Horsepower to hover a helicopter
pstan...@hotmail.com
helicopter. I can't seem to make much headway.
Make any reasonable assumptions, Eg 1000 lb helicopter, efficiency of
the blade = 80 % (ie, losses in lift due to blade/air interaction),
even size if necessary.
Any one with ideas?
Stan
Mark P
Perhaps something like the following. For the copter to hover it needs
to generate an upward force of w (its weight). Suppose that it achieves
this solely by the rotation of the overhead rotor whose blades are
tilted at an angle a relative to the horizontal plane of rotation. In
the simplest approximation, the force on each blade is directed
perpendicular to the angle of the blade. Thus if the total force on a
blade is f, then the vertical force is f_y = f cos a and the horizontal
force is f_x = f sin a. Thus f_x = f_y tan a = w tan a, assuming f_y = w.
The power to overcome this horizontal force is (torque)*(angular
velocity). To know the torque we need to know the distribution of the
force along the blade, and with reasonable assumptions you could relate
the force distribution to the velocity distribution (which is linear in
distance along the blade). But we're being simple, so let's just assume
all of the force is at the ends of the blade (of length L). So the
power is P = L * f_x * (ang. vel.) = L w tan a * ang. vel.
From here you can try to make a reasonable guess about these helicopter
parameters (blade length, blade angle, rotor velocity).
If you actually knew parameters to model air resistance then you could
do even better.
Mark Brader
> A friend asked me, how much horsepower is necessary to hover a
> helicopter. I can't seem to make much headway.
If you want to calculate it, I have nothing to add to the previous
article. But I think it's true that most helicopters when hovering
(with a full load) would have to be pretty near full power.
See, they fly more efficiently when moving forward, so that's when
climbing is normally done; they don't need to have the extra power
to be able to accelerate at any substantial rate straight up, and
an engine capable of doing so would just add weight and cost.
So if you look up a helicopter of the sort you're interested in and
find its maximum power rating, then reduce it a bit, you're probably
close to the right answer.
--
Mark Brader | "And so it went. Tens of thousands of messages,
Toronto | hundreds of points of view. It was not called the
m...@vex.net | Net of a Million Lies for nothing." --Vernor Vinge
My text in this article is in the public domain.
Bob
> A friend asked me, how much horsepower is necessary
Kepp in mind that every horse you add adds weight to the coptor which
means you'll need more horsepower to overcome the extra weight.
I'm also wondering how we physically convert the power of horses to
rotor speed. Best I can think of is to march the horses in a circle,
harnessed to it's hub. Connect the hub to the propellor/rotor via
gears that amplify the speed.
But how many horses will it take? If it takes too many the radius of
the circle becomes too large and we need to enlarge the helicoptor.
That adds weight, too.
I think this is why helicopters are a recent invention. You can't
power one with horses.
Lamice Daxeh
P.S. And then there's also a waste disposal problem to deal with.
Mark Brader
> I think this is why helicopters are a recent invention. You can't
> power one with horses.
Recent? Why, they're positively 20th century!
--
Mark Brader, Toronto | "If you want a 20th century solution, the
m...@vex.net | obvious answer is helicopters!" -- Bob Scheurle
Puppet_Sock
[snip]
> I think this is why helicopters are a recent invention. You can't
> power one with horses.
And yet, helicopters were not invented until after horses existed.
<big wide :^) >
You could probably render some part of horses into fuel that would
work in a helicopter.
Socks
dgates
You need a big, infinitely strong rubber band. The horses could use
their power to wind it up really tight and then it could power the
helicopter.
Patrick Hamlyn
>You need a big, infinitely strong rubber band. The horses could use
>their power to wind it up really tight and then it could power the
>helicopter.
ummmm.... why does it need to be big?
If it's infinitely strong you might as well make it just big enough to see so
you don't lose it.
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group (polyforms...@egroups.com)
riverman
"Mark P" <use...@fall2005REMOVE.fastmailCAPS.fm> wrote in message
news:KtWzf.12107$F_3....@newssvr29.news.prodigy.net...
> pstan...@hotmail.com wrote:
>> A friend asked me, how much horsepower is necessary to hover a
>> helicopter. I can't seem to make much headway.
>>
>> Make any reasonable assumptions, Eg 1000 lb helicopter, efficiency of
>> the blade = 80 % (ie, losses in lift due to blade/air interaction),
>> even size if necessary. Any one with ideas?
>>
>> Stan
>
>
Hmm, I googled 'helicopter horsepower' and found the answer right away. A
lot depends on the size of the chopper: Sikorsky's first successful copter
had a 65HP engine, while the MH-53E Dragon has a total of almost 14,000 HP.
