Traction Effort: 308...not that great.

[GP40-2]

The 308 number can only be reasonably applied to first generation DC-DC locomotives (DC generator and DC traction motors) That's because DC-DC drives are around 82% efficient. 308 is 82% of 375 from the formula. AC-DC drives are up to 88% efficient, so 330 would be a good approximation. AC-AC drive locomotives are around 92-93% (at points in their power output approaching 94%), so 348-350 is a good approximation. Of course, the only way to really know what a specific locomotive is doing would be an actual test, since the Traction HP rating is a nominal number only, and in fact may be a few hundred HP higher if the auxiliaries are not using the maximum HP allotted to them.

[MLG4812]

   Sorry about that guys... these are just standard geared and ballasted locomotives. I do recall that on a few of the western roads a few decades ago, many SD40-2's, SD45T's, etc. were high speed geared for their intermodal trains. Don't know if BNSF still practices this for their transcon stacktrains today. I suppose if mine were an actual railroad (with bean counters and such), the focus would be on mixing and matching locos for maximum fuel to tonnage ratio as opposed to optimum and flawless performance over the line itself.

MLG4'8.5"

[NorthWest]

Adding weight improves adhesion and increases tractive effort at lower speeds, which aids in starting the train and operating at near stalling speed. This, I think, was the OP's point about getting as close to stalling as possible.

My point about gear ratios involved slow speed operation as well.  

[Will Davis]

I'm looking around here to see if I can find a drawbar pull or horsepower vs. speed curve that will help with this. I have generator volts vs. amps for GE units with Automatic Power Matching. I also have a tonnage rating guide for MILW units. Maybe after sifting through this stuff I can more directly address this analysis.

[timz]

NorthWest

What are your locomotives ballasted to?

He has to assume they're not slipping, so he's assuming their weight doesn't affect their TE.

Gear ratio affects the speed range over which the engine puts out its full power-- i.e. the range of speeds in which the 308 applies. Outside that range we're guessing-- none of us knows what number should replace the 308 for a 62:15 GP40 with no field shunting at 70 mph.

[NorthWest]

I've been watching this with interest, but have seen no reason to reply until now.

Will brings up an interesting point. What are your locomotives ballasted to? That "308" figure will be different among diesels of the same model, depending on weight and traction motor gear ratio.

[Will Davis]

Should have given this earlier...

Weight range on the SD-45 is 368,000 to 420,000 lbs. Continuous effort with standard 62:15 gears (65 MPH) at 11.1 MPH MCS is 82,100 lbs.

Weight range on the U36B is 256,000 to 280,000 lbs. Continuous rating with standard 81:22 gears (75 MPH) is 54,100 lbs. at 17.3 MPH.

[Will Davis]

Timz is correct. The U36B will be limiting power due to the Automatic Power Matching system below about 20 or 25 MPH. The SD-45 will also be power limiting due to Performance Control. The power on the EMD is limited to 500 HP per axle at the 11.1MPH, rising to full output linearly with speed until full power is reached at something like 17 or 20 MPH.

[timz]

Slippery or not, the U36B probably doesn't produce full horsepower at 16 mph. Even the SD45 was only supposed to do full horsepower at 17 mph and above.

[MLG4812]

Mr. Oltmannd, your calc's let me know that I'm at least on the right track (no pun intended).

Here are mine so far:

Train Types and Characteristics:

TRAIN 1: (1) SD45R + (1) SD40-2 + (100) flatcars/containers + (1) caboose: 4,159.1 average total tons (4,759.1 maximum total tons)

TRAIN 2: (1) SD45R + (1) U36B + (100) flatcars/containers + (1) caboose: 4,113.6 average total tons (4,713.6 maximum total tons)

TRAIN 3: (1) U36B + (1) U36B + (70) flatcars/containers + (1) caboose: 2,939.5 average total tons (3,359.5 maximum total tons)

 

Train Handling (Equations):

Flat Level Track: 0.6 + 20/axle weight + 0.01 x (speed in mph) + Speed Coefficient x (speed in mph²) or Ru =0.6 + 20/w + 0.01V + KV²/wn

 

Grade Resistance In Pounds: (combined ruling grade %) x (20) x (total train tons)

 

(Ruling grade is 1.2% for westbound container trains) + (0.12% curve resistance): 1.32% Combined Ruling Grade-Westbound

(Ruling grade is 1.1% for eastbound container trains) + (0.16% curve resistance): 1.26% Combined Ruling Grade-Eastbound

 

Locomotive TE (traction effort in pounds) = Horsepower(308)/mph

 

