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#31
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No I didn't (duh).
Got it now. There's a lot there. What section has the flywheel info? If we keep this up we'll get to 25 posts soon. |
#32
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LOL!!! under the "Sorting Out The Mopar Small Blocks" link... and what i copied and pasted is close to bottom of the article.
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#33
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So, the author of that article is a "rebuilder", no more/less creditable than anybody else.
So, we still don't know for sure. |
#34
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Guess I'll be holding off on this until it's sorted out.
Mroldfart2u, I'm 1 up on you now with 9 posts. |
#35
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Quote:
I Have googled it every way i can think of for an EXACT definitive answer only to come up empty handed. I KNOW the answer is out there somewhere... (hey i've heard THAT before... lol) It shouldnt be that hard of an answer to find, unless there is no definitive answer. But thats very doubtful. So I suppose you are right, we still dont know for sure... |
#36
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I sent the question to Mopar guru Rick Ehrenberg, if he answers I'll post it.
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#37
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Quote:
My other thoughts are: For sure Mother Mopar knows... Mopar Performance knows... McLeod knows... Hopefully we won't have to resort to waterboarding... |
#38
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I've sent the question as well, with a small donation to encourage a response.
I'll try to contact the person who runs mopar1.us as well. |
#39
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Here's the reply from Rick:
Since the "certified" number seems to be a variable (!), I think someone is going to have to take a later converter drive plate, chop off the weights and weigh them (or measure them and do the math), then do more math to come up with the in/oz figure. I have seen both numbers. One thing's certain: The number is significantly less than the almost 20 in/oz of the earlier 360s. So... no further ahead. Doesn't make much sense to buy a flexplate or flywheel with detachable weight for this purpose. Perhaps someone out there has a flexplate they've removed from a truck?? We would also need to know precisely the distance from center. |
#40
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Idaho
As per the PM you sent me, I could not find the instruction sheet for the weight package I bought years ago. But I did find this description. If it is still available it might be worth the $20. This might be where I got the imbalance number I was using. P5249843 magnum 360 balance weight and positioning template package required for rebalancing performance torque converters. For use w/ externally balanced cast iron crankshafts in magnum 360 eng. from 1993-01, including magnum crate eng. magnum 360 balance weight and positioning template package $20.50 I do have a truck (big dia.) 360 magnum truck flywheel. If I get some time this week I will go get it (it is in storage) and will see if I can measure how much weight is missing and from what diameter. My plan with it was to cut it down to fit a 130 tooth ring gear and drill it to accept a 10.5 inch clutch. |
#41
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OK, I have a converter rebuilder's catalog and it states that the 360 Magnum uses a 91-gram (3.21 oz.) external weight and the 360LA uses a 120-gram (4.23 oz) weight. The catalog picture appears to show the weight on the same 5" radius as the drive lugs on the 11" converter.
If my math is correct, the 360LA would need 21.15 in. oz. of weight (5 X 4.23) which is close but conflicts with Mopar's tech drawing which states 19.79 in. oz. Using the 19.79 figure, the radius would be 4.68". In any case, it appears that the Magnun needs only about .76 the weight of the LA. Math guys correct me if I'm wrong. |
#42
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Using John's numbers from the rebuilder's catalogue (and ignoring the radius) 91/120 = 0.76 . Then multiplying 19.79 x 0.76 = 15 g.in which is very close to 14.65
I have not balanced a crankshaft, but I have balanced 30' long rolls for our paper machine so I do do know that getting close in balancing is important. You can chase the last little weight addition for hours to get it exact. So Idaho, reality is that by using 12.8 or 14.65 will get you to an acceptable balance for your situation. It is quite likely the crankshaft is not balanced that closely. I can pretty much guarantee that mopar did not assemble the crank, flywheel and harmonic balancer to balance them. They likely did the crank and components independantely each with a tolerance. I will try and get my 360 magnum flywheel tonight and see if I can calculate the imbalance. |
#43
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Here is a picture of the 360 magnum truck flywheel. I will attempt to measure the void to determine how much weight is missing and post that. Also looks like they did a final balance on it too judging by the drill hole.
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#44
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Thanks for your efforts. Much appreciated.
$20.50 is not bad to get the info. One would of course need to know distance from center, which may not be specified since I presume it bolts on to pre-existing mounting holes. I could also get a a Mcleod weight from a member on FABO but still would be missing distance from center. I've asked if he can weigh it with the mounting bolts so at least I'd have that part. Interesting that the Magnum flywheel is cast with a recess, then final balanced with drilling. Maybe putty in the recesses, then measure the putty volume? Re the rebuilder's catalogue info, if the converter weights are installed on the same location, then the .76 ratio applied to the known weight for an LA engine would seem reliable. .76 X 19.79 = 15.0 in-oz (not g-in) If 5" is specified as the known center of mass for the mounting location, then there is a discrepancy. If not, 4.68" is likely the correct radius. Then there is the discrepancy between your 14.65 and 15.0. I wonder if mounting bolts could make up the difference? I am assuming the weights specified are being sold for mounting on the same convertor with the same mounting location. If true, and 5" is not the specified center of mass, then I think we have an answer with 15 in-oz. Does this make sense to you? |
#45
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I spoke to the tech department at McLeod racing. He looked up the balance spec for the 360 Magnum and reports it at 22.67 in-oz (127.39 gm). I asked to double check this was not for an LA engine - he did, but I did not get the feeling he was dead certain this is the right number. I'm having a hard time believing it. The pistons are lighter than LA pistons. I think the rods and crank are the same... right?
