little notes on neutralápoint okay whatáis
neutralápoint?
weáhave seen that neutral point is that xcgálocation,
atáwhich the aircraft become neutrally stable
or dcm by dcl = 0 or dcm by d alpha =á0,
right.
and so far we have been talking about neutral
point where onlyáwing, tailáso and some
fuselageácontribution, althougháwe have
not computedáanything.
butáwe have never mentioned about the effect
ofáengine, theáeffect of type of engine.
the aircraft is flyingáwith.
so weáwill take the effect ofáengine, weáwill
take only 1 case propeller drivenáengine
right, whatáwill be the effect of the having
engine in the nose and moreáprecisely theáquestion
is, if this is the n 0 fixed that is stick
fixed case with poweráoff withápower off
then do you think n 0 will change with poweráon.
so doáyou think the n 0 will change, with
power on thatámeans, whenáyou switch on
the engine for a typicaláairplaneáof this
configuration.
supposeánose mountain likeáthis, foráthis
case i repeat theáproblem.
weáare trying to understand the effect of
thrust oráengine, onáthe neutral point this
is the case power off, that is no power is
there.
no engine is being rolled, and i am more focused
to this sort of nose mountain engine, without
this being on, if this is ná0. now the moment
the power on case come.
power on so what will happen? will the neutral
point still remain at some point let us say
here or it will try to go backward or forward
that is the question, or in indirectly.
what is theáquestion?
witháthe engine on, whether the aircraft
will become more stable or lessástable, becauseáwe
understand if ná0, let usásay n 0 1 and
this n 0 2 and another aircraft n 0 2, then
i know that this aircraft is more stable than
this aircraft becauseáit haságot a large
static margin for a given cg location isn't
it, for the given cg location this aircraft
has this much of staticámargin, and thisáaircraft
has larger starting margin okay.
so, i would like to know now what happens
if thrust is on, will there be any modification
on the n 0 stick fixed okay.
letásee one thing weáunderstand, ifágrinding
of this propeller, adds most stabilizingáeffect
to the airplane then our n 0 fixed should
go backwardáso itáwill become moreástable.
ifáadding this nose mounted engine, ifáit
contributes towards moreádestabilizingáthen
this n 0 point shift this way clear, because
it will become lessástable.
soánow what happenásee, ifáthis is the
propeller moving and some delta if it is seeing
then you know that by momentum exchange, there
will be a thrust generator small though in
this direction, because some momentum exchange
will be there one this direction another this
direction.
soáthis force which is proportionaláto alpha
is now acting ahead of cgáokay.
soáthis will give any component or any force,
which are ahead of cg that will giveádestabilizingáeffectáright.
so, this configuration will give aádestabilizingáeffect.
soáthe neutral point fixed will now with
power on n 0 power on will come forward here
instead of power off is here.
is it clear again iárepeat, becauseáof this
engine thrust or engine power this propeller
drivenáairplane.á
if there is angleáof attack, theámovement
of exchangeáin this direction also, that
will give a force roughly i canáwrite this,áfunction
of alpha and since this force is ahead of
cg, cg is somewhereáhere, foráany positive
alpha it will give nose up movement and nose
up movement means? it is destabilizingáfor
a positive disturbance of alpha.
so, this will contribute towardsádestabilizingáeffect,
and hence the aircraft will become lesser
stable compared to poweráoff.
soáthe neutral point will shiftáhere.
soáthis is the point where i write n 0 or
n 0 bar or xn 0 or xn 0 bar, this is power
on and power off is hereáclear.
soáwhen you are flying you have to ensure
that, cg should not go beyond power on, then
it will become unstable during power on right,
is this clear?
that is the definition neutralápoint, ifáthis
is neutral point whether it is power on and
poweráoff, pointáis if the cg goes beyond
the neutral point, the aircraft will become
staticallyáunstable.