I was startled to find out that up to 30% of the total HP of most
whirlybirds is used to drive the tail rotor.
--riverman
Mark Brader
> lot depends on the size of the chopper: Sikorsky's first successful copter
The question was how much power *to hover*, not how large the motor is.
However, as I have previously suggested, the two answers should be
fairly similar.
--
Mark Brader | "People tend to assume that things they don't know
Toronto | about are either safe or dangerous or useless,
m...@vex.net | depending on their prejudices." -- Tim Freeman
pstan...@hotmail.com
a diagram. You should teach.
But I'm stuck trying to make any further conclusions.
Let say that L = 20 feet, the length of the blade, and that the weight
of the helicopter is 1000 lbs.
Then using your formula P = L * w * tan a * ang vel gives me:
P = 20,000 * tan(a) * ang vel
Can I make any conclusions from this? P is the power necessary, I
presume, to hover with our assumptions. So if I want P to be a
minimum, then I need to get the term tan(a) * ang vel to also be a
minimum. If I double the velocity, it seems tan(a) should go down,
but how much?
Perhaps, since Lift of an airfoil = .5 * p * Area * Cl * V^2 where p
is density, A is area of the airfoil, both can be a constant. Cl is
the coefficient of lift. For a symetrical airfoil, the basic equation
for Cl is Cl=B*a, where B is a constant and a is the angle of attack,
the same angle a you used.
So if I double velocity, then in order to get the same lift, then Cl
must now be one quarter, so a is one quarter, so tan(a) is ....? one
quarter too?? I think so, for small angles. So now the term tan(a) *
ang vel is half what it was, which to good since that makes P to be
half what is was.
Ergo, in order to howver on the minimum horsepower, I want maximum
velocity of the blade??
Mark, or anyone, can you put my argument back on track?
Stan
On Fri, 20 Jan 2006 00:50:50 GMT, Mark P
osmium
Using round numbers and simplifying assumptions. . If an object with a
weight of one pound were dropped, it would fall 33 feet in the first second.
To recover from that situation we need to lift it 33 feet in that second.
One HP is 550 ft lbs/sec which we can use up as we wish. It would take
33/550 HP to lift one lb in one second, or .06 HP. If we assume the
helicopter and the passenger weigh 1000 lbs, it would take 60 hp to keep
airborne. Considering the practicalities of the real world, I can only
conclude that the weight was quite a bit less than 1,000 pounds.
I assume the fuel weight would be really close to zero, for safety and also
to get off the ground. I am reminded that Jenna Yeager on the round the
world flight weighed something like 100 lbs. My favorite weight for a
woman!
Mark P
> <pstan...@hotmail.com> wrote:
>
>> A friend asked me, how much horsepower is necessary to hover a
>> helicopter. I can't seem to make much headway.
>>
>> Make any reasonable assumptions, Eg 1000 lb helicopter, efficiency of
>> the blade = 80 % (ie, losses in lift due to blade/air interaction),
>> even size if necessary.
>>
>> Any one with ideas?
>
> Using round numbers and simplifying assumptions. . If an object with a
> weight of one pound were dropped, it would fall 33 feet in the first second.
Not true. Taking g = 32 ft/s^2, it falls 16 ft in one second.
> To recover from that situation we need to lift it 33 feet in that second.
> One HP is 550 ft lbs/sec which we can use up as we wish. It would take
> 33/550 HP to lift one lb in one second, or .06 HP. If we assume the
> helicopter and the passenger weigh 1000 lbs, it would take 60 hp to keep
> airborne. Considering the practicalities of the real world, I can only
> conclude that the weight was quite a bit less than 1,000 pounds.
This calculation doesn't really work. If you choose a different time
you'll get a different power, since d ~ t^2 and your power calculation
gives P ~ d/t ~ t.
Bob
> If [the rubber band is] infinitely strong you might as well make it just big
> enough to see so you don't lose it.
A good idea. But if it's infinitely strong I'm not sure how the horses
are going to twist it. At the very least, they'll only be able to
twist it until it stores enough energy to counteract the strength of
the horses.
osmium wrote:
> ... If we assume the helicopter and the passenger weigh 1000 lbs, it would
> take 60 hp to keep airborne. ...
ObPuzzle: If we replace the horses with a person riding a bicycle, how
fast would that person have to pedal to keep the coptor airborne.
OP Stan lamented:
> Mark, or anyone, can you put my argument back on track?
Not a chance. ;)
Bob H
Mark P
[Reply moved to undo top-posting. Please post below previous messages
to maintain readability.]
