I contend that this is not true. Pascal’s Principle shows that the downward force on the pistons is not the same. Burning the same amount of fuel/air in both engines results in identical cylinder pressure. Because (Force = Pressure * Area) the small bore engine has less force being exerted on its pistons.


Torque = Force * Distance or

Torque = Pressure * Area * distance

Two 8 cyl engines: both 602cu in, Bore/Stroke 4"/6" and 5.657"/3"

Cylinder pressure of 500psi

18855lb/in = 500psi * 12.57sq in * 3in

18848lb/in = 500psi * 25.13sq in * 1.5in

The 7lb/in difference is rounding error.

EDIT: forgot to divide stroke by 2

EDIT: Engines of equal displacement having different strokes and experiencing the same cylinder pressure will produce the same torque at the crank.

http://www.grc.nasa.gov/WWW/K-12/Win...principle.html

[This message has been edited by Brent (edited February 25, 2001).]

 

this article has good info on this subject starting with the last paragraph on the first page...check it out, it goes over a few things you didn't consider.


http://www.airflowresearch.com/Articles/A3-P1.htm

 

Brent,
You just calculated the torque at the piston, not at the crank.

If leverage did not multiply force then how do you explain the handle on your torque wrench?

 

ODB: There is no such thing as torque at the piston. The distance variable is the length of the crank arm (stroke/2). If you follow my example you can see that the smaller piston negates the increased stroke.

Torque @ Crank = Cyl Pressure * Piston Crown Area * Stroke/2



[This message has been edited by Brent (edited February 25, 2001).]

 

You're right. Technically there is only force at the piston.

So are you telling me that the diameter of your torque-wrench negates the torque multiplying of the length?

 

Or is it that a smaller diameter hand cannot generate the same force on the handle?

I still don't get what you're saying.

 

I'm saying the smaller piston exerts much less force on the crank than the larger piston.

PistonDownwardForce = CylPressure * PistonCrownArea

The small piston (4") exerts 6285lbs of force on the large crank (6").

The large piston (5.657") exerts 12565lbs of force on the small crank (3").

 

so the same charge of gunpowder would exert more force on a cannonball than a bullet?

You are severely confused I am thinking.

The angle at which the force is applied makes a huge difference in torque at the flywheel.
Force applied to a piston at TDC does little for torque. Force applied at half stroke makes the most difference. Now what does this have to do with piston size?

The LS1 piston is smaller, so I guess that makes less power and shrouds the valves to death. hmmm.. funny how they still run in the 11's STOCK with highway gears (2.73). They must cheat.

 

how are you quantifying the force here?
Do you mean just instantaneous? because an engine doesn't work like that.
The amount of time the force is applied means more than the peak force.

methanol burns very slowly and is still applying force at mid stroke. That's one reason it makes so much more torque than gasoline.

 

Please explain to me what Pascals Principles mean.

For example: Why use a humungous, heavy, large diameter hydraulic ram on ammusement park rides? Because Pascal tells us we would need unsafe amounts of oil pressure to develope the force we need to move the ride.

I believe ignition timing takes care of making sure the burn happens when it's going to make the most power. Timing is affected by the bore diameter in some cases.

I have not insulted your intelligence, please don't insult mine. The example you and I are using holds displacement and cyl pressure constant which means the cyls are filling the same. We are only comparing lowerend dimensions and the effect they have on torque output.


[This message has been edited by Brent (edited February 25, 2001).]

 

Brent, I'm sorry but there are a few discrepancies...
1. You are under a misconceptions if you believe this to be true...
If you burn the same amounts of fuel and air, the smaller piston will exert more force than the larger piston, because there is more cylinder pressure. Same material matter in a smaller volume yields a higher pressure.

2. Pascal's Principle. When most people refer to Pascal's Principle, they refer to the following...
Pascal's law states that when there is an increase in pressure at any point in a confined fluid, there is any equal increase at every other point in the container.'
You can't really use this in terms of cylinder fill/compression. True it is an overall true statement in fluid dynamics, however, this principle is generally used in hydraulic fluid, and assumes an uncompressable state of the fluid in which to transmit force.

