Differences between automotive & airplane engines

An automotive engine burn the same amount of gas than an airplane one


Bull****....

It can, too. The volumetric efficiency of a high-RPM engine
suffers at that higher RPM, and I have experience with a Soob to prove
it.

Dan

But when we DO use 'horsepower' we must be careful to
never use it in isolation, always identifing the rotational speed at
which that 'horsepower' is being produced.


Absolutely and utterly wrong.

It is *torque* which must always be associated with the rotational speed
at which it is being produced.

Read that first sentence again. He's not wrong; he just
didn't specify "torque" for those who don't know the relationship
between it and RPM and HP.
When you say "absolutely and utterly" it should be used
only where it applies. Clearly, that's not here.

On 10 Feb 2006 05:55:48 -0800, "Lou" wrote:

I'll agree with the automotive engine with PSRU being heavier, but are
you sure about your other statement "the lighter the better"?
I'm currently looking at an engine that is 100lbs lighter than the one
recommended for my plane. Although cutting 100lbs from the total
weight is a dream come true, it brings up the question of weight and
balance. I can move the engine forward to make up the difference in
balance, but I don't know how far or how to find out.
Lou

calculate it out yourself.
weight times distance from the datum for the original engine will need
to equal weight times distance for the new engine.

ie moment arm of the old engine needs to equal the moment arm of the
new engine.

for your calculation purpose you can select an arbitrary point on the
fuselage as your datum point.
the manufacturer's data for the engine should show the cg position of
the engine. measure the distance from that cg point to the datum.
multiply the engine weight by the distance you just measured.
the result is the moment arm.

now divide the answer above by the new engine weight and you will get
a number. that number is the distance from the datum to the cg
position of the new engine that maintains the original aircraft
balance.

if you dont have the cg position of the engine you can work it out
easily.
(think about the consequences of dropping the engine as you consider
this next bit. it needs to be done carefully!)
just hang the engine up on a rope from a part somewhere on the engine
and draw a vertical line from the rope down across the engine.
hang it up by a different position on the engine and draw another
vertical line. the vertical lines will intersect at the cg.

(btw hang it so that you can see the side elevation of the engine)

just to work out a hypothetical answer.
suppose you make the pilot's door knob the datum point.
measuring from there to the engine cg gives say 52 inches.
the engine weighs 180 lbs.
multiplying that gives a moment of 9,360 inch lbs.

the new engine is 100 lb lighter. say 80lbs.
9,360 divided by 80 = 117.

the CG of the new engine needs to be positioned 117 inches from our
datum of the the pilots doorknob to keep the existing aircraft cg.

there you go lou. that should make it easier.
btw take into account all the stuff that changes as well.
cowlings and engine mount if they have any significant weight.
and include the prop and spinner in the engine weight.
Stealth Pilot
Australia

Lou wrote:

I'll agree with the automotive engine with PSRU being heavier, but are
you sure about your other statement "the lighter the better"?
I'm currently looking at an engine that is 100lbs lighter than the one
recommended for my plane. Although cutting 100lbs from the total
weight is a dream come true, it brings up the question of weight and
balance. I can move the engine forward to make up the difference in
balance, but I don't know how far or how to find out.
Lou


Googling this group for weight and balance yields 25 pages...
So I picked up one of mine here (Dec 10, 2002) and included
Brian's note at the bottom...
(Unfortunately, there was no link attached, so here is the text).



There is a lot of smoke and mirror magic around weight and balance
because so many people understand it so poorly.

At the heart of all of it, though, is a rotational force about a
reference point. the rotational force is called a MOMENT, and
the reference point is called the DATUM.

Sometimes the datum is located at the tip of the spinner.
Sometimes it's located at the main gear axles.
Sometimes its located at the leading edge of the wing.
It doesn't particularly matter where it is located, as long
as you use the same location to work the problem.

You'll often see the term STATION. This is the distance from the
datum to a particular place on the aircraft. Say, for instance,
the instrument panel?

The station numbers change according to where the datum is placed.
But the instrument panel stays in the same physical location.
It's all an offset from a zero point.

One reason to place the datum at the tip of the spinner is because
all the station numbers are positive. No negative distances to
confuse things.

One reason to place the datum at the axles is because the datum
is station zero.zero. Multiply the weight on the wheel times
zero (the ARM is zero at the datum) and the moments for that
wheel come out to zero. Makes the arithmetic a little easier?

And, the reason to place the datum at the leading edge of the wing
it because that's where we are going to wind up anyway. The results
of our CG calculations will finally boil down to a point some given
distance aft of the leading edge.

CG range is often refereed to in terms of a percentage of the wing
chord. Say 25% would be the forward CG limit, maybe 33% would be the
aft limit. So our end number actually refers to a distance aft of
the leading edge. The actual numbers will be different, depending
on where the datum is located, but they all (hopefully) come out
at the same place on the airplane.

First rule:

weight x distance = moments pounds x inches = pound inches (!)
So,
moments / inches = pounds
and
moments / pounds = inches

Practical example:

A bowling ball, held at the chest, has a certain weight.
Held at arms length, it has exactly the same weight!

But due to the longer distance (called ARM) it has a much higher moment.
\
THAT's what feels so heavy.
That rotational force.

So, to solve your little weight and balance question.

The only distance from anything. that matters, is the
distance from the CG of the instrument to the DATUM
specified for that aircraft.

If you have a "before" weight and balance already done,
multiply the weight of the instrument times the distance
from the datum given in the "before" problem.

Then add that moment to the airplane's moment,
and the instrument weight to the airplane's weight.

Divide the new moments by the new weight and you get the
new CG location.

Does that help?

Or do you maybe feel like I sometimes do after some
of your answers???


