Yellow Bullet Forums banner

1 - 20 of 29 Posts

·
Registered
Joined
·
238 Posts
Discussion Starter #1
I’ve read and heard all kinds of ways to figure the maximum safe RPM of your engine. I’ve read about piston speed and inertia weight being the most commonly used. Does anybody have any preference on either of these theories or something different?

After running my new combination at the track it was apparent that it’s over geared. It has the following parts.

· It’s a BBC 2-bolt block with ARP main studs
· Velasco 4.25” steel crank, internal balance
· Eagle 6.385” long H-beam rods with ARP 8740 bolts
· SRP forged 18cc dome pistons with the standard wrist pin
· ATI balancer
· 10.2:1 compression

The bob weight is 2222 grams, recipicating weight is 992 grams.

· 220 grams recipicating rod
· 550 grams piston
· 150 grams wrist pin
· 56 grams rings
· 10 grams ring support
· 6 grams spiral locks

Using the information above on an online calculator it used 772 grams (didn’t count reciprocating rod weight). To figure inertia and gave me the following at 7300rpm and 7500rpm.

· 5475 average 5779
· 7297 upward 7702
· 3653 downward 3856

Other places use piston speeds which would figure 5171 at 7300 and 5313 at 7500.

So what do these numbers really mean and what is considered safe? I know the old GM books said piston speeds of 4800 for short burst, other books said 5000. Assuming my valve train is up to the task what would be considered the maximum safe rpm limit for my short block?

Thanks for reading
 

·
Registered
Joined
·
852 Posts
Using the information above on an online calculator it used 772 grams (didn’t count reciprocating rod weight). To figure inertia and gave me the following at 7300rpm and 7500rpm.

· 5475 average 5779
· 7297 upward 7702
· 3653 downward 3856

Thanks for reading
What's the unit of measure in the figures above?

A typical 7/16" rod bolt will have a shank diameter of .372" and an area of .1086in^2. So a standard 8740 ARP (190 KSI) rod bolt will be able to carry the following loads:

(assuming 30% reduction from ultimate to yield)

Ultimate Tensile (max. load before failure) - 20,634 lbs.
Yield Tensile (max. load before deformation) - 14,444 lbs.

So to maintain a margin of safety of 2X, then the max. load of the bolt should be below 7,222 lbs.

As an example, the same bolt with ARP 2000 (220 KSI) material would have a max. load of 8,362 lbs. and an L19 (260 KSI) bolt would be 9,882 lbs.



Cole
 

·
Registered
Joined
·
1,640 Posts
Note if your bore size is different it does not matter. I come up with:

Bore_=_4.25___Stroke_=_4.25___Rod_Length_=_6.385___RPM_=_7300
Wrist_Pin_Offset_=_0.0
Piston_Weight_=_772.0___Rod_Weight_=_660.0
Small_End_Rod_Weight_=_220.0___Big_End_Rod_Weight_=_440.0
Rod_CG_/_Distance_from_Small_End_=_4.256667______GAS_PRESSURE_=_0

____________Reciprocating____Total_____Piston_Side___Piston
Crank_Angle____Force_________Force________Force__Inertia_Force
__Degree_______Pounds________Pounds_______Pounds______Pounds
___.000______9375.393_____12495.445_________.000____7296.173


Bore_=_4.25___Stroke_=_4.25___Rod_Length_=_6.385___RPM_=_7500
Wrist_Pin_Offset_=_0.0
Piston_Weight_=_772.0___Rod_Weight_=_660.0
Small_End_Rod_Weight_=_220.0___Big_End_Rod_Weight_=_440.0
Rod_CG_/_Distance_from_Small_End_=_4.256667______GAS_PRESSURE_=_0

____________Reciprocating____Total_____Piston_Side___Piston
Crank_Angle____Force_________Force________Force__Inertia_Force
__Degree_______Pounds________Pounds_______Pounds______Pounds
___.000______9896.150_____13189.506_________.000____7701.440

Stan
 

·
Registered
Joined
·
16,879 Posts
Note if your bore size is different it does not matter. I come up with:

Bore_=_4.25___Stroke_=_4.25___Rod_Length_=_6.385___RPM_=_7300
Wrist_Pin_Offset_=_0.0
Piston_Weight_=_772.0___Rod_Weight_=_660.0
Small_End_Rod_Weight_=_220.0___Big_End_Rod_Weight_=_440.0
Rod_CG_/_Distance_from_Small_End_=_4.256667______GAS_PRESSURE_=_0

____________Reciprocating____Total_____Piston_Side___Piston
Crank_Angle____Force_________Force________Force__Inertia_Force
__Degree_______Pounds________Pounds_______Pounds______Pounds
___.000______9375.393_____12495.445_________.000____7296.173