TRAIN 1: (1) SD45R Locomotive+ (1) SD40-2 Locomotive + (100) Container/Flatcars + (1) Caboose=6,600 HP

 

(At 14 tons per container lading-4,159.1 tons average train weight)

Flat Level Track at 52.1 mph= 38,982.9 lbs train resistance                             Locomotives TE at 52.1 mph= 39,017.3 lbs (sufficient)

1.26% Grade Eastbound at 17 mph= 119,100.7 lbs train resistance                Locomotives TE at 17 mph= 119,576.5 lbs (sufficient)

1.32% Grade Westbound at 16.4 mph= 123,875.8 lbs train resistance            Locomotives TE at 16.4 mph= 123,951 lbs (sufficient)

 

(At 20 tons per container lading-4,759.1 tons average train weight)

Flat Level Track at 51.2 mph= 38,720.3 lbs train resistance                             Locomotives TE at 51.2 mph= 39,703.1 lbs (sufficient)

1.26% Grade Eastbound at 15.1 mph= 134,007.3 lbs train resistance             Locomotives TE at 15.1 mph= 134,622.5 lbs (sufficient)

1.32% Grade Westbound at 14.5 mph= 139,519.6 lbs train resistance            Locomotives TE at 14.5 mph= 140,193.1 lbs (sufficient)

 

TRAIN 2: (1) SD45R Locomotive+ (1) U36B Locomotive + (100) Container/Flatcars + (1) Caboose=7,200 HP

 

(At 14 tons per container lading-4,113.6 tons average train weight)

Flat Level Track at 54 mph= 41,059.9 lbs train resistance                                Locomotives TE at 54 mph= 41,066.6 lbs (sufficient)

1.26% Grade Eastbound at 18.6 mph= 118,651.8 lbs train resistance             Locomotives TE at 18.6 mph= 119,225.8 lbs (sufficient)

1.32% Grade Westbound at 17.9 mph= 123,313.9 lbs train resistance            Locomotives TE at 17.9 mph= 123,888.3 lbs (sufficient)

 

(At 20 tons per container lading-4,713.6 tons average train weight)

Flat Level Track at 53.6 mph= 41,311.7 lbs train resistance                             Locomotives TE at 53.6 mph= 41,373.1 lbs (sufficient)

1.26% Grade Eastbound at 16.6 mph= 133,474.0 lbs train resistance             Locomotives TE at 16.6 mph= 133,590.4 lbs (sufficient)

1.32% Grade Westbound at 15.9 mph= 138,877.9 lbs train resistance            Locomotives TE at 15.9 mph= 139,471.7 lbs (sufficient)

 

TRAIN 3: (2) U36B Locomotives + (70) Container/Flatcars + (1) Caboose=7,200 HP

 

(At 14 tons per container lading-2,939.5 tons average train weight)

Flat Level Track at 61.9 mph= 35,695.6 lbs train resistance                             Locomotives TE at 61.9 mph= 35,825.5 lbs (sufficient)

1.26% Grade Eastbound at 25.4 mph= 87,115.5 lbs train resistance               Locomotives TE at 25.4 mph= 87,307.1 lbs (sufficient)

1.32% Grade Westbound at 24.5 mph= 90,312.1 lbs train resistance              Locomotives TE at 24.5 mph= 90,514.3 lbs (sufficient)

 

(At 20 tons per container lading-3,359.5 tons average train weight)

Flat Level Track at 61.6 mph= 35,945.2 lbs train resistance                             Locomotives TE at 61.6 mph= 36,000.0 lbs (sufficient)

1.26% Grade Eastbound at 22.8 mph= 97,119.9 lbs train resistance               Locomotives TE at 22.8 mph= 97,263.2 lbs (sufficient)

1.32% Grade Westbound at 21.9 mph= 100,849.4 lbs train resistance            Locomotives TE at 21.9 mph= 101,260.3 lbs (sufficient)

 

 I'm rather wary of the U36B's calculations as they were known to be slippery.

MLG4'8.5"

 

[oltmannd]

A pair of SD40s (or a trio of GP38s)  will get you up the hill and get you in the upper 50 mph range on the flat and 15 mph up the 1.32%.  Subbing SD45s for SD40s gets you about 5 mph more speed on the flat and about 18 mph up the 1.32  %.

