No clear winner, powerwise..

[ISX]

So I did some digging around to try and round up the specs on the big 3 and was amazed to learn that there is no clear winner at all. Cummins seemed to have it's hayday with the 1st gen's, then powerstroke came on with the 99-02 powerstrokes, now dmax seems to be taking the lead. Now I am not talking reliability or anything, just power specs. I was barely able to come up with the 6.2/6.5 specs, I actually guessed on those a bit from the 983 different ratings I was seeing on 244 different forums. I used all the maximum power specs with all 3, so I used the HO 6spd specs for the 24V.. and ford didn't have a diesel in 1998, thats why theres a break in the line..

[dorkweed]

What doesn't get measured in your graph; is, LUGABILITY!!! INLINE 6 baby. They'll "lug" at low RPM's way better than a V engine will. That's why they make the best towing engine set up!!! Bigger diameter pistons with a longer stroke makes for a low RPM torque monster. I do realize the DuraMaxiPad makes similar numbers at the same rpm, but thats with 2 extra cylinders.........and it's made in Japan!!!!!!:banghead:

[ISX]

You know, I could get technical here and say a V8 will lug better than any I6 purely because it has 2 more cylinders, so to make things drastic, it would be the same as lugging a single cylinder and a V16, the V16 has so many power strokes going per revolution that it can idle very very low. I believe with these big 3, it all comes down to displacement. My ford had wonderful torque down low, but would fall on it's face up high. It was a 7.3. I haven't driven a 99-02 7.3 but I know everyone loves them and I am just wondering how torquey they are down low. I am actually thinking this was more just due to having more power strokes per revolution. It might have ran a little smoother down low because of it, but I'm not saying it made more power than a cummins down low. Obviously the huge stroke of the cummins makes the big torque numbers. Alright I'm thinking about this too much :lol:I did have the ISX at school holding its 800RPM idle speed using just 2 injectors though So maybe rotating mass of an I6 has something to do with it all too. I know a V8 has more friction since there are more pistons so maybe the more power strokes just cancels out the friction so it isn't a complete POS. Need more classes!

[flman]

You know, I could get technical here and say a V8 will lug better than any I6 purely because it has 2 more cylinders

Or you could say it lugs alot easier :lmao2:then you drop it down 2 more gears lower then the Cummins would need carrying the identical load.

[Gil2264]

I'm not an engineer, but I sure that there's a reason why almost every 18 wheeler out there uses an inline 6.

[cmns2500]

yup i agree....blame Cummins haha

[ISX]

I need to learn more on this because it's driving me crazy. In one hand you have a V8 having 2 more cylinders which equates to 2 more power strokes going per revolution. On the other hand you have 2 cylinders less yet having a long stroke making each power stroke longer. This is all on the topic of "Lugability", so obviously the I6 wins down low, but I'm just wondering why. What is it about a longer stroke makes it better than having 2 more power strokes!?

[flman]

I need to learn more on this because it's driving me crazy. In one hand you have a V8 having 2 more cylinders which equates to 2 more power strokes going per revolution. On the other hand you have 2 cylinders less yet having a long stroke making each power stroke longer. This is all on the topic of "Lugability", so obviously the I6 wins down low, but I'm just wondering why. What is it about a longer stroke makes it better than having 2 more power strokes!?

Easy, turn a hand crank with a long lever, and a short lever, which one turns easier? Or compare the short stroke of a child's big wheel or tricycle to that of an adult bicycle. It is as simple as a lever. You know, that is a good theory, I will have to use that one on the next pro V8 diesel guy. They never have a come back to the Semi or heavy equipment using inline diesels, and they are applications that require alot of grunt.

[ISX]

Easy, turn a hand crank with a long lever, and a short lever, which one turns easier? Or compare the short stroke of a child's big wheel or tricycle to that of an adult bicycle. It is as simple as a lever. You know, that is a good theory, I will have to use that one on the next pro V8 diesel guy. They never have a come back to the Semi or heavy equipment using inline diesels, and they are applications that require alot of grunt.

SOBBBBB! I wish I had read that sooner! Woulda clued me in to calculate the last thing I did on here, first. Nevertheless, I calculated and calculated trying to find just what in the world gives the cummins the edge. Finally found it, in green. I figured out how much volume there was at any given point in time, being allocated to solely a "power stroke". This also explains why it's harder to turn since there is that much volume also being compressed. If you all want to mess with it for yourselves this excel works in any program, no fancy formulas

Eh, so I thought the 5.9 was on top at first, had calculations screwed up. At least it is consistent with my 7.3 being very luggable, and my brothers 6.0 being junk down low. BUT, we are back on top by farrr with the 6.7.