Looking at airfoil lift is a different approach entirely. My heuristic
solution neglects this and instead looks only at the work done by the
rotor blades as they push through the air. (Last I heard there's still
debate over how much of an airplane's lift is generated by the airfoil
and how much by the angle of the wing relative to horizontal.)
You're right that making a smaller angle decreases the power. This is
because in my model that means that the blade is virtually slicing
through the air and encountering very little forward air resistance. In
a medium of infinite air resistance this would be sufficient (and
explains why it's easier to tread water than to "tread air"), but the
problem in reality is that this ends up generating very little upward
force.
My hope in my original post was that it would be easier to "guess"
realistic values for the rotor than for the engine. For example, I
might guess 10 feet for L, 10 degrees for a and 1500 rpm for the angular
velocity (the audio frequency of a helicopter might give a clue to its
rotor rotational frequency). This gives about 500 HP which seems order
of magnitude correct at least.
To really do this correctly you need to deal with the air resistance as
it relates both to the power required to overcome it and the force of
lift generated by it. I'm not sure I can provide much useful direction
on that approach though. :)
Sawney Beane
fixed-wing aircraft is where it imparts the least kinetic energy to
the air as it moves a mile. By getting lift without imparting much
kinetic energy to the air, a glider can fly with no engine.
Parasitic drag comes from stirring up the air. It increases with
speed. Induced drag comes from imparting downward motion to the
air to get lift. It decreases with speed. If you're going twice
as fast, you're pushing twice as much air down. If the wing is
pushing down with a given force against twice as much air, the air
will end up moving down half as fast. As kinetic energy is
M(V^2)/2, the wing going twice as fast will impart half as much
kinetic energy in downdraft.
A hovering helicopter will generate a big downdraft, so it uses a
lot of power. A hovercraft needs much less power because it has
skirts to block the draft. At very low height, the ground may slow
a helicopter's downdraft. If a helicopter were designed primarily
for hovering efficiency, the rotor would be as long as possible to
get lift with the minimum downdraft.
A worst-case scenario would be trying to rescue a climber on a
cliff on a high mountain. The air is thin, there's nothing to slow
the downdraft, and a big rotor could be hit the rocks.
Rich Grise
> pstan...@hotmail.com wrote:
>>> pstan...@hotmail.com wrote:
>>>> A friend asked me, how much horsepower is necessary to hover a
>>>> helicopter. I can't seem to make much headway.
>>>>
>>>> Make any reasonable assumptions, Eg 1000 lb helicopter, efficiency of
>>>> the blade = 80 % (ie, losses in lift due to blade/air interaction),
>>>> even size if necessary.
>>>>
>>>> Any one with ideas?
> To really do this correctly you need to deal with the air resistance as
> it relates both to the power required to overcome it and the force of
> lift generated by it. I'm not sure I can provide much useful direction
> on that approach though. :)
It seems like it shouldn't be so complicated. Take the weight of the
chopper, and figure out how much air you have to move, and how fast,
to exert a force equal to the chopper's weight, and figure out how
many HP it takes to move that much air.
I'm not about to try to do the math, because I don't know anything about
the physics/aerodynamics of moving air, but somebody in HVAC or something
might have some ideas.
Cheers!
Rich
Rich Grise
> Patrick Hamlyn wrote:
>> If [the rubber band is] infinitely strong you might as well make it just big
>> enough to see so you don't lose it.
>
> A good idea. But if it's infinitely strong I'm not sure how the horses
> are going to twist it. At the very least, they'll only be able to
> twist it until it stores enough energy to counteract the strength of
> the horses.
>
> osmium wrote:
>> ... If we assume the helicopter and the passenger weigh 1000 lbs, it would
>> take 60 hp to keep airborne. ...
>
> ObPuzzle: If we replace the horses with a person riding a bicycle, how
> fast would that person have to pedal to keep the coptor airborne.
So fast that you'd wind up with The Gossamer Brick. ;-)
I really doubt if it's possible for a person to generate enough power to
lift himself _and_ a chopper.
Cheers!
Rich
Sawney Beane
>
> >
> > osmium wrote:
> >> ... If we assume the helicopter and the passenger weigh 1000 lbs, it would
> >> take 60 hp to keep airborne. ...
> >
> > ObPuzzle: If we replace the horses with a person riding a bicycle, how
> > fast would that person have to pedal to keep the coptor airborne.
>
> So fast that you'd wind up with The Gossamer Brick. ;-)
>
> I really doubt if it's possible for a person to generate enough power to
> lift himself _and_ a chopper.
>
Here's a page about gliders. A sink rate of 1 knot would be 3 hp
per thousand pounds.
In their examples, a 1000-pound glider requires 4.8 hp at 31 knots.
That's only 8% of what a helicopter needs.