3. Assuming the same amount of produced power at the piston, the longer stroke will exert more force on the crankshaft centerline. A good example? Try moving a 100 lb. boulder with a 2x4x4 and then try with a 2x4x12. Pivot point being the same (crank centerline) which one will allow for easier movement of the boulder with the same amount of force?

Hope that cleared things up a bit

 

This is getting good, keep going... We will all be faster.

------------------
87 Formula 350 13.94@96.67
98 Z28 1LE 13.15@107.59
1LE Owners Association

 

I posted on this at NE3G.

If the stroke is longer, than to equal the displacement means you need to decrease your cylinder size (pi*r^2*h=V). Pi is constant so if h increases, only way for V to remain the same is if r decreases. That combined with Jesters velocity theory are what limit the perfect world that is mathematics.

Allow me to elaborate a little more. When you have a smaller cylinder than the valves have to be smaller and thus the airflow is reduced and now you have reduced airflow with an increased demand which multiplies the problems.

I think your problems are a little off. Any formula is best oringally proven if you come out with the correct units first.

500 N/m^2 * 60 m^2 * 3.75 m = # Nm which is a torque value.

I did it in metric because it's easier to see. Someone could use english measurments to displacy the same thing.



[This message has been edited by 84FTA (edited February 25, 2001).]

 

right we're not talking about liquids here. Gases are compressible.

you deducted power agaist the piston for being small diameter and gave it power simply for moving more (stroke)..
Neither case is true.
X-amount of energy will exert X-amount of force against the piston no matter what its size. The force will just be less per square inch with the larger piston.. doesn't matter one bit.

A 3.75" stroke crank has 25% more leverage than a 3" stroke crank.. period.

the final outputs are the result of several trade-offs in the realworld, but still the leverage advantage is there and makes a world of difference in practical application.

a word about tuning.
you cannot control your burn rate using timing.. period.

the fuels burn rate controls where your optimal timing needs to be..period.

 

84FTA,
valve size has nothing to do with what we're talking about.

also you can NEVER assume you have more airflow with bigger valves, and less with smaller valves. It doesn't work that way and I thought I'd explained why already.

We are talking about an equal mass charge of air and fuel burning at an equal rate.

 

Sorry for not conveying my ideas fully. I have tendencies to think through and post only one half of it. Yes, the valve thing is true, however, this whole post is based on generalities. If you want to debate how methane combusts as opposed to other fuels you're using generalities, there are fluxations to the enviroment in each case. The point I am trying to make has nothing to do with your debate but the oringal post of making the idea of something with a longer stroke will make more power, the problem with this is the attempt to make equal cylinder pressures, wether constant or varying, to be equal. Yes, if the above lab conditions are perfect it does work, however, achieveing that numerical cylinder pressure in each engine is going to be more challenging to the one with the longer stroke.

------------------
1984 WS6 Trans Am Hartop
Former L69 Car under restoration
1984 Trans Am T-tops
4-bolt main 350, performer intake, headers, Holley 650, T-5, hayes clutch, dual elec. fans and 3.23's.
Daily driver and restoration
13.98 @ 101

 

Sorry for the long post but Brent is right all other things being equal two engines of the same size will make the same horse power. Below is the equation for finding HP based on force*distance. MEP is the force on the piston. 33,000 is the constant. The equation is kind of long but it shows that a longer stroke does not make more power than a large bore. There are other things that effect hp like frictonal loss and other variables. The easier way of thinking of it is if you had a 5lb wieght on a 4' lever or a 10lb weight on a 2' lever.

Both engines below are 60.13ci per cyl.

MEP=100psi
Bore=4.125"
Stroke=4.5"
RPM=2,300
Rod angle factor 1.57

F= Force
D= Distance
d= stroke

Force on piston- 100psi
Area= 3.1416*2.062²= 13.36
Force= 100psi*13.36= 1,336 lbs force on pistons

Force at crankshaft with rod angle factor
F=1336/1.57=850lbs

Horse power per cylinder is found by F*D/33,000

Distance for one revolution- 3.1416*4.5/2
you dived by 2 to get circumfrance for power stroke.

