, rocket, or any unguided item has a zero glide ratio even any out of control aircraft that doesn't have a control to set it in flight to a glide position ?

Jim, probably a better way to look at this glide ratio thing is to consider that if the object can't produce lift then it doesn't have a glide ratio. Rocks, bullets, arrows etc. for the most part are ballistic objects. Although there may be a very small lift component for these items, it really isn't significant to influence the "flight path" to any degree. For the most part, the initial velocity and angle of elevation at launch has the greatest impact on its trajectory. Well I guess drag should be thrown into the mix to be more accurate.

Trostle · « **Reply #32 on:** February 10, 2009, 03:21:17 PM »

Quote from: jim gilmore on February 10, 2009, 02:20:23 PM

First off let me say that this thread has lived way longer than nI ever expected when I was only wondering if my method of scaling was correct and why It didn't scale the wiegjt as well as the size.

Second question and maybe I'm mis-understanding the actual definition of glide ratio.
So a bullet, arrow, rocket, or any unguided item has a zero glide ratio even any out of control aircraft that doesn't have a control to set it in flight to a glide position ?

Jim,

Maybe I am completely off base here, but from your questions, I think you are trying to read too much into the "glide ratio" term. The glide ratio of any aircraft is simply a measure of how far it will sink for a unit of forward direction in a zero power situation. It is a measure of performance, much like other terms used to describe the performance of an aircraft like climb rate, landing speed, maximum ceiling, maximum speed, stalling speed, takeoff distance, landing distance, terminal velocity, endurance, cargo capacity, and so on. The glide ratio is a function of its design just as all of these other parameters. The glide ratio is constant for any given design. To mention the glide ratio of a spinning or out of control aircraft is really meaningless.

Keith

jim gilmore · « **Reply #33 on:** February 10, 2009, 03:41:39 PM »

OK So Like I am Saying arrows do not glide and it / Glide ratio assumes that the airplane is put into a position to glide...

Trostle · « **Reply #34 on:** February 10, 2009, 04:14:42 PM »

Quote from: jim gilmore on February 10, 2009, 03:41:39 PM

OK So Like I am Saying arrows do not glide and it / Glide ratio assumes that the airplane is put into a position to glide...

You are starting to get the concept. Rocks and such do not glide unles they are shaped in some way to create some aerodynamic lift as they travel through the air.

In a way, you are correct about what you say about glide ratio, but you still seem to want to think of it differently than what the term really represents. The glide ratio of an aircraft is a number. The aircraft glide ratio is what it is, whether the airplane is taking off, or cruising, or climbing, or landing, or diving, or looping or setting on the ground or whatever. In order to measure its glide ratio (as opposed to its design glide ratio, indeed the airplane would have to have some altitude, attain some forward velocity and be in a gliding/no power situation. For a given configuration, the glide ratio for that aircraft will always be the same. Again, glide ratio is a measure of an aircraft's ability to glide just as takeoff speed is a measure of how fast an aircraft must fly to takeoff. Takeoff speed sort of "assumes that the airplane is put into a position to" takoff.

Keith

Dick Fowler · « **Reply #35 on:** February 10, 2009, 08:25:21 PM »

Jim, another thing to think about when considering the glide ratio of an aircraft is that the forward velocity and the sink rate are set up to be constant. So a properly set up "power off" airplane will sink a predictable distance for each unit of distance it moves forward. These two values give you the "ratio" of how many distance units forward it will fly for every unit of distance it drops.

"Sully" had it down to a a fine point... he knew exactly how far he could fly given all the variables... what a great job that guy did!

Larry Renger · « **Reply #36 on:** February 24, 2009, 10:20:56 PM »

If you consider everything, the dropped bullet will hit the ground first. The earth is curved, so the fired bullet will have farther to fall, depending on its velocity. Fast enough, and it will never come down, but exit to space and beyond. Of course it will melt first ....

minnesotamodeler · « **Reply #37 on:** February 25, 2009, 05:47:53 AM »

Quote from: Larry Renger on February 24, 2009, 10:20:56 PM

If you consider everything, the dropped bullet will hit the ground first. The earth is curved, so the fired bullet will have farther to fall, depending on its velocity. Fast enough, and it will never come down, but exit to space and beyond. Of course it will melt first ....

Surely by just a miniscule fraction of a second, being we're talking about what, a couple hundred yards of Earth's curvature? Standing on a 3" anthill while dropping the bullet would make more difference, right?

Al Rabe · « **Reply #38 on:** March 01, 2009, 07:27:56 AM »

Glide angle is determined by lift and drag. If the airplane is flown at its best lift/drag speed it will glide farther. Here is a quote from the F-51D Pilot's Manual (I flew them too) on a chart for power off gliding distances. "You will glide approximately 14.5 statute miles for each 5000 feet of altitude lost by holding 175 mph-no-wind condition." That is almost three miles of glide for each thousand feet of altitude lost or, a glide ratio of 15.312' for each 1' of altitude lost.