With a gross weight of 500 pounds and 2/3 prop efficiency, a person
would need to produce 3.6 hp to cruise at the least taxing speed.
So even peddling a glider is beyond the pale.
The second chart shows how much more power is required at lower
speeds. The Wright Brothers were incredible to build a glider,
engine, and propeller that could lift off.
sewiv
Hover it where? 3 feet off the ground is a lot cheaper than 1000 feet
off the ground,
due to the ground effect. See
http://www.masquito.be/helicopter_aerodynamics.htm
--
Sandy
Narasimham
Weight of chopper * downward wind velocity = Power * efficiency etc.
Stephen Riley
pstan...@hotmail.com writes
>A friend asked me, how much horsepower is necessary to hover a
>helicopter. I can't seem to make much headway.
That's because you're trying to equate force and energy. You need to
provide a downward force equal to the weight of the helicopter, but how
much energy that takes depends on your constraints, mass and velocity of
air pushed down and physical constraints including such things as
material strength, fluid dynamics, etc.
At most basic the more mass you can push down the less energy it takes
to provide a force, but then blades need to be stronger and increase the
mass of the helicopter, etc. So perhaps find more mass to push against,
e.g. parachute, wings,...all the way to rigid pole connecting earth
directly to helicopter. The latter would only be energy efficient for
'hovering' purposes though :)
>Make any reasonable assumptions, Eg 1000 lb helicopter, efficiency of
>the blade = 80 % (ie, losses in lift due to blade/air interaction),
>even size if necessary.
>
>Any one with ideas?
Blade connecting hull of helicopter directly to ground, or blade shaped
like a big floaty thing made of something light :)
--
Stephen Riley
Rich Grise, but drunk
> In message <3i50t1162s5q510e7...@4ax.com>,
> pstan...@hotmail.com writes
>
>>A friend asked me, how much horsepower is necessary to hover a
>>helicopter. I can't seem to make much headway.
>
> That's because you're trying to equate force and energy. You need to
> provide a downward force equal to the weight of the helicopter, but how
> much energy that takes depends on your constraints, mass and velocity of
> air pushed down and physical constraints including such things as material
> strength, fluid dynamics, etc.
>
> At most basic the more mass you can push down the less energy it takes to
> provide a force, but then blades need to be stronger and increase the mass
> of the helicopter, etc. So perhaps find more mass to push against, e.g.
> parachute, wings,...all the way to rigid pole connecting earth directly to
> helicopter. The latter would only be energy efficient for 'hovering'
> purposes though :)
For some reason, this brings to mind the rocket belt (anybody remember
that?), and later, the jet belt. James bond took a 45-second ride once in
the famous rocket belt, IIRC. 45 seconds was as long as it could last with
all of the fuel that it could carry and still be hand-carried to the
takeoff point. (and, presumably, from the landing point. ;-) ). ISTR
seeing something later, which used the same principle, but with jets - or
_a_ jet, with ducted exhausts, which, theoretically, didn't burn fuel
quite as fast, but failed for practical reasons, like it costs a billion
dollars to build one.
Or, it could be because jet engines don't like to ingest their own
exhaust, which, if you're hovering at 6 feet (2m), what else is there
to breathe? ;-)
Cheers!
Rich
Stephen Riley
drunk" <yahr...@example.net> writes
>For some reason, this brings to mind the rocket belt (anybody remember
>that?), and later, the jet belt. James bond took a 45-second ride once in
>the famous rocket belt, IIRC. 45 seconds was as long as it could last with
>all of the fuel that it could carry and still be hand-carried to the
>takeoff point. (and, presumably, from the landing point. ;-) ). ISTR
>seeing something later, which used the same principle, but with jets - or
>_a_ jet, with ducted exhausts, which, theoretically, didn't burn fuel
>quite as fast, but failed for practical reasons, like it costs a billion
>dollars to build one.
Yep, and as demonstrated at the 1984 Olympic games. Jet packs being less
energy efficient than a helicopter since less mass (at higher velocity)
is used to provide impulse. A jet (sucking in air) would theoretically
be more efficient than a pure rocket design for the same reason.
The makers of the Solo Trek (a very windy looking device!, if you follow
the link below) beat that 45 seconds with a claimed airborne time of 2
hours.
http://www.21stcentury.co.uk/technology/solotrek_xfv.asp
--
Stephen Riley
Sawney Beane
Airliners got much greater range and economy when they went to
bypass jets, which produce a bigger stream at lower speed.
Matthew Russotto
Hmm, that site keeps redirecting me to
http://www.mosquito.iq/brick_aerodynamics.htm
--
There's no such thing as a free lunch, but certain accounting practices can
result in a fully-depreciated one.