3.1416*4.5/2= 7.068"/12 for feet

7.068/12= .589'

Feet per min traveled= RPM*circumfrence/2

2,300*.589= 677.35 ft/min

850*677.35/33,000= 17.44HP per cylinder

You would use the same formula as above for the next engine.

Bore-4.375"
Stroke-4.00"

Area=15.03
Force=100psi*15.03= 1,503lbs of force on the crank

F=15.03/1.57=957lbs

3.1416*4/2= 6.283/12= .523'

2,300*.523/2= 601.45 ft/min

957*601.45/33,000=17.44HP per cylinder

 

I hate to break this to you this way,
but the engine doesn't use calculations to make engine power.

maybe you do, and if you really believe all of what you said then please
put it to use to prove it in the real world.

I'll call it:
3000 RPM's
the engine with 25% more stroke will make much more cylinder pressure,
and do it EASILY.

Those equations for horsepower are meaningless. They ignore the true behavior of air in an engine.

And finally (forgot to say this earlier), an engine is NOT an air pump...


The power output of an engine is based closer to how much air it can TRAP in the cylinders for combustion, NOT on the volume it can push through.
You can push twice the gases out of the exhaust but if it doesn't get trapped in the cylinder and combusts efficiently then you make no more power than before.

one more time.
power is not DIRECTLY related to airflow.

 

as unimportant as horsepower really is,

aaaye nevermind.. will pick this up later.
ODB


[This message has been edited by The ODB (edited February 25, 2001).]

 

more later, I'm tired..
didn't mean to post this

[This message has been edited by The ODB (edited February 25, 2001).]

 

ODB so what your saying is bore means nothing right so a 305 should make the same power as a 350 because they have the same stroke. Dosen't work that way. Why is it so hard for you to grasp that a larger piston puts more force on the crank. And yes power is directly related to air flow thats why blowers work. ODB you're wrong.

 

Well, here is what I know:
say you have 2 rounds of ammo of different sizes, but the amount of gun powder in the cartridges is the same (therefore you have an explosion inside of equal force) then ODB's reasoning would be right, because the gun powder would be able to push the smaller bullet with more force due to weight and more "concentrated force" (smaller diameter or bore). This same theory however, does not work for an internal combustion engine. The bigger piston has the advantage on this case because it is more able to suck (vacuum) larger quantities of air on the intake stroke that would mix with even more gas, consequently having a much bigger explosion, much more gases expanding and pushing this bigger piston down with way more force than the smaller diameter piston.

Just put it this way: a maxed out 305 will never maked more power than a maxed out 350.

To put all this on perspective, a 500 ci prostock engine (just like warren Johnson's) will suck so much air, that all this air could fill up a hot air balloon by the time it goes through the lights in a 1/4 mile. I read this on a magazine and I'm pretty sure they said it could fill up the Goodyear blimp (which is way bigger than a hot air balloon), but I'm not 100% sure (don't call me a liar, that's what I read). I told this to my friends and they laughed and said "really?! so it sucks all the air in?! How do you breathe if you are close to it?!
It took me a second, but I answer back " it comes out the exhaust you idiots, is not like is gonna suck the whole world and keep it in". They all shut up and mentioned nothing about it anymore. I'm pretty sure they were very embarrassed to say such a thing.
Rick

 

From what i've read here, brent is right, the rest of you arguing seem to be misunderstanding him.
I picked this example from red devil, becuase it perfectly exemplifies whats wrong. If you burn the same amount of fuel/air mix in 2 engines with the same diplacement, you will get the same cylinder pressure. Why? Simple becuase you are blowing up the fuel/air in the same volume. The example started with the same displacement engines, let's keep it that way. Yes if you have the same fuel air mix in a smaller engine the cylinder pressure would be higher, but we weren't talking about a smaller engine vs. a large engine. And bore size has nothing to do with displacement in our example cuz we are varying the stroke to compensate.
I too don't really like seeing pascals principle applied here, but the idea he's getting across is the same no matter what you call it. pressure x area, since pressure is the same, and area is smaller we have less force pushing down on the crank with the smaller bore. Since the smaller bore has a longer crank throw it will make up that force difference and produce the same torque. Of course difference in rod stroke ratio play into this. And then theres the differeing frictional losses of the small pistion vs. large piston, and long stroke vs. short stroke. And then there's the airflow differences between a small bore and a large bore. And differences in timing and detonation management in a large vs. a small bore.
It's impossible to build a real world comparison of two identical displacement engines with differeing strokes and get ANY type of meaninful results. All those 'extra' factors (that i barely scratched the surface on' would skew the results, and the best attempts to 'level' the field would only make the experimenter give his bias to which combo he liked to prove his point. That is why this post is focusing on equations. If equations did not dictate EVERY aspect of the universe, we would not have engineers. And i wouldn't be getting paid to surf the web all day
...ed

