by 351tom
 P: 14 I am trying to determine rod bearing load when a 90 foot pound torque starter motor stalls. I have drawn the geometry in the picture below. I am figuring using a 10 tooth starter gear & a 153 tooth flywheel, so I calculate 15.3 x 90 ft lbs = 1377 ft lbs torque at the flywheel. So if I consider the distance traveled by the flywheel at 12 inches from it's center during a transition from 5 degrees BTC to TDC I get 1.04687 inches. During this same transition, the piston moves .00864. To calculate the effective leverage, I divide 1.04687 by .00864 to get 121.165, then I take 121.165 x 1377 lbs at 12 inches to get 166.844 lbs, and divide it by 2.704 square inches or rod bearing area to get 61,703 PSI on a rod bearing. Where did I go wrong? is the load really that high? A soft metal bearing could not endure that load could it? [IMG][/IMG]
 P: 1,109 I think you are wrongly assuming the starter motor is stalled by an infinite compression ratio. The maximum cylinder pressure = (atmospheric pressure * compression ratio). That limits the peak bearing pressure, = (peak pressure * piston area). A “hydraulic lock” results when water gets into the cylinder, being incompressible, something will then fail.
 P: 14 I wasn't limiting the stall to just compression. I think the most common occurrence is due to ignition BTC causing the expanding burning charge to force the piston down while the starter motor is trying to force the piston up. The results of my calculations seem off somewhere however, because even the compressive strength of the rod would seem to be exceeded at 166,844 lbs.
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Thanks
P: 5,212

I think this is where teeth get stripped from the flywheel.
 P: 14 Ok, let's try this another way, a stock starter motor on a Chevy 350 has 90 ft lbs of torque. For modified high compression engines, and other hard starting engines, higher torque starters are available in 160,180,200,or even 250 ft lb levels. If an engine needs more than the stock 90 ft lb starter, how much load is being put on the bearings by the fully loaded starter?
P: 1,109
 Quote by 351tom ...how much load is being put on the bearings by the fully loaded starter?
The obvious theoretical answer is “infinite load”. That is because as the crank passes TDC, it reverses the direction of piston movement, at which point it has an infinite advantage.

There are two things here. The torque on the starter motor and the cylinder area * pressure at TDC. Those forces are opposed through the ring gear, crank and rod.
The load applied to the rod bearings will be the smaller of those two.

You must also consider the load characteristics of the battery, starter cables and switch. They may limit starter current and hence the available torque to below the starter rating.

A starter motor can destroy a motor when hydraulic lock prevents rotation. Repeated pre-ignition can also result in damage.
 P: 14 Disregarding the battery, cables , etc, and assuming the cylinder area * pressure at TDC is just enough to load the starter at 90 ft lbs, are my original calculations correct?
 P: 1,109 No. Your calculations are wrong. You fail to take the crank advantage at the instant before TDC. Instead you consider a broad average of the 5% before TDC.
 P: 14 Ok, would I be fairly close if I took the distance from 0.5 degrees BTC TO TDC? In this case, the flywheel moves 0.10472" and the piston moves 90 millionths of an inch, so the advantage is 1163.5 and multiplied by the 1377 lbs is 1,602,139 lbs on the rod and 592,507 psi on the rod bearings.
P: 1,109
 Quote by 351tom Ok, would I be fairly close if I took the distance from 0.5 degrees BTC TO TDC?
No. In short, your simplistic model and analysis is inapplicable to the situation.
Any attempt to approach a limit without actually getting there involves a false assumption if the mechanism must then pass that limit. Ask yourself why you do not analyse it from 0.25% before to 0.25% after TDC.

You should also consider the expulsion of the bearing oil film, stretching of the main bearing bolts and stretching of the cylinder head bolts. Along with piston pressure, those will also limit the maximum bearing pressure.