so, nowáyou know when the engine power is
on for this sort of a configuration then actuallyáthe
neutral point is not what is i am writing
here with power of case, it will give power
on little ahead ofáit so, iámust ensure
that the cg of the aircraft should never go,
beyond.n0 the power on so, this will limit
the case how much backward i can take the
cg okay.
so, the cg backwardáand the cgáforward so,
howámuch backward i can take that is dictated
by primarily by n0 power on clear okay.
nowánext question comes, how further forwardáiácan
take the cg because,áiáknow if n 0 is here,áiáput
cg here i put cgáhere, iáput cgáhere, iáput
cg here in all the cases it will be statically
stable, only think if it isá1, 2áandá3,
thirdácase will be the most stable case,
this is little less than the third case, it
is less than the second case, but third case
will be most stableácase.
nowáwho decides, that can i put the cg anywhere
in the front because i give aálogic, ifáit
is cg is forward of neutral point, it will
be statically stable.
so howádoes it matter to me?
it will be stable let it be moreástable.
doáyou know fundamentally, if an aircraft
becomes highly stable, then i need to put
large control deflection to trim it for different,ádifferentácl,
howádoáiáknowáthat?
oneáis by simple common sense, if something
is highly stable, it will always resist any
change, which try to change its equilibrium
to another equilibrium so, we need more effortáright,
butámathematically or expression wise whatever
we have developed we could see hereáagain,
thisáexpression comes to our rescue, you
know that delta e trim or delta required is
delta e 0, + d delta e by dcl into cl trim
this is also cl trim okay.
andáwhat is thisáthisáis delta e 0á+ -áxcg
- xn 0 bar divided by cm delta e into clátrim,
thisáis = delta eárequired.
so, asái amátaking cg forwardásuppose this
is theáairplane,áand cg i am taking forward
in thisádirection, so this gap xcg - xn 0
is going on increasingáokay.
negative, negative, positive and negative,
so what is happening as it goes on increasingáthe
delta e required also will go on increasing
right and there will be limit beyond which
you may not have realistic delta e available
because delta e also have a maximum value
of deflection, may be + - 25ádegrees.
beyondáthat the flow will stall over the
deflection this is the elevator, let usásay,
so beyondáfifteen degree of beyond 25 degree,
depending upon what sort of elevator you are
using, the flow may separate that is why everyáairplaneáis
characterize with delta e max is + -,álet
usásay 25 degree, some high fidelity aircraft
may have 30 degree, 45 degrees but almost
of our normaláairplane,áit is 15 to 25ádegree
depending upon, what sort of technology you
haveáused.
so theápoint is there is the delta e max,
you cannot exceedáthat, okábut if you go
on making it this separation larger and larger,
or the making the aircraft more or more stable.
this value may exceed the limit given byáthere.
so, you have a restriction on how far you
can put the cg forward because beyond point,
thisávalue is become so large that delta
e will may become more than the delta e maxárequired,
typicallyáthe delta e up requiredáokay.
ifái amámoving the cg towards nose of the
airplane thatáis, we are making it more and
more stable as i moving cg forward pictorially
ifáiádrawádeltaáe vs.cl trimáright thisáis
one case highly stable anotherácase, thisáis
another case right here okay.
what is the understanding here, in which case
cg is most forward this is the cg, xcg is
most forward because the slope is the highest
here most negative, then this is xcg 2 i call
it xcgá3.
now, seeásuppose you want to trim at this
cl trim value so for cg location let me draw
it foráyour, thisáis xcg 3 this is 2 this
isá1 rightáso this is the most forward right
this is 1 hereáright.
mostáforward one that means neutral point
may be somewhere here so, this is highlyástable,
mostástable in thisáconfiguration.
so, now if you want to trim a aircraft for
the particular cl trim, you see if am using
this 3 location, where the starting point
is least i need only this much of delta e
negative that is delta e isáup, okayábut
if changeáthe cg to cg from cg 3 to cg 2,
and the stability hasáchanged, we say now
i require from here to here this much of delta
e required, which is again negative right
this much it is much more compared to thisávalue.