 

I think you guys are all talking about the same thing. I dont think odb is trying to say that a smaller bore of the same displacement generates more power or downforce with the same intake charge.
I think all he is trying to say is with the power from the combustion being equal, the longer rod can exert more force due to the lenght of the rod and its angle on the centerline of the crankshaft.
I am not quite awake yet though and I may have missed this completely. If so I apologize.

 

bzzztt.. wrong.
blowers, turbos, or nitrous does not increase the rate of airflow into the engine. They simply modify the density, and more importantly the oxygen content.

 

not exactly. I'm saying that the leverage advantage of a longer stroke means way more than the increase in piston area in a real world application.

 

Brent is right, any loss in stroke is made up with the extra area of the piston, and vice-versa. Assuming both engines remain the same displacment.

OBD do you disagree with a larger area piston having more force applied to it?

 

Not in the example, but in the real world yeah I disagree.
Brent has the calculation down right. I checked it, but it doesn't work out like that in actual function of an engine.
He was completely wrong about the timing controlling burn rate.

As far as the engine comparison goes, there is something wrong with that calculation, but I'm stumped right now finding it.
Someone help..

 

If you re-read my post it says nothing about controlling burn rate. Burn rate is a function of mixture, chamber design, quench, etc etc. Ign timing is for starting the burn at the proper time.

 

I don't know about the burn rates, but I'm quite sure that in theory, (not taking into account valve flow, piston to cylinder friction, rotating mass, etc.) the two engines should produce the same torque, on a given revolution.

 

ODB, anyone responding to this post already knows the limitations and idocies of attempting to use a simple equation (and these are simple!) to estimate the power of something that is a complex machine. Again, but under ideal, restricted, perfect, conditions his equations are true. Are they useful? Doubtful.

The best way to model this is using DD2K. They have taken the time to make as accurate as possible equations to model real life behavior of engines. Quick Math here.
V = (pi*r^2*d) * 8
349.8 = (3.14*r^2*3.75)*8
D(diameter) = 3.853

Make an engine with a stroke of 3.75 and a bore of 3.853 and it will have the same displacement as a normal 350. If my webstie cooperates tonight I'll post some graphs.

 

http://84transam.8m.com/DD2KGRAPH.jpg

Dyno 1 is a 'normal' 350. Has a stock mufflers and manifolds, 600 cfm dual plane carb, on a Sum-K1101 (close to stock for a 350) along with stock heads and stock manifolds

Dyno 2 is our stroked 350. It has all the above except smaller valves to clear the cylinder walls.

As you can see the regular 350 (dotted line) outperforme the new 350 very easily.

------------------
1984 WS6 Trans Am Hartop
Former L69 Car under restoration
1984 Trans Am T-tops
4-bolt main 350, performer intake, headers, Holley 650, T-5, hayes clutch, dual elec. fans and 3.23's.
Daily driver and restoration
13.98 @ 101

 

thanks 84FTA ( i think)
I couldn't see the graphs, but you said the larger bore engine did better.

could you punch up my combo and see what it says. I've used a couple of desktop dragstrip programs and none of them were even within a second of my actual times.

My next time running my stroker 383 should be in the low 10's N/A... pick a number. Then just change to a bigger bore and less stroke to equal 383 and tell me what it says. I would consider trying anything that I thought might work and needs to be tested.
thanks again.