[MLG4812]

Thank you gentlemen,

   My imaginary route is a 70's or early 80's era bridge line that carries TOFC/COFC between port A and port B. Sort of a Panama Canal that crosses two mountain ranges (1.2% and 1.32% ruling grades respectively) with a 50 or 60 mile "high plains" run in between that is relatively flat. Trains with locos weigh around 4,000 tons average (I like a nice challenge in powering my trains.) HP/ton isn't too big a deal as trains are dragging through the mountains more than speeding across the plain. Any double-header combination of SD40-2, SD45 have enough starting TE for these trains even from a stop on the 1.32%. I threw the U36B in to reduce the weight and fuel a bit and see how close I can get to slipping the train :)     

[oltmannd]

MLG4812

Greetings,

   I enjoy the mathmatics involved in my hypothetical railroad that is complete with grades, level track running, and adverse weather. Trying to decide which type of motive power to use for certain prototype trains in these conditions is a hobby of mine along with my own N-scale layout. One thing that has always bothered me however are certain equations that are either out of date or too vague to be of any use. One such example is the ubiquitous Traction effort equation: 

TE=HP x 308/SPEED

It is the 308 in this formula that confuses me. I have no problem with this 308 because I understand it is a derivative of the 82.3% of the 375 (374.15) efficiency coefficient for TE:

308/374.15=.823(100)= 82.3% (viola...) The long accepted average efficiency of a standard locomotive.

Having stated that, I also realize that the 308 coefficient cannot apply to every locomotive because some locos are more/less efficient than others in their power output. My question (finally) is where can I find the true efficiency coefficient for particular locomotives to plug into the TE formula to get a "proper" traction effort result. I have searched for hours..days..with less than satisfactory results. I have found information from another forum that stated that this coefficient can be anywhere from the mid-upper 200's to the mid 300's...quite a swing from 308. Any help or resources will be greatly appreciated. (P.S. The locos I am currently calculating are the SD40-2, SD45, and the very popular ???U36B).

Thank You Again

MLG4'8.5"

 

 

 

308 works well for those models you're thinking about.   As timz pointed out, it won't be flat over the whole speed range, but it doesn't  really matter for what you're trying to do - which is locomotive application.

There are a few things to consider.  One is that the basic technology getting shaft HP down to the rails is pretty much the same between those models.  A traction alternator with diode rectification, big cables to series wound traction motors, spur gears running in goopy grease to steel wheels pushing against steel rails.  Same range of volts and amps.  Same gear ratios.  So, not much difference.

Another thing to consider is how RRs do locomotive application.  If you are powering for drag service, all you need to know is the maxiumum continuous tractive effort.  That's going to depend on the minimum continuous speed more than the "308".  At what speed do you reach the "red" on the ammeter  - or lose adhesion?  If the "308" was really "290" is just means that the speed at which the load meter reaches the red is a bit lower.

If you are dispatching for a "schedule" and using HP/ton being off a bit on how much HP actually makes it to the couple isn't a killer.  If the "308" is actually "290" and you fall below your dispatch minimum HP/ton, you'll just wind up taking a bit longer to get where you're going - a minor impact to on time perf.  

Where RRs really depend on the "308" is when doing fuel consumption comparisons.  If one loco model is more efficeint than another - even a little bit, that's important.  It can sway a purchase decision.

[timz]

When I was clocking trains in the 1980s I found that new units would do 85% or better at the rail; let's say 82% was a slightly-above-average figure for units that weren't new. That's at 25 mph and up; can't expect them to do that well at 12 mph.

[MLG4812]

Greetings,

   I enjoy the mathmatics involved in my hypothetical railroad that is complete with grades, level track running, and adverse weather. Trying to decide which type of motive power to use for certain prototype trains in these conditions is a hobby of mine along with my own N-scale layout. One thing that has always bothered me however are certain equations that are either out of date or too vague to be of any use. One such example is the ubiquitous Traction effort equation: 

TE=HP x 308/SPEED

It is the 308 in this formula that confuses me. I have no problem with this 308 because I understand it is a derivative of the 82.3% of the 375 (374.15) efficiency coefficient for TE:

308/374.15=.823(100)= 82.3% (viola...) The long accepted average efficiency of a standard locomotive.

Having stated that, I also realize that the 308 coefficient cannot apply to every locomotive because some locos are more/less efficient than others in their power output. My question (finally) is where can I find the true efficiency coefficient for particular locomotives to plug into the TE formula to get a "proper" traction effort result. I have searched for hours..days..with less than satisfactory results. I have found information from another forum that stated that this coefficient can be anywhere from the mid-upper 200's to the mid 300's...quite a swing from 308. Any help or resources will be greatly appreciated. (P.S. The locos I am currently calculating are the SD40-2, SD45, and the very popular ???U36B).

Thank You Again

MLG4'8.5"