From: Brian Anderson - view profile
Date: Tues, Dec 10 2002 12:22 am

Jim,

EE's are the brightest of the lot - - - they can measure and calculate
things you can't even see.

The revised CG calculation is straightforward, but you need to calculate the
original moment first, i.e. the total weight [W] x the arm from the datum
[D]. Add to this the additional moment for the instrument, i.e. 8 lbs x the
distance of the instrument CG from the datum [d]. The resulting moment is
[W*D] + [8*d]. Divide this by the new total weight [W+8], and the result is
the new CG location from the datum.

Hence, new CG location = [[W x D] + [8 x d]]/[W+8]

I know even an elderly EE can follow that. After all, I is one too.

Brian

In article ,
"Kyle Boatright" wrote:

"Alan Baker" wrote in message
...
In article .com,
wrote:

A by-product of that lack of education is how Americans view
'horsepower,' typically insisting that 50hp (at 5000rpm) is EXACTLY THE
SAME as 50hp (at 1000rpm). Indeed, most will whip out their calculator
and 'prove' they are identical :-) But as the Wright brothers
discovered more than a hundred years ago, horsepower is not a factor in
the equation of flight. With powered flight, the factor we must
concern ourselves with most is thrust. Working back through the
equation, for a given propeller efficiency & rpm we will eventually
arrive at a given quanta of torque which then may be converted into
units of 'horsepower,' should we wish to do so, although it serves no
useful purpose. But when we DO use 'horsepower' we must be careful to
never use it in isolation, always identifing the rotational speed at
which that 'horsepower' is being produced.

Absolutely and utterly wrong.

It is *torque* which must always be associated with the rotational speed
at which it is being produced.

--
Alan Baker

Alan,

You do understand that if you know the HP at a given rpm, you can easily
calculate torque.

KB

Yes. But if you know torque without knowing RPM, you can't tell what
performance you can get out of an engine, but if you know horsepower,
you do know what performance you can get.

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

In article .com,
wrote:

But when we DO use 'horsepower' we must be careful to
never use it in isolation, always identifing the rotational speed at
which that 'horsepower' is being produced.


Absolutely and utterly wrong.

It is *torque* which must always be associated with the rotational speed
at which it is being produced.

Read that first sentence again. He's not wrong; he just
didn't specify "torque" for those who don't know the relationship
between it and RPM and HP.
When you say "absolutely and utterly" it should be used
only where it applies. Clearly, that's not here.

But that's my point. He is absolutely and utterly wrong, when he says
that you need to know the rotational speed before you know all you need
to know when you know the horsepower.

With horsepower, you can use gearing to get any rotational speed you
want; the horsepower remains constant. Torque changes with gearing.

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

"Chris Wells" wrote in message
...

How are "normal" airplane engines tuned to run at a lower rpm? What
changes would have to be made to an automotive engine to shift the
power band down accordingly?

Lengthen the stroke. High RPM engines have a large bore, and short stroke.
Low RPM engines have a longer stroke, and smaller bore, all else remaining
equal.

Al

wrote in message
oups.com...

Lou wrote:
I'll agree with the automotive engine with PSRU being heavier, but are
you sure about your other statement "the lighter the better"?
I'm currently looking at an engine that is 100lbs lighter than the one
recommended for my plane. Although cutting 100lbs from the total
weight is a dream come true, it brings up the question of weight and
balance. I can move the engine forward to make up the difference in
balance, but I don't know how far or how to find out.
Lou

You weigh the airplane without the engine installed and
calculate a balance point for it. Knowing the weight of the engine, you
then figure the arm at which it needs to be located to bring the
airplane's empty CG to the point the designer calls for it. Not a big
deal at all. Pages 134 and 135 of William Kerschner's Advanced Pilot's
Flight Manual shows how.

Dan

However, you will also be changing the area and arm relationships of the
side view of the aircraft (there is a name for this which I cannot recall)
and the size of the verticall fin will need to be increased if you are to
retain the same yaw stability. Then, because of the increased area of the
vertical stabilizer, a larger rudder would be needed to retain the original
crosswind landing capability. In addition, due to the increased planform
area forward of the CG, a larger horizontal stabilizer may well be required
to prevent any sort of deep stall or flat spin tendency. Finally, just as a
larger vertival stabilizer requires a larger rudder, a larger horizontal
stabilizer will very likely require a larger elevator.

To put it another way: Engineering is the science of compromise, and an
airplane is a series of compromises flying in close formation.

Peter

In article ,
Chris Wells wrote:

How are "normal" airplane engines tuned to run at a lower rpm? What
changes would have to be made to an automotive engine to shift the
power band down accordingly?

You are entering an engineering thicket when you decide to convert
automobile engines to aeronautical use.

One item nobody has yet mentioned is the matter of thrust bearings. An
automobile engine is designed to deliver its power through a clutch, to
a gearbox, with relatively low axial forces imparted to the crankshaft.

A direct-drive aircraft engine, however, delivers its power to a
propeller, which pulls (or pushes) axially on the crankshaft. If you
took an automobile engine and hung a prop on the end of the crank, you
amy or may not have enough thrust bearing to take the loads.

A properly-designed PSRU will have a thrust bearing to take those loads.
Some PSRUs, however, may impart side loads to the power end of the crank
and result in wear and fatigue issues.

There is an axiom for homebuilders: If you want to develop engines,
convert automobile engines; if you wish to fly, use aircraft engines.

I'm well aware of the purpose of the PSRU, I'd like to know if it's feasible to convert an automobile (or other) engine to run at an RPM low enough so that a PSRU wouldn't be necessary. I'm thinking a custom camshaft would be needed, and different ignition timing, what else?