 

info on my combo



------------------
*I do custom performance mods on Edlebrock Performer carburetors (dualplane intake mods in the works),
White 1986 Irocz, 305 with iron #416 heads,
383 with aluminum TFS heads,
Edlebrock Performer-RPM intake and Performer #1407 carburetor, +110hp shot of crack, 700R-4 tranny, Vigilante 2400 lockup converter, 3.25:1 Ford 9" rear, Mcreary Road-Stars, SLP-stainless 1.75" shortie headers & Y-pipe, single 3" Borla exhaust, Linginfelter-TPI camshaft part number 74216 pulls 17" vacuum solid. Cam specs 213/219 @.050 114-LSA, sometimes advertised at 216/219 @.050 112-LSA .462/.470 lift @1.5:1 ratio. Using Harland Sharp 1.65:1 roller rockers. MSD-6AL, billet distributor, multi-retard, blaster-3 coil, and RPM switch. SouthSide machine subframe connectors, SSM lift-bars, Moroso 4" underdrive crank pulley.

N/A runs 10.9 @124,
Crack-runs 10.3 @135
haven't run at track since Oct-99

 

I do not appreciate your implication that I am an idiot. ODB's original statement was restricted to same CID and cylinder pressure (IMEP). All the induction, cam, heads combustion efficiency etc factors that influence cylinder filling and emptying do not matter at this point because IMEP is the direct result of them.

The fact is, breaking down the components of engine operation is useful in understanding them. How do you think they model engines in DD2K? They compute the factors, then add them up.

I agree his example has no meaning in the real world because stroke and bore are dictated, to an extent, by the blocks we use. Still, his assertion about stroke Vs bore in equal CID engines is not true. In fact, its backwards once you add in friction, because a the longer stroke engine develops more friction.

Cyl Pressure or IMEP (Indicated Mean Effective Pressure) is measurable using a pressure transducer inserted into the combustion chamber. Indicated HP = IMEP * CID * RPM / 792000. Notice how it does not depend on bore or stroke. Friction HP = FMEP * CID * RPM / 792000.

BHP = IHP - FHP

A better explanation is here: http://www.racer-x.f2s.com/proc/stroker.htm

 

I didn't imply that you were an idiot. It wasn't aimed towards anyone. Just a general statement. (This is my last comment on this because I'd rather see the facts come to conclusion and a useful debate arise rather than it getting locked because of flaming).

Anyhow. the difference in fiction would be neglible. Even though the stroke is longer, the circumference around the circle is smaller.

3.75 x 3.855 x 3.14 = 45.4 sq. in of cylinder wall

3.48 x 4.000 x 3.14 = 43.7 sq in of cylinder wall

Steel to steel (lubricated) as something like a CoF of around .02 if I recall. So the difference in friction is negligble.

------------------
1984 WS6 Trans Am Hartop
Former L69 Car under restoration
1984 Trans Am T-tops
4-bolt main 350, performer intake, headers, Holley 650, T-5, hayes clutch, dual elec. fans and 3.23's.
Daily driver and restoration
13.98 @ 101

 

you're right, the friction would be very close to the same.
strokers have higher piston speeds,
but large pistons would tip side-to-side more.


I think that if an engine were hydraulic, then your equation holds true, and I'm still not exactly sure why it doesn't work real-world.
It may have something to do with piston speed and the burn rate of gasoline.

 

I think the problem arises with airflow which is discussed on Parson's site. Because if I create either a 1.90/1.60 fake valved 3.855 or a small valved 350 with 1.80/1.50 (what I used to for my 3.855 headers). DD2K would give them both the same airflow reguardless of stroke (one of it's downfalls IMHO). When I do this they have nearly exact curves, which now gets them even, but doesn't concern the fact that the stroke should make more. The two other factors I've put into consideration for this phenomenon after the airflow are these.

Rod angle realationship to the crank. When the Force down is at it's maximum, then to create the most rotational engery the the rod would have to be closest to tangential on the crank.

Another idea I've come up with is the fact that the piston on it's down stroke now has to push 7 more pistons with a higher speed.