Bore_=_4.25___Stroke_=_4.25___Rod_Length_=_6.385___RPM_=_7500
Wrist_Pin_Offset_=_0.0
Piston_Weight_=_772.0___Rod_Weight_=_660.0
Small_End_Rod_Weight_=_220.0___Big_End_Rod_Weight_=_440.0
Rod_CG_/_Distance_from_Small_End_=_4.256667______GAS_PRESSURE_=_0

____________Reciprocating____Total_____Piston_Side___Piston
Crank_Angle____Force_________Force________Force__Inertia_Force
__Degree_______Pounds________Pounds_______Pounds______Pounds
___.000______9896.150_____13189.506_________.000____7701.440

Stan
WTF ^^^^^

300 rpm before it stops pulling :p
 

·
Registered
Joined
·
238 Posts
Discussion Starter #7
What's the unit of measure in the figures above?

A typical 7/16" rod bolt will have a shank diameter of .372" and an area of .1086in^2. So a standard 8740 ARP (190 KSI) rod bolt will be able to carry the following loads:

(assuming 30% reduction from ultimate to yield)

Ultimate Tensile (max. load before failure) - 20,634 lbs.
Yield Tensile (max. load before deformation) - 14,444 lbs.

So to maintain a margin of safety of 2X, then the max. load of the bolt should be below 7,222 lbs.

As an example, the same bolt with ARP 2000 (220 KSI) material would have a max. load of 8,362 lbs. and an L19 (260 KSI) bolt would be 9,882 lbs.


Cole
Thank you very much that seems to make some sense to me. However if the online inertia calculator was right your figures show the 7200-7250 I'm spinning now is probably about max for the current rod bolts.
 

·
Registered
Joined
·
238 Posts
Discussion Starter #8
Note if your bore size is different it does not matter. I come up with:

Bore_=_4.25___Stroke_=_4.25___Rod_Length_=_6.385___RPM_=_7300
Wrist_Pin_Offset_=_0.0
Piston_Weight_=_772.0___Rod_Weight_=_660.0
Small_End_Rod_Weight_=_220.0___Big_End_Rod_Weight_=_440.0
Rod_CG_/_Distance_from_Small_End_=_4.256667______GAS_PRESSURE_=_0

____________Reciprocating____Total_____Piston_Side___Piston
Crank_Angle____Force_________Force________Force__Inertia_Force
__Degree_______Pounds________Pounds_______Pounds______Pounds
___.000______9375.393_____12495.445_________.000____7296.173


Bore_=_4.25___Stroke_=_4.25___Rod_Length_=_6.385___RPM_=_7500
Wrist_Pin_Offset_=_0.0
Piston_Weight_=_772.0___Rod_Weight_=_660.0
Small_End_Rod_Weight_=_220.0___Big_End_Rod_Weight_=_440.0
Rod_CG_/_Distance_from_Small_End_=_4.256667______GAS_PRESSURE_=_0

____________Reciprocating____Total_____Piston_Side___Piston
Crank_Angle____Force_________Force________Force__Inertia_Force
__Degree_______Pounds________Pounds_______Pounds______Pounds
___.000______9896.150_____13189.506_________.000____7701.440

Stan
Thanks Stan, I'm not sure I'm understanding the formulas well enough to understand what you have. I believe the total weight of my rods was 784 grams and the big end was 564 grams without the bearing. So are you seeing the 7300 were my rev limiter is currently set about my maximum safe rpm?
 

·
Registered
Joined
·
1,640 Posts
Bore_=_4.25___Stroke_=_4.25___Rod_Length_=_6.385___RPM_=_7300
Wrist_Pin_Offset_=_0.0
Piston_Weight_=_772.0___Rod_Weight_=_784.0
Small_End_Rod_Weight_=_220.0___Big_End_Rod_Weight_=_564.0
Rod_CG_/_Distance_from_Small_End_=_4.593291______GAS_PRESSURE_=_0

____________Reciprocating____Total_____Piston_Side___Piston
Crank_Angle____Force_________Force________Force__Inertia_Force
__Degree_______Pounds________Pounds_______Pounds______Pounds
___.000______9375.393_____13374.733_________.000____7296.173

If we use the RPM, stroke, rod length and piston weight I calculate Piston Inertia Force in Pounds

If you include the small end weight I calculate Reciprocating Force in Pounds

If you include total rod weight I calculate Total Force in Pounds

All 3 of these forces will be highest @ TDC because that is where acceleration is the highest

Stan

 

·
Registered
Joined
·
3,173 Posts
Thank you very much that seems to make some sense to me. However if the online inertia calculator was right your figures show the 7200-7250 I'm spinning now is probably about max for the current rod bolts.
That 7,222 lbs is per bolt. Hopefully your rods have 2 bolts in them.

As long as your housing bore is concentric, your bearing clearance is correct and oil delivery is not an issue, rod bolts are stretched properly your going to be fine at 7500

We run an 8740 bolt in a hemi that goes 8000 with a piston/pin/ring combo close to 1000 grams with a 4.25 stroke and it runs for 300 passes before getting looked at.
 