whatáwill happen for this case, ifáiádraw
it like this value, will be exceptionallyáhigh.
so whatáis theámeaning, youámean somehow
generated clátrim, okayábut how much cl
trim you can trim actuallyáby elevatorádeflection
is decided by what is the delta e max available
and hence, directly it also tells you what
is the maximum forward cg you can put for
so that your aircraft isátrimmable, okay.
so, the aircraft is trimmable so, that i can
trim theáairplaneáfurther cl okay. that
is very,áveryáimportant so the most forward
cg is primarilyádictated by most forward
cg + delta e maxáavailable, they areádecides
what should be the most forward cg and most
fcg is divided by n0 which is power onácase,
if poweráoff somewhere here power off we
have alreadyáseen, n0 poweráoff.
so, nowátwo different things are here one
is fcg is governed n 0 power on stick fixed
neutral point power on and forward cg is dictated
by what is the delta e max available for all
practical purpose right.
you may also wonder, what happen during landing
when i am actually flying at a higher cl but
the propeller is also rotating you may wonder,
what is happening if i am trying for the landing
where cl is also maximum.
but power is not on in a true sense is not
a full power in theoretical sense i can say
propeller wind milling right.
then what happens how do i take care of that
so, as per as this course is concerned let
me tell you shouldáiáproceed with how will
you proceed so, far what you have seen this
is the n0 stick fixed here i am talking about,
stick fixed right and power off.
if the engine is nose mounted engine it will
giveádestabilizingáeffect so n0 power on
will come here, power on will be ahead of
n0 power off.
so, i should plan that cg of the airplane
should never cross this it will become statically
unstable during power or more.
the next question came, how much forward cgáiácan
allow the layout designer toáput, thenáiárealize
that forward cg is dictated by the delta e
available.
so delta e max that will decide how much reallyáiácan
how much forward cgáiácan trim because after
it will give more cm soáiáhave to have any
elevator power to control it,áitáwill nullify
it.
this part we have a same, now second question
whicháiáwas of course whispering is propeller
wind milling,áwindámilling that is the special
do landingáitĺsánot a full power it is
propeller just wind milling.
so,áthat case if you see propeller wind milling
it is neither power on but not power off right.ásoáit
is between these two so n0 fall wind milling
should come somewhere here,ápropelleráwind
milling, should come somewhere here, the especially
doing landingáiáshould be careful about
it,ábecauseáthat time n0 is somewhere here,
n0 is not power on it will behind, but ahead
of n0 stick fixed do you understand this or
not?á
n0 power on that time, it is contribute the
full this stabilizing effect for a propeller
engineáináthe nose of the air plane.
but wind milling it is in between power on
and power off for somewhere here it will be
thereáright.
soáfor all practical purpose during landing
this will be the limit of cg, beyond this
cg it will become unstable.
but for forward, most forward cg this is the
most forward cg when the airplane is at approaching
like that it is at high cl max value and we
need to know thatáiáhave enough delta e
max, which should be able to trim that cl
okay.ábecauseáwhat will happen?
asáiámove the cg forward and forward delta
e max requirement will go on an increasing
we have seen that, and since there is a limit
on delta e max + - 25 degree.
so it will be more cg forward cg more governed
byáthis, foráthis is a limit air putting,
here and if you know these three things, three,
four things you will be able to handle this
at all situation, as we progress where when
you for a little bit on as a stickáfree we
willáagain come back to thisáneutral point
and áagain address this things at this from
designers prospective we will solve the problem
and will try to get field for this.
asáiáhave been promising that once some
theory, base background, is ready then we
will be solving some numerical problem to
get better in sight of those expressions,
one of the derivations made was dcm by dcl
= - static margin, right staticámargin, many
books may also refer this as stability margin
andáwe know meaning of stability margin is
if this is the neutral point.
if this is the cg,ái am measuring from some
reference here, and this is static margin,
of course this is expressed as a percentage
of mean aerodynamic chordáokay.
weáare not trying to say this how to utilize
this dcm by dcl is = - static margin and get
some insight about design.