------------------
1984 WS6 Trans Am Hartop
Former L69 Car under restoration
1984 Trans Am T-tops
4-bolt main 350, performer intake, headers, Holley 650, T-5, hayes clutch, dual elec. fans and 3.23's.
Daily driver and restoration
13.98 @ 101

 

Ok, lets simplify this a tad, to try and clarify this, the same displacement thing can be boiled down to this...
It is agreed that in both instances the potential energy in the combustion chamber from the combustion process is the same. We then move out of fluid dynamics altogether and move into some other aspects. I do not have too much time, so I'll just do 1 clip for review right now...

Examples of levers of the First Class are see-saws and crowbars.
For the see-saw:
part A part C
-------------------------------------------------------
^ part B
If nothing else, part B modifies the direction of the applied force;
for example, if part A is pushed downwards, part C will go upwards.
But depending on where part B (the fulcrum) is located, it can also
modify the quantity of the force, which is applied at part A and
received at Part C:
A C
-------------------------------------------------------
^ fulcrum
In this variation, when A is pressed downwards by any force, the
fulcrum will cause it to be multiplied, and a larger force will
cause C to move upwards.
A C
-------------------------------------------------------
^ fulcrum
In this variation, when A is pressed downwards by any force, the
fulcrum will cause it to be divided, and a weaker force will
cause C to move upwards.

There is a precise mathematical relationship here, involving only
four things:
1) the distance between A and the fulcrum (D1)
2) the force applied at A (F1)
3) the distance between C and the fulcrum (D2)
4) the force received at C (F2)
The relationship is simply this: (F1)(D1) = (F2)(D2)
(Multiply F1 and D1, and their product will be the same as the
product of multiplying F2 and D2.)
If we know any three, we can calculate the fourth, but the item most
often sought is F2, the force received at C: F2 = ((D1)/(D2))(F1)
(Divide by D1 by D2 to discover a "force multiplier", and then use
that on F1. This may look ugly on paper, but let's think about it.)

In the original see-saw, the distances from A and C to the fulcrum
are same (there are 27 hyphens on each side of the ^), so D1 = D2.
Mathematically these cancel each other out, and give us a force-
multipler of just 1. That means the F2 force received at C is the
same as the F1 force applied at A (except for the changed direction).

In the first variation of the see-saw sketched above, the distance
from A to the fulcrum (D1) is rather larger (45 hyphens) than the
distance D2 from C to the fulcrum (5 hyphens). When D2 is divided
into D1, we get a force-multiplication of 5. Therefore the force
received at C will be 5 times stronger than the force applied at A.
(NOTE: There is always a price to pay! If A is moved downwards a
distance of 5 units, then C will only move 1 unit of distance upwards.
The "mechanical advantage" of being able to apply a small force at A
is balanced by the disadvantage of having to apply it across a large
distance, to obtain a decent amount of movement at C.)

The second variation of the see-saw, sketched above, is just the
opposite: D1 is 5 hyphens and D2 is 45 hyphens. Therefore the force-
multiplier is 1/5 (a mechanical disadvantage); the force received at
C is 1/5 the force applied at A. Here the price of having to apply
a large force is balanced by the benefit of having to move A only 1/5
the distance that C will move.

The way you arrange Part A, the fulcrum, and Part C, in a First Class
lever, depends on what force you have available, and what you need
to accomplish with it. The ancient Greek scholar, Archimedes, who
discovered the Law of the Lever, is supposed to have said, "Give me a
lever long enough, and a place to stand, and I will move the world."
This statement is only partly correct, because he would also have
needed a place to put the fulcrum.

Now let's look at the crowbar, which is generally non-adjustable:
\ A
\
\
\
\
\
\
\
\/ C
fulcrum
This sketch shows a distance of 9 for D1, and a distance of 1 for D2.
The same mathematical relationship applies (indeed; it applies to ALL
levers), so any force applied at this crowbar's A will acquire a
mechanical advantage of 9 when prying something at C. The bent shape
of the crowbar affects the way the force changes direction, and that
is the ONLY significant difference between it and the see-saw. It is
still a First Class lever because of the overall arrangement of A,
fulcrum, and C.
============

An example of a lever of the Second Class is the wheelbarrow. As you
can see from from this greatly simplified sketch, the arrangement of
A, fulcrum, and C is different from the First Class.