The long connecting rod has a cross-section that is less than the crank bearing area. You should expect elastic compression of the connecting rod. If that compression exceeds the yield point, then the rod will bend and so begin the catastrophic failure of the engine. Again, elastic compression will reduce peak bearing pressure.
 P: 14 Your responses keep leading me to very high forces being developed. Would it be safe to say that if a stock 90 ft lb starter is not sufficient to crank the engine, then a higher torque starter is not a solution because at 90 ft lbs, the engine is already incurring damage?
 P: 1,109 Not necessarily. There are two things here. Is the starter motor capable of turning the motor, and how fast can it turn it. Power = torque * RPM. The starter must turn the engine fast enough to start after some oil has drained from worn piston rings. (This is especially true of diesel engines). During starting, the ignition timing should prevent an advanced spark occurring, so we can ignore pre-ignition. The greatest starter motor torque will be required somewhere in the last half of the compression stroke. I define BDC as -90° and TDC as +90°. The cylinder pressure rises as the reciprocal of the remaining cylinder volume. P1*V1 = P2*V2. Advantage follows a 1/Cos function. The crank arm lever advantage has a minimum of unity near 0° and increases as it approaches TDC. Starter torque will be a maximum where the combination of those two effects peaks. That peak pressure will occurs close to +60° and will be about 87 psi. If you work out the pressure * area and then force to the crank, you can apply the crank throw and pinion to ring gear tooth count ratio to compute maximum starter motor torque required to turn the motor.
 P: 1,109 Here is a first try at computing torque needed to turn the engine against compression. I don't trust it yet. It needs to be checked.  deg x y h vol psi force advantage ft.lb T.D.C 90.0 0.000 1.750 7.500 4.398 161.7 2032.0 1.633E+16 0.000 80.0 0.304 1.723 7.465 4.833 147.1 1849.1 5.759E+00 3.060 70.0 0.599 1.644 7.363 6.117 116.3 1461.0 2.924E+00 4.763 60.0 0.875 1.516 7.199 8.186 86.9 1091.8 2.000E+00 5.203 50.0 1.125 1.341 6.979 10.939 65.0 817.0 1.556E+00 5.005 40.0 1.341 1.125 6.716 14.245 49.9 627.4 1.305E+00 4.581 30.0 1.516 0.875 6.422 17.949 39.6 497.9 1.155E+00 4.110 20.0 1.644 0.599 6.108 21.886 32.5 408.3 1.064E+00 3.657 10.0 1.723 0.304 5.790 25.893 27.5 345.2 1.015E+00 3.240 0.0 1.750 0.000 5.477 29.817 23.9 299.7 1.000E+00 2.857 -10.0 1.723 -0.304 5.182 33.530 21.2 266.5 1.015E+00 2.502 -20.0 1.644 -0.599 4.911 36.929 19.3 242.0 1.064E+00 2.168 -30.0 1.516 -0.875 4.672 39.940 17.8 223.8 1.155E+00 1.847 -40.0 1.341 -1.125 4.467 42.516 16.7 210.2 1.305E+00 1.535 -50.0 1.125 -1.341 4.298 44.632 15.9 200.2 1.556E+00 1.227 -60.0 0.875 -1.516 4.167 46.276 15.4 193.1 2.000E+00 0.920 -70.0 0.599 -1.644 4.074 47.447 15.0 188.4 2.924E+00 0.614 -80.0 0.304 -1.723 4.019 48.147 14.8 185.6 5.759E+00 0.307 B.D.C -90.0 0.000 -1.750 4.000 48.381 14.7 184.7 1.633E+16 0.000
 P: 14 After seeing those small torque requirement numbers, it's hard to imagine how compression could cause a starter motor to stall. Even using an extreme 14:1 compression ratio, a stock starter could easily handle it; 14:1 deg____x______y______h_____vol___psi_____force___ft. lbs 70 __ 0.599__1.644__ 7.363 __ 5.177__120.0 __1508.0__ 4.92 65__ 0.756 __1.578__ 7.278 __ 6.246__ 111.6__1402.9__ 5.77 60__ 0.875___1.516__7.199 __7.238___81.6 ___1025.4__4.89 11:1 65__ 0.756___1.578__7.278___7.188__98.9___1243.4 __ 5.12
 P: 1,109 I still don't believe it. I must have made a mistake somewhere. Your numbers come out close to mine, and I know you did not copy my equations, that's why I did not post them. The conversion from inch_pound to foot_pound appears not to be the problem. It seems very logical that the peak starter torque should be somewhere in the 60° to 70° range. Because the starter is a DC motor, the maximum torque is when stalled. Without load it will have a maximum RPM determined by supply voltage. In between those extremes, torque will fall linearly as RPM rises. Direct injection diesels without turbo chargers can have a compression ratio up to 22:1. The starters used on diesel engines often have an additional gear reduction stage within the starter. Maybe that compensates for their higher compression ratios and permits standard starter components to be used.
