F-16 Take off distance
ram...@mit.edu
Hi,
I am interested in finding out the minimum take off run for a F-16 A/B
(Block 15). How can I find this?
Please reply by email : ram...@mit.edu
Thanks,
Ramana
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José Barahona da Fonseca
Hi,
The thrust of F-16 engine is about 79.2kN, it weights about 7000Kg so
we will have that the acceleration is a=79.2/7*9.81=111 m/s^2. Now we
must know the minimum takeoff speed, say 250Km/h (I'm not sure, It's
only a guess). So the time to takeoff will be, neglecting the Total drag
which is much less than 79.2kN, t=250/3.6/111=0.626 s and the takeoff
distance will be 1/2 a t^2=22 meters(!!)...the real value is a litlle
bit greater than 22 meters because the Total Drag of F-16 increases
with V^2: Drag(V)=1/2 rho Sx Cd V^2+InducedDrag(V), rho=1.2256
Kg/m^3...to calculate the exact takeoff distance we must solve a
complicated differential equation: dv/dt=a-gsin alpha-Drag(v)/m.
To obtain the landing distance we must solve a similar differential
equation: dv/dt=amin+gsin alpha-Drag(v) where alpha is the AOA.
BTW I'm interested to calculate the total drag of F-16 and A-7 at
subsonic and supersonic speeds to estimate the cost of
a given F-16 or A-7 flight to help FAP(Portuguese Air Force)
on flight planning and budget estimations. If you can help me
or know somebody that can help me, please tell me!
Thank you for your attention.
Bye. Adeus! Chau!
Zé Fonseca
Lisboa, Portugal.
PS: Please reply. You are welcome!
Donny CHAN
In article <19971005222...@ladder01.news.aol.com>, fub...@aol.com
(Fubar2X) wrote:
]The actual thrust of a PW F100-P-220 is about 104 kN in afterburner, and
]the maximum takeoff weight of a F-16C is about 19,200 kg. This would yield
]an acceleration of 5.42 m/s^2. If the takeoff speed is around 300 km/hr (
]162 knots ), the time to takeoff would be 15.4 sec and the takeoff distance
]640 m, or about 2100 ft. My _Encyclopedia of World Military Aircraft_ says
]the takeoff roll at MTOW is around 2500 ft, so this is pretty close.
Fascinating. Just wondering: By how much does the catapult of a carrier
typically reduce the take-off distance of an aircraft?
--
"Otomodati to site yone. Sore dattara, watasi mo
Sinzyou-san no koto suki yo." - Nakamoto Sizuka,
Sotugyou Crossworld SS (1997)
José Barahona da Fonseca
In article <61b2c8$c...@news.inforamp.net>,
You are right, Fubar2X had considered Force= weight acceleration
but the right equation is Force= weight/g acceleration...
so with Fubar2X data we will have, considering a takeoff speed
of 300Km/h,
a=104/19.2 g=5.4 g m/s^2
t=300/3.6/(5.4 g)=1.57 secs
take off distance=0.5 a t^2=66m, that is about 1/10 of Fubar2X value.
With catapult I think we will have a takeoff distance of about 40m
or less than 40m!
Thank you for your attention.
Bye. Adeus! Chau!
Zé Fonseca
Lisboa, Portugal.
PS: Please reply. You're welcome!
Phil and Mia Waters
Donny CHAN wrote:
>
> In article <19971005222...@ladder01.news.aol.com>, fub...@aol.com
> (Fubar2X) wrote:
> ]The actual thrust of a PW F100-P-220 is about 104 kN in afterburner, and
> ]the maximum takeoff weight of a F-16C is about 19,200 kg. This would yield
> ]an acceleration of 5.42 m/s^2. If the takeoff speed is around 300 km/hr (
> ]162 knots ), the time to takeoff would be 15.4 sec and the takeoff distance
> ]640 m, or about 2100 ft. My _Encyclopedia of World Military Aircraft_ says
> ]the takeoff roll at MTOW is around 2500 ft, so this is pretty close.
>
> Fascinating. Just wondering: By how much does the catapult of a carrier
> typically reduce the take-off distance of an aircraft?