| |
A (handle) | body coNtains C |
-------------------------------------------------( ) fulcrum
V (ground support) (wheel)

In this sketch, the number of hyphens only approximates the actual
distances from A and C to the fulcrum, because the true fulcrum is the
axle of the wheel. Nevertheless, the Law of the Lever still works:
(F1)(D1) = (F2)(D2). In this approximation, if we only count the
hyphens, then D1 is 49 and D2 depends on where in the body C is located
(7 here). That would give us a mechanical advantage of 7, but usually
a wheelbarrow gives much less, because often the load in the body is a
lot closer to A. For example, the above phrase "body coNtains C" is
balanced aound the capital 'N'. If that whole phrase was the load,
a physicist would say the 'N' represents the "center of mass" of this
load. And since the number of hyphens from fulcrum to 'N' is 14, when
this D2 is divided into the D1 of 49, the mechanical advantage is 3.5.
It should be noted that for a Second Class lever, the particular
arrangement of A, fulcrum, and C will always yield a mechanical
advantage, and will never yield a mechanical disadvantage.

One frequently-overlooked fact about wheelbarrows concerns the ground
support, indicated by 'V' in the sketch. It too can be used as a
fulcrum! The user can press downwards at A, and the wheel will lift
upwards, perhaps to clear a curb at a construction site. This usage
is simple First Class leverage, already described, but the load in
this case is not simply C (or even N); it also includes the body of
the wheelbarrow and the wheel! Even so, it is reasonably obvious
from the sketch that the user is at a mechanical disadvantage; a
considerable amount of force has to be applied at A, to get the wheel
to lift. This is why someone who uses a wheelbarrow a lot will
install a ramp at the curb.

Another example of a Second Class lever is the simple nutcracker.
If you imagine one handle of the device to be stationary while the
other does all the moving, then you can easily envision how the
arrangement of A, fulcrum, and C is equivalent to that of an
upside-down wheelbarrow (with the wheel attached to the ground, and
the body removed).

============
An example of a lever of the Third Class is the human fore-arm:
|
muscles attach to |
C (fingers) fore-arm bones at A |
-------------------------------------- fulcrum (elbow)

As you can see from this sketch, A and C are reversed, relative to
a Second Class lever. A Third Class lever is always at a mechanical
disadvantage. In a typical human fore-arm, the force received at the
fingertips may be about 1/8 of the force applied by the muscles.
However, as mentioned with respect to First Class levers, there is a
trade-off of force for distance. The muscles attached at A only have
to move a small distance to make the fingers move a large distance.
(If the mechanical disadvantage is 1/8, then this distance factor is
8 times, of course.) Again referring to that Second Class lever, the
wheelbarrow, the fact that it always has a mechanical advantage, and
makes a load easier to lift, is balanced by the fact that the load
doesn't lift very far. But it is enough to let one move the load.

Another example of a Third Class lever is a type of ancient catapult,
the onager. It had ropes wrapping a fulcrum/axle; the location where
the wooden catapult-arm escaped this mesh is equivalent to the place
where human muscles attach to the fore-arm. The ropes were twisted
tighter and tighter to provide force to move the catapult-arm, which
was first locked into place. When freed, the catapult-arm moved VERY
quickly: Enormous force was traded for the ability to toss various
objects onto a distant enemy. (This type of catapult became obsolete
partly because the ropes were too easily damaged, and partly because
a superior catapult based on the First Class lever was invented, the
trebuchet.)


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"A lawyer with his briefcase can steal more than a hundred men with guns."

-Mario Puzo

 

Quick addendum before bed...
Short levers will act faster, but weaker, than systems with long levers (which are slower but stronger. This is essentially what we all overlooked in this discussion, we focused on the problem too narrowly, rather than generally.

 

I know it seems that everyone these days wants to build a 383 stroker because of the high torque, makes a great street and strip engine, but as far as strickly race engines go I thought the 377 destroker was the hot engine. The larger bores unschrouds the valves making for better high rpm air flow and also the shorter stroke allowing it to rev quicker.

I just want you to know I have no personal experiance in building either engine but I do know I would reather have a 377 in a drag car fer shur.