I just did the ford 300 I6, its 66.69, or only 1.37 short of the 6.0

powerstroke.xls

[dorkweed]

I sure wish I had all the time that you spend digging all this stuff up!!!! Good info though!!!

[ISX]

I sure wish I had all the time that you spend digging all this stuff up!!!! Good info though!!!

I spend more time making it look nice for you all than I do finding the info The formulas don't take very long, I think I spent 20 min on that whole thing, which includes me getting mad since nothing I put in was answering my question of why the I6 is king of luggability. I just kept thinking of anything that even remotely had to do with possibly showing an advantage. Some of those things are really neat, I couldn't believe each piston was completing the entire 4 stroke cycle that many times just idling Makes you understand why dirty oil or fuel can wreck havoc in short order. If there is anyone who doesn't have a clue about what the green part is, which I really didn't know what to call it on the spreadsheet so here's my explanation. You have the firing order of the engine which takes 2 revolutions of the crankshaft to complete the 4 stroke cycle for every single piston. So in other words, only half the pistons are on the power/compression stroke per crankshaft revolution. So to make it easy I just determined everything based on ONE revolution of the crankshaft, so on our cummins I just used 3 pistons since they are the only ones firing. Since it takes one revolution to complete the cycle for 3 of the pistons, it also equates to 360 degrees and because the pistons must be evenly spaced, 360/3=120 degrees out of phase from each other (this is on my spreadsheet), this applies to all pistons since it's not like the engine takes a break after 3 pistons, so all 6 are 120 degrees out of phase, meaning since the firing order is 153624, piston 1 injector fires, and 120 crankshaft degrees later piston 5 injector fires, then 3 and so on. Alright, now that everyone should be on the same level I can get into the green part in specific. If one of the pistons was on the power stroke and reached BDC (bottom dead center), that means the following piston is 120 degrees behind, and the 3rd piston is 120 degrees behind that one. Remember that the power stroke for each piston only lasts for half a crankshaft rotation, since when the crankshaft revolves a half turn, it starts pushing the piston back up and is therefore on the exhaust or compression stroke. So if the first piston is at BDC and the next piston is 120 behind, that isn't over 180 degrees behind therefore that piston must ALSO be on it's power stroke. So if we said the 1st piston at BDC was at 0 degrees then the second piston would be at 120, the third piston would be at 240 degrees which means it is being pushed up and therefore on it's compression stroke. So the only pistons in the equation are the 1st and 2nd, we can add the entire volume of piston 1 since it is as BDC and since piston 2 is only 60 degrees below TDC (being 120 behind #1), we can calculate it's percentage based on 60/180=1/3 this means piston 2 has only gone down 1/3 of the way from TDC on it's power stroke. So you do 1/3 of total cylinder volume, add it to piston 1's volume, and you have the total area that is being used for the power stroke. This area NEVER changes no matter how the engine turns, I just used BDC to make it easy. On a V8 the pistons are 90 degrees out of phase, so when one is at BDC, the next is halfway down the cylinder (since 90/180= 1/2) and the third is at TDC so there is no area to calculate since it isn't exposing any cylinder area for the power stroke. As you gain pistons, the total power stroke area goes up less and less. Meaning 3 pistons might be 10, 4 pistons might be 20, 5 pistons might be 25, 6 pistons might be 27.... It is an exponential rate. If you were to graph the areas based on cylinder number, the line would initially change drastically and then flatten out, it will never touch 0 but when you get up to 100 pistons to 101 pistons you hardly notice anything because the line is so flat. It's very interesting. Alright got a pic up showing the exponential curve. I just used the 5.9 cummins bore and stroke to get areas. The bottom numbers are number of pistons and the numbers on the line are the total area of all the power strokes at any given time. You can see the more pistons you have the less difference it makes. Dang I screwed up, its showing engines with pistons 3-16, not 1-14, you can't do 1 or 2 piston engines. I'd fix it but I deleted it before I noticed If you didn't understand it, hope you do now, if I made it more confusing let me know and I will get some pics up that will make it obvious.

[Wild and Free]

You need to research a little know term called "Revgain" and throw that into your equation now. It is a term used by performance engine builders. Has to do with how fast an engine can and or will accelerate. I have seen deep technical discussions on this subject while comparing v-8 and I6 diesels before.

[Mopar1973Man]

A way to validate post #11 is just look at the valvetrain when your doing your valve lash... First batch is done at TDC #1 (153) then your rotate the crank 360* and now your at TDC #6 (624)...

Firing order 153624...

[ISX]

A way to validate post #11 is just look at the valvetrain when your doing your valve lash... First batch is done at TDC #1 (153) then your rotate the crank 360* and now your at TDC #6 (624)...

Firing order 153624...

