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I believe it is 383

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I have a conversion tool on my computer at work and it comes out at about 372.25 cubic inches

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Alley said:Just curious, how many cubes does that equal on the 6.1 ltr?

Crazy Luck is correct.

Conversion tool or not, take any size liter and mutiply by 61 and you will get a very close cubic inch.

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Bore x stroke: 4.06 x 3.58 in, 103.0 x 90.9mm

Displacement: 370 cu in, 6059cc

Displacement: 370 cu in, 6059cc

More majic numbers in this way to figure required HP.

If you ever wondered how much HP is required to pull your vehicle along whether it be an 18 wheeler or motorcycle here's a method that is fairly accurate. There are two very small error conditions which tend to offset each other and are not addressed. You must know as close as possible the vehicle weight and decide at what speed in mph you want to test for and mph must be converted to fps(feet/sec) by multiplying mph X 1.46. Example: if testing for 60 mph we calculate the KE(kinetic energy) at 65mph and at 55 mph and then coast down from 65 to 55 mph and use a stopwatch timer to determine the number of seconds it took for the coastdown. From the KE at 65mph we subtract the KE at 55mph then divide by T(coastdown time in sec) and then divide by 550. The final number will be HP required at 60 mph. Now the formula to calculate KE is as follows: KE = W(weight in lbs) X V squared(V is ft/sec) divided by 2G(64.32).

As an example consider a 3000 lb car at 60 mph and a coastdown time of 10 sec from 65 down to 55.

KE at 65 is 3000 X 9006/64.32 = 420,056 ft/lbs

KE at 55 is 3000 X 6448/64.32 = 300,750 ft/lbs

The difference is 119,305 ft lbs divided by 10sec = 11930 ft lbs/sec divided by 550 ft lbs/sec = 21.69 HP.

In testing the coastdown time choose a fairly level road with as little wind as possible and for the above example drive at 70 mph, shift to neutral, start timer as speedometer goes by 65 and the stop at 55; you may want to do about 3 and take the average.

One of the small errors is that the flywheel effect of the tires, wheels, brake discs/drums, axles and drive shaft all help to keep the car from coasting down, but do not help to propel it along. If we knew the weight and "radius of gyration" of these and then calculate the power between 55 and 65 mph it could be deducted from the final number. In the above example for 1 wheel/tire/brake weighing 30 lbs and with a ROG of 1 ft the power would amount to only about .07 hp.

It seems that yet another small error is due to calculating the KE at 55 and 65 mph the V is squared and perhaps it would be a little more accurate if we used the "root mean square" of these rather that the common mathmetical average. The RMS of 55-65 is 60.2.

So in the first error we would need to deduct a little hp and in the second we add a little to the final number. And these amounts are very small and tend to offset each other.

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Where are you located now?Moparitis said:maneval69 from: Apex, NC

In 1951 I was in NC near Fort Bragg on ARMY manuevers. Near Lillington, Dunn; walking distance from a village named Linden.

________

MARIJUANA CARD

About 40 miles South of Tyler, Tx.maneval69 said:Where are you located now?

In 1951 we were bivouaced about 1/2 mile South of Linden on the bank of a fairly large running stream that dumped into the Cape Fear river about 3 miles West of Linden. We could walk South and hit what was then a dirt road and is now 2027 and walk West on it to a RR track and the walk South on it across the stream and into Linden where we could buy drinks etc. Don't recall if they had beer or not, but I found a bootlegger soon as we got there. Oh, by the way I drove a GMC 6 X 6, (duece and a half) both ways across the Smokies from Ft. Campbell, Ky and then drove a Jeep round trip from Campbell to Ft. Hood, Tx.

Here's one I did about Top Fuel Dragster Horsepower:

Some years ago I was sorta skeptical when reading about 6-7,000 HP + in the Top Fuel dragster engines in cars that can cover the standing 1/4 mile in 4.44 seconds at 333 MPH. Some time ago I came up with a rough formula to try and confirm this HP rating.

To determine the force reqd. to cause a certain acceleration I used the age old formula F = M X A where F = force in lbs, M = mass & A = acceleration in ft per sec per sec.

Given that the car weighs about 2250 lbs for a mass of 70.

The tire is 36" tall for a radius of 1.5 ft.

Assume a coefficient of traction of at least 1.

The rear end gear ratio is 3.2 to 1 and there is no transmission used.

A slipper type clutch is used with variable lock up rpm/time.

Timing devices indicate that after 1 sec. the car is at 125 mph or 182 ft per sec.

The car has accelerated at a rate of 182 ft per sec per sec and may exceed this.

We will assume 6,000 rpm at the 1 sec time, but the clutch is not yet locked up.

Max. engine size is 500 CI.

