Differences between automotive & airplane engines

Also level the plane as it would fly though the air.Only my $0.02.
LJ

Richard Lamb wrote:
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

"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?

Longer stroke.

--
Geoff
the sea hawk at wow way d0t com
remove spaces and make the obvious substitutions to reply by mail
Spell checking is left as an excercise for the reader.

"Peter Dohm" wrote

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.

It should not change all that much, I'll bet. If you look at that heavy
engine moving a few inches, and the increased cowl area in front of
aerodynamic center pressure, then look at that long, long arm back to the
fin and rudder, it should only take about a third of the area the engine
added to make it all work out. Increase the fin/rudder height a couple
inches, or add a small dorsal fin, and all will be well in the world. :-

Reminder: all usenet advice is worth what you pay for the advise.

To the OP; what are you using that is 100 lbs lighter, and what was the
original? That is a nice weight savings!
--
Jim in NC

"Chris Wells" wrote in message
...

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?


--
Chris Wells

Changing a cam has *some* impact, but it doesn't change the amount of
fuel/air mixture the engine can burn on each combustion stroke, which is
what really determines the output of an engine. That's why small engines
have to spin fast to develop, say 200 hp, while larger engines can run
considerably slower. Changing the timing? Well, the slower an engine
turns, the more the timing must be retarded to prevent knocking. Most
engines (auto or aviation) have their ignition timing set to produce the
most power possible while leaving a margin against knock or pre-ignition.
So changing the timing may not be practical, nor is it likely to turn a
pig's ear into a silk purse.

The bottom line is that unless you're willing to turn a relatively large,
heavy engine slowly and with a disappointing power/wt ratio, there really
isn't a good way to take an automobile engine, bolt it onto an airplane, and
go flying.

Auto engines are designed for what they do, as are airplane engines. Same
thing for horses and camels. You won't win the Kentucky Derby with a camel,
but neither is a thoroughbred going to survive long in the desert...

KB

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?

You need much longer stroke, which would mean a crank with longer
throws which won't fit in the crankcase anymore, and a longer cylinder
to accomodate the longer piston travel, which you're not going to get,
either. Take a look inside any auto engine crankcase sometime, and see
how close all that stuff is running; there's not much extra room. Chevy
took the 283 and made the subsequent 305/307/327/350/400 engines out of
it by boring larger diameter cylinders (which required just a bit more
casting thickness) and a crank with a tiny bit more throw, which they
managed to squeeze into the case. See
http://www.aces.edu/~gparmer/sbc.html
The Chev 350 V8 has a bore of 4" and a stroke of 3.48, while my
old Gypsy Major, a four-banger that had about the same displacement,
had a 4 5/8" bore and 6 1/2" stroke. Big torque. Redlined about 2650
rpm. Modern aircraft engines like the O-320 are more oversquare like
the auto engines, but still have longer strokes of about 4".
The car engine's crank, as someone else pointed out, won't take
prop thrust loads well, and certainly can't handle the gyroscopic loads
the prop places on it. Even the direct-drive conversions usually have
some sort of extension and bearings to take those loads; those that
don't, like many of the VW and Subaru conversions, have had crankshafts
break in flight. Special forged cranks are required, but the bearings
in the case are still too light.
Try picking up a Lycoming 0-320 sometime: 280 lbs or so. Then
try picking up the Chev 327, almost the same displacement, and see what
it weighs: 575 lbs. Then pile on the radiator and some water, too. Your
airplane has to lift all that.

Dan

It should not change all that much, I'll bet. If you look at that heavy
engine moving a few inches, and the increased cowl area in front of
aerodynamic center pressure, then look at that long, long arm back to the
fin and rudder, it should only take about a third of the area the engine
added to make it all work out. Increase the fin/rudder height a couple
inches, or add a small dorsal fin, and all will be well in the world. :-

It will change it some. My Jodel came out considerably heavier
than designed, mostly to the use of birch instead of the mahogany
specified in the original French drawings, fabric over all ply surfaces
to meet Canadian requirements, a tailwheel insterad of a skid, and so
on. Since most of the added weight is behind the CG, it was tailheavy
and the engine had to go 11" further forward.
The longer nose side area results in a little less directional
stability with the same tail, and I won't spin it because I don't know
just what the effect of the extra weight in the tail (and the longer
nose arm to balance it) might do to the spin; it might flatten into an
unrecoverable situation. Sure does an awesome slip, though.

Dan

ORVAL FAIRAIRN wrote:
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.


Well put.. Converting an auto engine for aircraft use is not for the
novice to try. As Orval says. If ya want to fly now,, install a
Lyc/Cont/Rotax. My Zenith 801 is coming up on 100 hours and so far it
hasn't killed me yet. In fact I am headed out first thing in the mornin
in my auto engine powered toy to Idaho for breakfast to take advantage
of this good thick cold air. It's -6f now and headed to -20 by morning.
That always helps when one is based at 6600 MSl...
Ben
Jackson Hole Wy
www.haaspowerair.com

On Fri, 10 Feb 2006 06:12:41 GMT, Alan Baker
wrote:

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.

You are both arguing the same thing, since horsepower is the product
of torque and speed. With the commonly understood units of RPM and ft
lbs, the product needs to be devided by the constant 5252 to provide
horsepower.

I can say my engine produces 91 HP at 3000 RPM, or I can say it
produces 160 ft lbs torque at 3000 rpm, and I am saying exactly the
same thing. The fact the engine may also produce 140 hp at 5000 rpm is
totally immaterial except to indicate it MIGHT be able to stand up to
producing 90 hp for a significant amount of time without overheating.

All the torque in the world won't move anything if it is not allowed
to cause motion, or speed.

On Fri, 10 Feb 2006 16:28:10 GMT, Alan Baker
wrote:

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.

Yes, you CAN use gearing, at the expense of complexity.And efficiency.
Much better to design the engine to produce the power you need at the
speed you need it. However, sometimes you trade efficiency and
durability for weight - and a geared 1.2 liter 80 hp engine running
at 6000 RPM can weigh significantly less than a direct drive 2.7 liter
engine providing the same power at 2800 rpm. (well, about 40 lbs less,
anyway)

Alan Baker wrote:
...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.


Fantastic! I've been trying to solve the following problem, and
clearly you know how to do it. Please help: I have a 40 horsepower
motor but no idea what the rotational speed is. What gearing should I
use? Thanks in advance.

Daniel