Where did I say the firing order wasn't that? I did start referring to the pistons as being 1 2 and 3 instead of saying 1 5 and 3, maybe I shouldn't have.

W&F, I have been looking all over for more factors on this, I can't think of everything so thank you! I will look it up and get it all into the spreadsheet. I never like to assume I am at the end of the road with things, I know there has to be more to every equation and you just proved it.

---------- Post added at 11:43 AM ---------- Previous post was at 11:21 AM ----------

Alright from what I have gathered, rev-gain basically means the rate of acceleration of an engine overcoming it's own rotating mass. So sled pullers could care less since they start at whatever insane speed they want, but dragsters praise rev-gain since it needs to be gaining quick.

I am not exactly sure how to calculate this based on just engine specs. I have some sketchy ideas on how but not sure. I will keep thinking.

---------- Post added at 08:55 PM ---------- Previous post was at 11:43 AM ----------

Would it be more plausible to calculate internal rotating mass, and get a number telling how much force is required just to turn the engine over or do a certain RPM? Rev gain seems very tricky to figure out but the former seems a little easier. I would like to do rev gain but there is just nothing on the internet concerning calculating it, just stuff talking about what it is.

[ISX]

I am a little bewildered now. When I said my 7.3 was luggable, I meant at very load speeds, like <10mph but, I just drove a 1999 7.3 with a mild chip and it pretty well threw my whole chart out the window. According to my chart, the 7.3 is supposed to be on top, but this thing was an absolute pile. To test things I always go up a hill and try and hold whatever speed it has at the base of the hill, if it is an auto (like this one was) I press down on the pedal until it is about to downshift. So I was doing 55 at the bottom of the hill, start going up the hill, I start going down until it is damn near to the floor and I know it is going to downshift any second. It was so gutless that I was down to 50mph before it finally was able to hold the speed. Now I know everyone praises these things for their reliability and I do need to do some more research as I believe they have too big of turbo to be trying to go up a hill at 1600-1700RPM. I didn't hear whistle at all and this thing was straight piped. I don't want to base boost numbers off what I can hear but damn, it was worthless. I really think it was the turbo's fault but it's just kind of hard to believe how crappy it was. It definitely had more power under 2000 than the 6.0, which has none, but it was just unreal. I found out a couple other variables to the chart I just have to work on getting them all into the equation.

---------- Post added 06-24-2010 at 12:47 PM ---------- Previous post was 06-23-2010 at 10:36 PM ----------

Got some more statistics going, thinking piston speed has something else to do with it. The HP at the end is theoretical with all the engines being equal pressure and everything, it is hardly a real world value but does have some significance.

So the pistons moving faster does something but I have to think about it more.

[Mopar1973Man]

Something to think about on the luggablity... How big is the flywheel and how heavy is it per each engine??? I think the fact amount of weight these engines are throwing around (cranks, flywheel, etc) tends to allow much lower RPM's under load... Like the old pic of the connecting rods... Cummins being the biggest and Ford being the smallest...

[ISX]

Hmmm, that's definitely something to factor in. I don't know how big of difference there is between them all. From what I have read, V8 diesels should actually have heavier things because they need them to counteract their imbalances, whereas I6 doesn't need them since they are the only perfectly primary and secondary balanced engine. I read that, that was one of the reasons for using I6 in all semi's, costs less when you don't have to produce all the counterweights. I will have to research it though.

[flman]

Hmmm, that's definitely something to factor in. I don't know how big of difference there is between them all. From what I have read, V8 diesels should actually have heavier things because they need them to counteract their imbalances, whereas I6 doesn't need them since they are the only perfectly primary and secondary balanced engine. I read that, that was one of the reasons for using I6 in all semi's, costs less when you don't have to produce all the counterweights. I will have to research it though.

The few V8 diesels I have seen in medium duty trucks were all dogs and had to be screaming in low gear to pull a slight grade, so I think the I6 is used just because it can produce so much more torque at lower RPMS = Efficiency. I kind of doubt the counter weight theory.

[ISX]

The few V8 diesels I have seen in medium duty trucks were all dogs and had to be screaming in low gear to pull a slight grade, so I think the I6 is used just because it can produce so much more torque at lower RPMS = Efficiency. I kind of doubt the counter weight theory.

The power factor is another reason they are used, but not the same subject. I am talking about from a balance standpoint. The I6 is balanced by design, V8 is not, it needs things to balance it back out which add cost.

[ISX]

Got some more info. The lower the ratio between surface area and volume of the cylinder, the more efficient it will be because heat is lost through the cylinder walls, so by reducing the cylinder surface area, there is less area absorbing the heat, and of course, heat=power. As usual, cummins comes out on top.