Nearly all jet airliners glide about three miles for each thousand feet of altitude lost. For example, it takes about a hundred miles to lose 33,000 feet at idle thrust and this is true for the B-707, B-727, B-747, DC-8, A-320 and L1011s that I flew, and this worked OK for a normal high speed descent. We always used this for a rough estimate for beginning descent. Idealy, we would stay at cruise altitude as long as possible where miles per pound of fuel burn were much greater than at low altitude. The ideal would be to fly the descent and approach at flight idle, spooling up only when configuring for landing. The Lear Jet doesn't manage quite 3 miles/1000, more like 2 3/4 miles/1000.

Now, instead of using "rules of thumb" for calculating performance, flight director computers do it. One simply tells the computer the "cleared to" low altitude fix and the altitude and speed needed to cross it. The computer tells the crew when to begin the descent. The flight director is then followed which commands the pitch attitude to give optimum Mach and Indicated Airspeeds to make the crossing as cleared with minimum burn.

5,280' (one mile) times 3 equals 15,840' (three miles) for each 1000' lost for a glide ratio of 15.840 to 1. And yes, this is a heck of a lot better than a C-172 because Jets and Mustangs are so much cleaner aerodynamically. Still, to put this into better perspective, Jet airliner gliding ratios are only 1/3 that of high performance sailplanes.

Al

Ted Fancher · « **Reply #39 on:** March 08, 2009, 10:44:57 PM »

Quote from: Dick Fowler on February 10, 2009, 07:24:56 AM

Yes that's true. Most times this classic problem is conducted in a vacuum to eliminate drag.

What is really happening here is that the force of gravity begins to act on the fired bullet and the dropped bullet at the same time. Both bullets will accelerate toward the ground at the same rate with no regard for the forward velocity of the fired bullet. The gun didn't impart any force toward or away from the ground so the only force acting in a vertical direction is gravity. Both bullets are acted upon by gravity in the same manner and thus travel the same distance to the ground at the same rate and arrive at the same time.

Probably more than you want to know!

Dick,

Edit: Oops, I see Larry already brought this subject up. Fuggedit.

Just being a devil's advocate here since I've never really considered the question before.

Technically, since the earth is a sphere you can't really fire a bullet parallel to its surface, only tangent to it. By definition I believe the initial velocity of a projectile fired tangent to a sphere will be "away" from the "receding" surface, not parallel to it. Doesn't it, therefore, climb as a result of the initial impulse and, thus, have further to "fall" once drag reduces the velocity to a point that gravity overcomes the old Newtonian desire to continue in a straight line, etc. (That was Newton, wasn't it? Or is Oldtimers raising its ugly head again). Better yet, what would be the effect of the same experiment on a sphere with no atmosphere ... say, the moon?

Ted

Brian Hampton · « **Reply #40 on:** March 09, 2009, 12:52:58 AM »

Just to add to this hypothetical rock and bullet thing, if you fire a bullet parallel/tangent with the ground (let's call it at right angles to the gravitational direction at the muzzle) then the bullet is always influenced by gravity just as a dropped rock is. With one minor complication. Because the bullet has a forward velocity it's following a curved path which adds a tiny bit of centrifugal force that counteracts a tiny bit of the Earth's gravity so it will take fractionally longer to hit the ground. A satellite in orbit feels a centrifugal force that exactly equals gravity (the centripetal force) at whatever height it may be orbiting. A bullet travels about 10 times slower than a Low Earth Orbit satellite so the centrifugal force it feels will be about 100 times less so effectively it feels the force of gravity to be about 99% what it actually is at the surface. So it falls slower by around 1% than the stone.

If I remember my high school physics correctly

.

jim gilmore · « **Reply #41 on:** March 09, 2009, 07:15:59 PM »

Ihope this dies, soon it's getting deep in here . 1% hmmmm maybe .01 % doesn't the space shuttle go something like 219 miles min `or somthing obscene to remain in orbit .
1 here are a few questions to finish this thread that might actually point out she scale of things.
If a bullet were to go at 1200 feet per second was fired level from 6 feet how far would the bullet fall before it hits thhe groung ?
2 What is the diameter of the earth(appromatly because it is not round) ?
3 How long would it take the bullet to drop 6 feet.
4 How far would the bullet travel horizontally?
5 compute the diamter of the earth in inches and determin the distance from a tanget to the earth the distance in #4 .

