95' Junker Drag Truck

Tonight I shaved some material from the ends of the piston pins. Approximately 4.8mm of material was removed on an 18* angle. This dropped the pin weight from 570 grams a piece to 530 grams, a drop of 40 grams per cylinder or 240 gram reduction in reciprocating weight in the engine. When you factor the heavy crank, rods, and pistons, I'm not sure it's worth the effort but it was free!!! No price is too cheap for the Junker Drag Truck.

LightenedPin4_zpse87f578d.jpg


Stock non-intercooled piston pin:
LightenedPin3_zpsc19889cb.jpg


Lightened pin:
LightenedPin2_zpsfaf78cf1.jpg


LightenedPin1_zps485a93eb.jpg
 
Interesting, any worries about strength? I am not sure how much engagement the pin has to piston.
 
I finally ran the numbers on horsepower gain vs. weight loss on the piston, pin, and rod. I did not factor-in friction loss from the removed weight which would free up a little more power but the calculations would be difficult and likely provide little useful data.

HP gain is directly related to crank speed because this determines the average acceleration of the piston, pin, and connecting rod assemblies. Stroke length is extremely important in this calculation so the numbers below were based on 4.72" stroke length, a longer stroke motor (6.7 Cummins) would experience even more HP gain than the numbers below.

At 2500 rpm, every 138.15 grams is equal to 1 hp.
At 3000 rpm, every 95.9 grams is equal to 1 hp.
At 3500 rpm, every 70.5grams is equal to 1 hp.
At 4000 rpm, every 53.97 grams is equal to 1 hp.


The motor for the Junker Drag Truck has had a few different weight reductions:
-157 grams changing to a non-intercooled piston vs. stock
-16.5 grams from removing aluminum for piston bowl modification
-40 grams from machining the piston pin
-16.5 grams from de-burring and balancing the connecting rods
------------------------------------------------------------

230 grams X 6 piston/rod assemblies = Total reduction of 1380 grams.

Weight reduction based free horsepower gains:

10 HP @ 2500 rpm
14.4 HP @ 3000 rpm
19.6 HP @ 3500 rpm
25.6 HP @ 4000 rpm
 
Ha, I was wrong! That's a nice little bump at 4k which is totally do able on a 12 valve.

So was it worth 8 hours of work? Haha
 
Big Blue24 said:
I finally ran the numbers on horsepower gain vs. weight loss on the piston, pin, and rod. I did not factor-in friction loss from the removed weight which would free up a little more power but the calculations would be difficult and likely provide little useful data.

HP gain is directly related to crank speed because this determines the average acceleration of the piston, pin, and connecting rod assemblies. Stroke length is extremely important in this calculation so the numbers below were based on 4.72" stroke length, a longer stroke motor (6.7 Cummins) would experience even more HP gain than the numbers below.

At 2500 rpm, every 138.15 grams is equal to 1 hp.
At 3000 rpm, every 95.9 grams is equal to 1 hp.
At 3500 rpm, every 70.5grams is equal to 1 hp.
At 4000 rpm, every 53.97 grams is equal to 1 hp.


The motor for the Junker Drag Truck has had a few different weight reductions:
-157 grams changing to a non-intercooled piston vs. stock
-16.5 grams from removing aluminum for piston bowl modification
-40 grams from machining the piston pin
-16.5 grams from de-burring and balancing the connecting rods
------------------------------------------------------------

230 grams X 6 piston/rod assemblies = Total reduction of 1380 grams.

Weight reduction based free horsepower gains:

10 HP @ 2500 rpm
14.4 HP @ 3000 rpm
19.6 HP @ 3500 rpm
25.6 HP @ 4000 rpm
Click to expand...

Do you mind sharing your math for this? I am definitely interested.
 
CorneliusRox said:
Do you mind sharing your math for this? I am definitely interested.
Click to expand...


It all goes back to F=ma. The difficult part is calculating the average piston acceleration: [ame="http://en.wikipedia.org/wiki/Piston_motion_equations"]Piston motion equations - Wikipedia, the free encyclopedia[/ame]

The stroke of 4.72" with rod 7.559" @ 4500 rpm calculates to an average acceleration of 57,296 ft/ sec^2 or 17,463 meters/sec^2.

F(Newtons) = m(kilograms) x a(m/s^2)

100 grams for example:

F = .100 x 17463, F=1743.6 Nm/s

100 grams @ 4500 rpm equates to 1746.3 Newton Meters per second which equates to 2.34 HP

100g/2.34 tells us 42.72g is equal to 1 hp at 4500 rpm crank speed.
 
SASDakota said:
Interesting, any worries about strength? I am not sure how much engagement the pin has to piston.
Click to expand...

The greatest bending moment in the piston pin is in the center beneath the connecting rod. At that location, the pin is still stock sized. The greatest shear stress is at the two sides of the connecting rod where the pin first contacts the inner edges of the piston. This force decreases proportionally as the pin gets closer to the outside edge of the piston as seen in this diagram.

My cut on the end of the piston extends inward approximately 16.5mm or slightly short of where the pin is exposed to full shear forces at the edge of the connecting rod or the greatest bending moment in the center of the rod.

Here's a simple diagram

PinForce1_zpsb3aa0f6d.jpg


PinForce2_zps88491fbc.jpg


The cut or thinned portion of the pin stops short of the B1 & B2 marks. The load on the pin is proportional to the amount removed from the pin so that now the pin has equal strength relative to the amount of load it is subjected to.

In layman's terms, the metal was removed from portions of the pin that were overbuilt compared to the actual load they must endure in the motor.

To answer the question, no, I am not worried about strength. If I was greedy on removing weight, I would have bored the center of the entire pin .250" larger and removed a bunch of weight in addition to tapering the ends. Then I would be concerned about strength since I don't know how "overbuilt" the pin is from the factory.
 
Last edited:
Damn will, why don't you just redesign these things and go into business building fast stock bottom end motors LOL And the math is still too much for me, glad you gave me a rough idea of what weight saving will do though.

Since the crank is the central point, do you know how much a gram would help on it?
 
i enjoy reading this thread, for the in depth responses provided, and the math/calculations added... great thread!
 
I know you're doing this on the cheap, but once it's done, I don't think that the term "Junker" is going to apply.
 
Thanks blue! That is a very helpful and informative response.

It is incredible what you are achieving by just "tinkering".
 
Big Blue24 said:
It all goes back to F=ma. The difficult part is calculating the average piston acceleration: Piston motion equations - Wikipedia, the free encyclopedia

The stroke of 4.72" with rod 7.559" @ 4500 rpm calculates to an average acceleration of 57,296 ft/ sec^2 or 17,463 meters/sec^2.

F(Newtons) = m(kilograms) x a(m/s^2)

100 grams for example:

F = .100 x 17463, F=1743.6 Nm/s

100 grams @ 4500 rpm equates to 1746.3 Newton Meters per second which equates to 2.34 HP

100g/2.34 tells us 42.72g is equal to 1 hp at 4500 rpm crank speed.
Click to expand...

Thanks! Thank makes a lot of sense. Way to simplify it and make it easy, I'll definitely be retaining this bit of info.
 
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