·
Registered
Joined
·
6,716 Posts
winner....Go a few hundred RPM past peak power on the shift, it will live.

That 7,222 lbs is per bolt. Hopefully your rods have 2 bolts in them.

As long as your housing bore is concentric, your bearing clearance is correct and oil delivery is not an issue, rod bolts are stretched properly your going to be fine at 7500

We run an 8740 bolt in a hemi that goes 8000 with a piston/pin/ring combo close to 1000 grams with a 4.25 stroke and it runs for 300 passes before getting looked at.
 

·
Registered
Joined
·
2,761 Posts
Pay attention to the obvious, such as camshaft profile, valve springs, the whole fuel system (i.e. tank, pump(s), lines, regulator, carburetor), intake design, exhaust, ignition capabilities, etc., etc.
The advice from 61 4-speed is viable. Once all this stuff is dialed in, THEN take the vehicle out and note the RPM at which the engine quits pulling and back off 300 to 500 RPM and consider this as your red-line.
 

·
Registered
Joined
·
238 Posts
Discussion Starter #15
Pay attention to the obvious, such as camshaft profile, valve springs, the whole fuel system (i.e. tank, pump(s), lines, regulator, carburetor), intake design, exhaust, ignition capabilities, etc., etc.
The advice from 61 4-speed is viable. Once all this stuff is dialed in, THEN take the vehicle out and note the RPM at which the engine quits pulling and back off 300 to 500 RPM and consider this as your red-line.
Bill,

I have paid attention to these things and it is dialed in except for final gearing. The tunnel ram, heads, cam, springs and ignition will all handle more rpm than the shortblock IMO. The reason I asked this question is because I finally got this combination on a timed course and it continues to pull mph well past the end of the 1/4 mile. If I drop the gearing to quicken it up it will most likely pick up another 300rpm on my flash and pull down and then another 200rpm through the lights based off my previous gear change. My original target was to spin this combination 7200 through the lights and not exceed 7300 but it seems to want to rpm further than expected.

Here is what it did a few weeks ago.

Loren
 

·
Registered
Joined
·
14,776 Posts
The problem is if the force on the bolt exceeds it's clamping load, then the bolt may not fail per se, but it will stretch each time around. It may be able to handle the load, but can it do it without stretching or flexing? Bolts stretching and flexing is what causes them to break.
 

·
Registered
Joined
·
852 Posts
The problem is if the force on the bolt exceeds it's clamping load, then the bolt may not fail per se, but it will stretch each time around. It may be able to handle the load, but can it do it without stretching or flexing? Bolts stretching and flexing is what causes them to break.
Yes, this is called a fatigue failure. We fatigue test bolts everyday at my work and I wonder how the hell a rod bolt survives millions of cycles and still goes back in the same motor time and time again. One thing to remember is that if the bolts is installed to the proper stretch/clamp load, it should never fail because the yield point (point at which permanent deformation begins) will never be reached. This is why stretching is the best method to install a fastener. Bolt torque is a variable and is influenced by many different parts of the clamped system.
 

·
Registered
Joined
·
14,776 Posts
Yes, this is called a fatigue failure. We fatigue test bolts everyday at my work and I wonder how the hell a rod bolt survives millions of cycles and still goes back in the same motor time and time again. One thing to remember is that if the bolts is installed to the proper stretch/clamp load, it should never fail because the yield point (point at which permanent deformation begins) will never be reached. This is why stretching is the best method to install a fastener. Bolt torque is a variable and is influenced by many different parts of the clamped system.
You acknowledge fatigue failure, but then fall a little short with your following statement (IMO).
"One thing to remember is that if the bolts is installed to the proper stretch/clamp load, it should never fail"- as long as it's clamping force is greater than the load it sees.
If you put a grade 5 bolt in your rods and stretch them properly I guarantee you it will fail as a rod bolt. ;)
 

·
Registered
Joined
·
852 Posts
You acknowledge fatigue failure, but then fall a little short with your following statement (IMO).
"One thing to remember is that if the bolts is installed to the proper stretch/clamp load, it should never fail"- as long as it's clamping force is greater than the load it sees.
If you put a grade 5 bolt in your rods and stretch them properly I guarantee you it will fail as a rod bolt. ;)
Yes, that is correct. The statement was meant to apply when using the proper bolt for the application. I highly doubt you can go to Home Depot and find a rod bolt. I guess I should have put some fine print in my post and ran it across the lawyers desk before hitting enter. Ha ha.


Cole
 

·
Registered
Joined
·
7,230 Posts
The greatest load on the pin occurs at TDC on the exhaust stroke. Over time, you can rate the quality of your components to piston speed and come up with a fatigue life span for the parts you use that is very accurate. Build your power for that range. :)
 
1 - 20 of 29 Posts
Top