>
> --
>
> "Otomodati to site yone. Sore dattara, watasi mo
> Sinzyou-san no koto suki yo." - Nakamoto Sizuka,
> Sotugyou Crossworld SS (1997)
José Barahona da Fonseca
Hi again,
I forgot to say that I had neglected the Total Drag of F-16 which is
given by:
Drag(V)=1/2 rho Cd Sx V^2 + InducedDrag(V) + WaveDrag(V)
where rho=1.2256 Kg/m^3
and so the real takeoff distance is a little bit greater than 66m. The
exact F-16 takeoff distance is obtained through the solution of the
following complicated differential equation:
dv/dt=a-Drag(v)/mF16,
where mF16=19200/g
and integrating between v=0 and v=300/3.6 m/s.
Let's calculate a pessimistic estimation of Take off distance. I will
assume that L/D=8, so inducedDrag(v)=19200/8 and WaveDrag(v) is
negligible, Cd=0.15 and Sx=12 m^2, so
a'=(104,000-19,200/8-0.5 1.2256 12 (300/3.6)^2)/19,200 g=2.63 g
t'=300/3.6/(2.63 g)=3.23 secs
pessimistic takeoff distance estimation=0.5 a' t'^2=134 meters(!).
The exact value is somewhere between 66 and 134 meters, perhaps
(66+134)/2=100 meters!
Thank you for your attention.
Bye. Adeus! Chau!
Zé Fonseca
Lisboa, Portugal.
PS: Please reply. You are welcome!
José Barahona da Fonseca
Fubar2X wrote:
>
> José Barahona da Fonseca <j...@mail.telenet.pt> writes:
>
> >The thrust of F-16 engine is about 79.2kN, it weights about 7000Kg so
> >we will have that the acceleration is a=79.2/7*9.81=111 m/s^2. Now we
> >must know the minimum takeoff speed, say 250Km/h (I'm not sure, It's
> >only a guess). So the time to takeoff will be, neglecting the Total drag
> >which is much less than 79.2kN, t=250/3.6/111=0.626 s and the takeoff
> >distance will be 1/2 a t^2=22 meters(!!)...the real value is a litlle
> >bit greater than 22 meters because the Total Drag of F-16 increases
> >with V^2: Drag(V)=1/2 rho Sx Cd V^2+InducedDrag(V), rho=1.2256
> >Kg/m^3...to calculate the exact takeoff distance we must solve a
> >complicated differential equation: dv/dt=a-gsin alpha-Drag(v)/m.
>
> the maximum takeoff weight of a F-16C is about 19,200 kg. This would yield
> an acceleration of 5.42 m/s^2. If the takeoff speed is around 300 km/hr (
> 162 knots ), the time to takeoff would be 15.4 sec and the takeoff distance
> 640 m, or about 2100 ft. My _Encyclopedia of World Military Aircraft_ says
> the takeoff roll at MTOW is around 2500 ft, so this is pretty close.
>
> of thrust ), and no F-16 weighs 7000 kg at takeoff ( that's more like the
> empty weight ). Even so, using your numbers would predict an acceleration
> of 79.2 / 7= 11.31 m/s^2, or just a little over 1 g, as you'd expect (
> 17,800 pounds thrust vs 15,400 pounds weight ). The time to 250 km/hr then
> would be 6.14 sec, and the takeoff roll 213 m, or about 700 ft.
>
> You went wrong in your numbers by inserting the factor of 9.81 into the
> equation for acceleration, so your acceleration was a factor of 9.81 too
> high ( 111 m/s^2 is more than 11 g ). This resulted in your takeoff time
> and roll being too small by the same factor of 9.81.
>
> Fubar2X
Hi,
You are wrong: Force= Mass Acceleration
and weight=Mass g so
Force= weight/g Acceleration.
I think a weight of about 9000Kg for a litlle flight is
a realistic figure for FAP F-16s.
Fubar2X
José Barahona da Fonseca <j...@mail.telenet.pt> writes:
>You are right, Fubar2X had considered Force= weight acceleration
>but the right equation is Force= weight/g acceleration...
>so with Fubar2X data we will have, considering a takeoff speed
>of 300Km/h,
>
>a=104/19.2 g=5.4 g m/s^2
>
>t=300/3.6/(5.4 g)=1.57 secs
>
>take off distance=0.5 a t^2=66m, that is about 1/10 of Fubar2X value.