onceáiáam writing like thisáiáunderstand
the cm is expressed the cm at cl = 0 + dcm
by dcl into cl,áokay?
thisáis the expansion.
for a symmetric aerofoil how do you write?
we prefer ifái want write in terms of alpha
i write cm at alpha = 0 + dcm by d alpha into
alpha. this is workable for symmetric aerofoil,
but when it have been cambered aerofoil thenáiáreferred
to the cm as cm at cl is = 0 + dcm by dcl
into cl, okay.
now once we understand let us do a problem,
problem is this let me write slope of cm versus
cl is point 0 8.
then cm at cl = 0 is let me correct slope
of cm versus cl is ľ 0.15, because initiallyáiáwrote
point 0 8 is positive which is statically
unstable right?
so, minus sign okay yes its statically stable
and what is telling is cm at cl = 0 is point
0 8 so you could see that cm0 is greater than
0 and dcm by dcl is less than 0 so statically
stable and since the cm is positive it can
be trimmed at a positive angle of attackáokay.
the question is determine, the trim lift coefficient
that is cl if cg is xcg by c bar is 0.3álet
me again to repeat the problem for you to
understand the slope of the cm versus cl is
- point 1 5 that is dcm by dcl is - point
1 5, soáiáwrite dcm by dcl is ľ 0.15 so
this isástatically stableácase and cm0 that
is cm as cl is = 0 is point 0 8 so it is possible
to trim at positive alpha.
question is determining the trim lift coefficient
what is the cl for which it can be trimmed
easily withoutádoing anything else.
so whatáiáam trying to draw is cm cl the
slope dcm by dcl is ľ point 15 and cm this
value cm0 or cm at cl = 0 this value is point
0 8 what is this values cl trim the question,
very simple problem.
you know the slope of this line you know this
so the cl trimáiácan easily find out like
point 0 8 by.
let usásay this is x by this x is point 15
the slope is negative we have taken out the
negative sign so xáiáwill get as point 0
8 by point 15 and which is approximately equal
to you can check around point 5 2 or point
53.
soáthis is the value of cl trimáiáwill
need not solve onceái draw it like this know
from triangle right angle triangle you can
easily find out if this is point 0 8 the slope
is - point 1 5 slope is nothing but perpendicular
by base and negative sign you have to take
and that negative value of so straight forward
okay.
and see how efficient is this expression dcm
by dcl = - static margin with this problem
we will now go little further and try to extend
its value when you are designing an aircraft
typically if you designing a civil airplane
you will design for a particular static margin
may be 10% static margin that is 10% of the
linear dynamic chord okay.á10ápercent of
thatásoánow if want to select what is the
wing area required or.
what is the cláiáshould design for you have
a criteria likeáfor cl typicallyáiácan
write l = half row v square scl okay? so and
this = weight the lift = weight you know what
at which would have to fly and you know what
speed you want to fly you have your wing area.
we typically get is some sort of the cl validate
let us say cl point 6 let as point 6 you have
get so what is the meaning of this 10% static
margin right.
and cl point 6áiáwill write as point 6 and
draw a line - point 1 get 10 percent so dcm
by dcl is - point 1 because it is 10% and
then to find out what is the intercept so
this will be point 0 6 now whatáiádoáiádoáiáknow
cm0 = cm0 of wing + cm0 of tail + cm0 of fuselage,
soáiáensure the wing tail locations the
area are such that the sum of this three quantity
as of to give point 0 6 and the neutral point
and cg they are located such a way that their
static margin that is neutral point and cg
this gap is 10% of new mean aerodynamic chord
so how simplified.
it you approach becomes once you used that
relationship you will know this when we do
the design course may be we will be doing
in one of the books butáitĺsábetter that
you should not only know how to find deltaáe
foráa cl trim you should also know why we
are doing this.
soáthese examples are given to excite your
mind that you start thinking in terms of utilizing
this relationship for designing anáaircraft,
okay.áthank you very much.