Serge_Krauss · « **Reply #42 on:** March 10, 2009, 08:41:12 AM »

I'll let someone else do the HS physics for Jim's last questions - doctor's apt. in an hour. However, I need to differ with Brian. Centrifugal force does not act on the bullet. Centripetal force is the force (gravitational) in this case, that acts on the bullet to change its direction. Changing direction is one type of acceleration; changing speed is the other. Any combination of the two is an acceleration. So gravitational pull is the centripetal force that accelerates the bullet.

Centrifugal force is the reaction force of the bullet on the earth. Every object exerts a gravitational force. For any pair of objects (called a reaction pair) there is gravitational attraction between them - equal in magnitude but opposite in direction. Obviously the Earth does not accelerate much compared to the bullet, since the magnitude of acceleration is inversely proportional to their masses (Newton's second law).

So centrifugal and centripetal forces are equal in magnitude, opposite in direction, and act on different objects. Contrary to what someone took it upon himself to type into the Wikapedia article, Centrifugal force is NOT an imaginary force. It is very real and easily measured. It just does not act on the bodies that person assigns to it; in fact his very own first reference, Halliday and Resnick, asserts that these forces are real and act between reaction pairs of objects, and Newton's Third Law requires them both to exist between different objects.

In a CL context, not getting overly nit picky about nuclear binding forces, weak subatomic attractions, other electromagnetic forces, etc, the pilot pulls on the model with a centripetal force and in accordance with Newton's third law, the plane pulls on the pilot with an equal but opposite centrifugal force. Required by Newton's third law of motion, these forces are also actually named for their directions.

So this is all according to Newtonian physics theory and subject to revision according to relativistic effects and any further refinement to our understanding of "how things work", but it gets satellites into orbit without much correction and certainly designs good advanced aircraft. Now THAT was more than anyone asked for or wanted..."but I'm feeling much better now." Ha!

SK

Dick Fowler · « **Reply #43 on:** March 10, 2009, 09:01:07 AM »

Serge, still have my original college two text set of the Halliday and Resnick text books from the 60's. Nice for review and refreshing once in awhile... especially when old age starts to creep in!

Serge_Krauss · « **Reply #44 on:** March 10, 2009, 04:30:09 PM »

Dick-

You bet! ...and there are some fond memories of that exciting, if sometimes scary time.

SK

Pinecone · « **Reply #45 on:** March 15, 2009, 06:19:40 PM »

Quote from: Trostle on April 15, 2008, 03:05:06 AM

Gee thanks Mr. Rabe. I am sure that there are any number of different specifications that have been published for the P-51 that will list different numbers. I probably have at least 10 references in my library that give some numbers on the P-51D.

The pilot's flight manual trumps all other references.

LOTS of "authoritative" sources are wrong. Sometimes because the authors ask the wrong question, or use the wrong number.

One of my favorites is the "top speed." I flew the A-10. Almost every source quotes a top speed of 450 knots. NO way will an A-10 do 450 knots in level flight for any length of time. Maybe straight down from high enough.

450 is the barber pole (red line), or never exceed speed. Not the faster it will actually go.

BTW 365 KIAS is the fastest I ever saw.

jim gilmore · « **Reply #46 on:** March 15, 2009, 08:46:24 PM »

Hey, Nice to know of somebody that flew the warthog. I built them.

Pinecone · « **Reply #47 on:** March 16, 2009, 04:59:11 PM »

Where?

I was with MDANG and we had most of our aircraft fresh off the assembly line.

Ted Fancher · « **Reply #48 on:** March 19, 2009, 01:57:06 PM »

Quote from: Pinecone on March 15, 2009, 06:19:40 PM

The pilot's flight manual trumps all other references.

LOTS of "authoritative" sources are wrong. Sometimes because the authors ask the wrong question, or use the wrong number.

One of my favorites is the "top speed." I flew the A-10. Almost every source quotes a top speed of 450 knots. NO way will an A-10 do 450 knots in level flight for any length of time. Maybe straight down from high enough.

450 is the barber pole (red line), or never exceed speed. Not the faster it will actually go.

BTW 365 KIAS is the fastest I ever saw.

Just a thought, Pinecone. 365 KIAS at 10K feet and 50F = about 445 KTAS (Knots True Air Speed). I imagine performance data for public consumption generally is expressed in TAS because it is more straight forward from the layperson's perspective. As we know, Indicated Air Speed is nothing more than the rate at which the pitot tube impacts air molecules ... of which there are fewer as pressure decreases (altitude or temperature increase). Ergo your "true" speed through the air mass is greater than is indicated by the air speed indicator.

Ted Fancher

jim gilmore · « **Reply #49 on:** March 19, 2009, 04:30:28 PM »

Farmingdale, Long Island. New York. 1977-1987 .
Work on 747 control surfaces
A10
T46