You're again confused about units here. You know that T=ma, and W=mg (
where T=thrust, m=mass, a =acceleration, W=weight, and g=gravitational
acceleration ). We can rearrange this to ( a / g ) = ( T / W ), i.e. the
acceleration in g's is equal to the thrust to weight ratio. Hence if the
thrust and weight are in the SAME units ( kilograms, pounds, or newtons )
the resulting ratio is the acceleration in g's.
So, if you use newtons for the thrust ( 104,000 newtons ) and weight (
19,200 kilograms = 188,350 newtons ), the thrust to weight ratio is 0.55,
so the plane accelerates at 0.55 g = 5.42 m/s^2. If you want to use
kilograms, the thrust is 10,600 kilograms and the weight 19,200 kilograms,
so the acceleration is again 0.55 g = 5.42 m/s^2. In pounds, the thrust is
23,370 pounds and the weight 42,300 pounds, so the acceleration is again
0.55 g = 5.42 m/s^2.
The acceleration is 5.42 m/s^2, the takeoff time is 15.4 sec, and the
takeoff roll 640 m, or 2100 feet. This neglects the drag on the plane. If
the average drag is around 3750 pounds, the net acceleration is 0.46 g =
4.55 m/s^2, the takeoff time 18.3 sec, and the takeoff roll about 763 m, or
2500 feet, in agreement with the book values ( obviously I just chose the
drag value to get this to work out ).
Fubar2X
Maury Markowitz
In article <8761661...@dejanews.com>, j...@mail.telenet.pt wrote:
> > ]The actual thrust of a PW F100-P-220 is about 104 kN in afterburner, and
> > ]the maximum takeoff weight of a F-16C is about 19,200 kg. This would yield
> > ]an acceleration of 5.42 m/s^2. If the takeoff speed is around 300 km/hr (
> > ]162 knots ), the time to takeoff would be 15.4 sec and the takeoff distance
> > ]640 m, or about 2100 ft. My _Encyclopedia of World Military Aircraft_ says
> > ]the takeoff roll at MTOW is around 2500 ft, so this is pretty close.
The only considerations that would seem to be obvious here would be
rolling resistance, air drag, and
> You are right, Fubar2X had considered Force= weight acceleration
> but the right equation is Force= weight/g acceleration...
Airplane mass is what was posted, and the calculation appears to be
correct to me - you make a similar "mistake" when you supply engine thrust
in kN, but then convert the mass in kg via division (which doesn't make
any sense). Given F=ma, you converted the mass into weight, and were
attempting a F=Fa, which is incorrect. If you do wish to use the weight
of the plane you should look at a momentum transfer instead (after all,
it's really F = m (dP/dt)
You'll notice posters from the US tend to use the terms weight and mass
interchangably. I personally believe this is because the weight measure
is also used as a mass measure in many cases (ie, I've seen the term
"pounds-mass" in many spaceflight references) and the unit of weight, the
"slug", is rarely if ever used.
Maury
Maury Markowitz
In article <maury-07109...@199.166.204.230>, ma...@softarc.com
(Maury Markowitz) wrote:
> is also used as a mass measure in many cases (ie, I've seen the term
> "pounds-mass" in many spaceflight references) and the unit of weight, the
> "slug", is rarely if ever used.
Ooops, the slug is a unit of mass of course.
Maury
Gene Nygaard
In article <19971007004...@ladder02.news.aol.com>,
> >but the right equation is Force= weight/g acceleration...
> >of 300Km/h,
> >
> >a=104/19.2 g=5.4 g m/s^2
> >
> >t=300/3.6/(5.4 g)=1.57 secs
> >
> >take off distance=0.5 a t^2=66m, that is about 1/10 of Fubar2X value.
>
> You're again confused about units here. You know that T=ma, and W=mg (
> where T=thrust, m=mass, a =acceleration, W=weight, and g=gravitational
> acceleration ). We can rearrange this to ( a / g ) = ( T / W ), i.e. the
> acceleration in g's is equal to the thrust to weight ratio. Hence if the
> thrust and weight are in the SAME units ( kilograms, pounds, or newtons )
> the resulting ratio is the acceleration in g's.