 P: 1,231 Hey guys. Just a few points to add. Thermodynamics PV = C would be for an isothermal process, where the temperature of the gas as it is compressed remains constant. PV^gamma = C is for an adiabatic process where the compression of the gas is quick enough so that the heat produced does not have time to transfer from gas to the cylinder walls. For air gamma = 1.4 If you use model the compression as adaibatic rather than isothemal, the maximum cylinder pressure at TDC would be about 25 times that at BDC for a 10:1 compression ratio. Starter Motor torque. I do not know how that starter motors are rated but let us assume that maximum torque is at stall minimum velocity=0, and minimum torque=0 at free running maximum velocity. One can plot these two points on a graph of torque vs velocity( angular ) and join them with a straight line to obtain a simple motor curve that we innocently will assume the motor to follow. Assuming the starter motor maximum velocity is 1000 rpm as an example and that 90 ft-lbs is required to overcome stiction and all that to initiate turning the petrol engine over, and that the turning friction requires 45 ft-lbs. Matching 45 ft-lbs on the curve and we see that this starter will rotate at 500 rpm while cranking the engine. If we now replace the starter with a beefier model capabable of twice the starting torque of 180 ft-lbs at stall but still having a free running velocity of 1000rpm, graph that and match the turning torque of 45 ft-lbs, we find a different rotational speed while cranking greater than 500 rpm. We conclude that larger starters are not necessary chosen to overcome the torque of the petrol engine but could be selected to turn it over at a faster rate while cranking. Multiple cylinder engines Calculating the torque necessary to overcome one cylinder does not translate into a multiple of the cylinders. It could, but friring and timing issues reduce the multplication. If you have a 4 cylinder engine and all cylinders are arranged so that all compress the air and fire at the same time then the torque to turn that engine over would be the multiple. Most engine are not built that way which adds another complication to multiple engines. I hope that adds yo the analysis. So far it has been interesting. thanks
 Sci Advisor PF Gold P: 9,097 Consider shock loading, a common cause of bearing failure. Loads are usually averaged in engineering calculations which does not always reflect actual conditions.
 P: 1,109 256bits. I agree with all your points. While starting a multi-cylinder engine, the expanding air after TDC might be expected to cancel the compression on another cylinder at that time. This would be true if not for gas leakage past the rings, or with beta = 1.4 for air that would be a thermal loss during both the compression and expansion strokes. A compression testing gauge on a good engine gives a pressure that is very close to the compression ratio * atmospheric pressure. That suggests to me that beta is closer to 1.0 during starting, probably because the compressed air is losing heat to the cold cylinder walls faster than the engine is being turned. Chronos. Yes, shock loading can do funny things. With the exception of failure such as hydraulic lock, I see no likely cause for shock to the connecting rod bearings. Lack of lubrication might be a problem initially. The original fear suggested in this thread was that the infinite advantage at TDC could generate an infinite load on the rod bearings. I think it is now clear that the gas pressure limits that bearing load.

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