From F = M X A we get 12,740 lbs force reqd. which we must multiply by tire radius of 1.5 ft = 19,110 which we divide by the gear ratio of 3.2 to obtain 5,972 ft lbs torque on the drive shaft.

At 6,000 rpm and 5,972 ft lbs the HP is 6,822 and let's not forget while this is going on at least 10 % of the total engine BHP is being lost as heat from the slipper clutch which can reach temps. of 1,000 degrees F +.

It is no longer a problem for me to believe the 7,000 HP they're talking about and if you see them live you KNOW it's true.

As a curiosity we find from the BMEP formula: (BMEP = Torque X 150/CI) that the effective average pressure on the power stroke is 1,792 psi. I don't think the Max. combustion pressure is as proportionally high as a high performance gasoline engine or we could expect around 5,000 psi and it couldn't stand that; so they run the engines extremely rich to hold the Max. pressure down. Actually the combustion may be closer to a "constant pressure" as in a theoretical diesel that an actual diesel.

This indicates a tremendous pressure in the cylinders and a very good reason for the Hemi engine to be best suited for this type of work. The combustion chamber is a slice from a sphere. Years ago while working in pressure vessel design I learned that a spherical shape is by far the most efficient design for pressure retaining parts.

The displacement of a hemi 6.1L is calculated as follows (if you want an excel spreadsheet I made in 6.1 seconds to help with this let me know an I'll email it to you):

cylinders 8

bore 4.06 inches

stroke 3.58 inches

displacement / cyl 46.35 cubic inches

displacement 370.78 cubic inches

Later

Rakort

cylinders 8

bore 4.06 inches

stroke 3.58 inches

displacement / cyl 46.35 cubic inches

displacement 370.78 cubic inches""""""""

You must have the bore or stroke wrong; 370.78 is not 6.1 L.

Here's a little one about HP and Torque

Reading magazine articles indicates TORQUE and HORSEPOWER are probably the most misunderstood aspects of an engine. Torque is a twisting force and usually is expressed in FT LBS. Consider a torque wrench with a 1 ft. long handle and assume it's on a bolt head and the handle is in the horizontal position. Assume a 50 lb. weight is suspended 1 ft. out from the nut and we note that the reading is 50 FT LBS.

OK, now lets assume we can turn the wrench a complete turn and then the distance traveled would be 2 Pi X 1ft.= 6.28 ft. OK, now we say 50 X 6.28 = 314 ft lbs of work.

Horsepower is expressed as 33,000 ft lbs of work per minute or 550 ft lbs per second. If we had put about 87.5 lbs force on the torque wrench and made a complete turn in 1 sec. then we would have been working at the "rate" of 1 HP.

Keep in mind that torque is a force and horsepower is only a calculation. It is possible to have torque and zero HP, but you must have torque to develop HP.

There is one and only one RPM where the torque and the HP will be the same numerical number. That number is 5252 RPM. The reason being is when you divide 33,000 by 2Pi you will get 5252. At less than 5252 the torque will always be more than HP. At greater than 5252 the HP will always be more than the torque.

Some Air Cond. Electric motors are marked in ft. lbs. of torque. If a 1750 rpm motor is marked for 60 ft lbs then you know it's 20HP. 1750/5252 X 60 = 20.

BMEP=TORQUE X 150/CUBIC INCHES for a 4 stroke engine.

Also it is possible to develop Torque without HP; but impossible to develop HP without Torque.

Umm, no, I'm right and your not...Moparitis said:rakort>>>""""

You must have the bore or stroke wrong; 370.78 is not 6.1 L.

Ok, so they rounded it up.rakort said:Umm, no, I'm right and your not...

Thats a little like Ford did some years ago with their 4" X 3" smallblock; they called it a 5.0 and reason being there were 1000s of 283 SBCs that had been bored out to make them 301.59 and called 301s. I had several myself.

It would have looked even more like Ford copied the SBC if they had called it a 301.

This time let's go to Daytona:

In case you ever wondered about the power reqd. to pull you car at a specified speed; here's some unsophisticated formulae:

Pressure in pounds per Sq FT of Air resistance=.003 X MPH squared.

Horsepower reqd. =.000008 X MPH cubed per Sq. Ft. for air resistance.

Horsepower reqd. for rolling resistance = .0001 X Wt. of car X MPH.

Using the formula to find the HP for a car to run 200 MPH at Daytona; assume the car weighs 3500# and is 30 sq.ft with a drag coeffient of .25.

200 cubed = 8000000 X 7.5 sq.ft.= 60000000 X .000008 = 480 HP for air resistance plus 3500 X 200 = 700000 X .0001 = 70 HP for rolling resistance = a total of 550 HP.

I think we have heard the nascar teams talking about 550 or so HP with their restrictor plate motors.

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