>
> So, if you use newtons for the thrust ( 104,000 newtons ) and weight (
> 19,200 kilograms = 188,350 newtons ), the thrust to weight ratio is 0.55,
> so the plane accelerates at 0.55 g = 5.42 m/s^2. If you want to use
> kilograms, the thrust is 10,600 kilograms and the weight 19,200 kilograms,
> so the acceleration is again 0.55 g = 5.42 m/s^2. In pounds, the thrust is
> 23,370 pounds and the weight 42,300 pounds, so the acceleration is again
> 0.55 g = 5.42 m/s^2.
This is a stupid, ridiculous, and confusing example. "Weight" in this
context means mass, plain and simple. Force is mass times acceleration.
If you already have the thrust in newtons and the mass in kilograms, why
go to all the trouble of converting one to the other and getting an
unnecessary acceleration of gravity factor involved in your calculations.
Why multiply or divide one by the acceleration of gravity, so that you
can get the acceleration of the plane as a multiple of the acceleration
of gravity, then multiply by the acceleration of gravity again to get the
acceleration in meters per second squared? What happens when you divide
104,000 newtons by 19,200 kilograms? A newton is a kilogram meter per
second squared, so the answer is 5.42 m/s^2, and you don't need to know
the acceleration of gravity to get it.
Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/weight.htm
Fubar2X
gnyg...@crosby.ndak.net (Gene Nygaard) writes:
> < Snip > What happens when you divide
>104,000 newtons by 19,200 kilograms? A newton is a kilogram meter per
>second squared, so the answer is 5.42 m/s^2, and you don't need to know
>the acceleration of gravity to get it.
Exactly, but this is exactly what José Barahona da Fonseca refuses to
believe. He thinks when you divide 104,000 newtons by 19,200 kilograms you
get 5.42 g, not 5.42 m/sec^2. The first two sentences of my first post were :
>The actual thrust of a PW F100-P-220 is about 104 kN in afterburner, and
>the maximum takeoff weight of a F-16C is about 19,200 kg. This would yield
>an acceleration of 5.42 m/s^2. < Snip >
but so far this hasn't sunk in. Hence, the "stupid, ridiculous, and
confusing example" was an effort to show him how you actually calculate the
acceleration in g's. It apparently hasn't worked either.
Fubar2X
Gene Nygaard
In article <19971008050...@ladder02.news.aol.com>,
fub...@aol.com (Fubar2X) wrote:
>
>
> Exactly, but this is exactly what José Barahona da Fonseca refuses to
> believe. He thinks when you divide 104,000 newtons by 19,200 kilograms you
> get 5.42 g, not 5.42 m/sec^2. The first two sentences of my first post were :
>
> >The actual thrust of a PW F100-P-220 is about 104 kN in afterburner, and
> >the maximum takeoff weight of a F-16C is about 19,200 kg. This would yield
> >an acceleration of 5.42 m/s^2. < Snip >
>
> but so far this hasn't sunk in. Hence, the "stupid, ridiculous, and
> confusing example" was an effort to show him how you actually calculate the
> acceleration in g's. It apparently hasn't worked either.
>
> Fubar2X
Ah, now I see why you were doing it in such a complicated way. I knew
that your calculations were correct and those of José Barahona da Fonseca
were wrong, but I didn't understand why you were going though the
calculation of g's.
Gene Nygaard
Tarver Engineering
On Tue, 07 Oct 1997 15:41:41 -0500, ma...@softarc.com (Maury
Markowitz) wrote:
<snip>
> You'll notice posters from the US tend to use the terms weight and mass
>interchangably. I personally believe this is because the weight measure
>is also used as a mass measure in many cases (ie, I've seen the term
>"pounds-mass" in many spaceflight references) and the unit of weight, the
>"slug", is rarely if ever used.
Slug mass was very useful when the horsepower is the unit of power.
John
Maury Markowitz
In article <8762832...@dejanews.com>, gnyg...@crosby.ndak.net (Gene
Nygaard) wrote:
> This is a stupid, ridiculous, and confusing example. "Weight" in this
> context means mass, plain and simple.
Before we start yelling at each other, please keep in mind that the
"standard" definition of weight means something rather different depending
on whether you learned metric or imperial measures as a child. I'm pretty
sure that's where this confusion came from.
> Force is mass times acceleration.
Actually in Newtonian terms it's defined as a change in momentum, which
can be simplified out in linear cases (ie, if the mass is constant and the
acceleration linear).
Maury