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#46
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Hmmm after all the discussion here on the subject, dont ya find it odd that the OP of thread has not said a word sense the first page..... We have no way knowing if he has resolved his issue, or gotten the right flywheel to use on his mag motor with the 3sp he was wanting to use... Not to disregard anyone wanting answers, but I am of the thought that maybe just maybe there is NO definative answer because of all the differences on how IMBALANCE is measured... IF you are building/thinking of building one of these elusive beasts, the smart thing to do would be to balance the whole shooting match @ once, even if it means having either a flex plate hanging on the wall or a flywheel hanging on the wall doubt it would cost 20 bux more to have the 'extra' piece balanced... Just M2C.
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#47
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Quote:
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#48
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2003 is a long time ago. No surprise the OP is not around.
Sure I could pay to have everything balanced. I could also just cough up the dough for a MP or Mclead unit - likely would cost less. If I were in a hurry I would. A goal for me in this build is bang for the buck. The right number exists, its just elusive at this point. 340Duster1 has already proven it can be done. There is satisfaction in the quest, and others will benefit when they put the info to use. I am kinda curious how they balance a complete rotating assembly and what the typical cost is. John, in my last reply, i had not noticed it was you, not 340 responding in one of the posts. Thanks for your input. Does your catalogue specify a 5" radius? |
#49
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Cost me about a grand iirc. As I understand it, the piston/rod/ring assemblies are first weighed (including bearings) and brought to the same weight. Then a set of bob-weights equal to the piston/rod assys are bolted to the crank which is then spun up on the actual balancing machine, and weight is added or removed as required. Of course it's all computerised but there's still manual labour involved, not to mention the skillset.
I agree with you trying to do the flywheel yourself, I'm the same way. Makes my wife angry sometimes!! |
#50
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I imagine the equipment for balancing is not cheap, hence the cost.
I would put it in the category of "nice but not necessary" for this project. I purchased a used 360 Magnum for 600 and likely will get half that back selling heads and other parts. I'll be installing new aftermarket heads, cam, intake, carb, ignition system and headers. I'll be somewhere around $2500 total for an engine in the 400 HP range. Finding ways to keep the cost down is part of the challenge. My wife is almost never disappointed when I spend less on toys (I found a cool balancing video but cant post the link as my count is under 25 - there's lots of them on youtube) |
#51
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No, I measured the drive lug radius off of a converter I have...the area of the 11" converter where the drive lugs are located is the area on that size converter where the weight(s) are usually located.
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#52
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I finally measured the volume of the "void" missing on the 360 flywheel. It is 12.95 cc's. Converted to in3 that is .7902 in3. Cast iron is 0.254 lb/in3. So that equals 0.2007 lbs. Now to Ozs (x16) equals 3.21 oz. This is at a 3.5" raduis. This equals 11.24 in ozs. That is another one to ponder. I might have figured out what I did with the instruction sheet for the converter weights. I put it in my mopar magnum engine manual, now I just have to find the book.
Don't be afraid to try and balance the flywheel yourself. The factory tolerances were large and for a 400HP street engine you will be more than fine. if it were me I used the 12.84 one as it worked for me! Balancing isn't rocket science! |
#53
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This thread back from the grave again. With all the uncertainty about the right balance number I decided to buy a mopar performance Magnum flywheel. Since I have it I might as well solve the riddle once and for all.
I used modeling clay and syringes to measure the volume. You can see, each drilled hole is 6 cc. In the 6 cc syringe its slightly over, but two of them measure 12 cc in the larger syringe. It's because of a little air that gets in while loading them. The three holes measure 1.043 diameter. identical in depth (0.660") and the total volume is 18 cc. Using 0.265 lb per cubic inch as the weight of cast iron, here's the math. 0.265 lb/cubic inch x 16 = 4.24 oz per cubic inch. 1 cubic inch = 16.387 cc The balance hole volume is 18 cc / 16.387 = 1.098 cubic inches 1.098 x 4.24 = 4.65 oz. at 3.9" Center of flywheel to center of holes measured using a caliper = 3.90". I suspect the intended measurement is 3.88", the same as the LA 360 but I will use my measurement. 4.65 oz x 3.9" = 18.13 inch oz. This is much closer to the LA flywheel which has a known balance of 5.1 oz at 3.88" = 19.79 inch oz. There is a shallow drill spot closer to the outside with a volume of about 0.2 cc. I suspect this is a final balance adjustment. There is an angled hole through the thinner center portion that has a volume of about 0.3 cc. I doubt these are significant. In my view the 3 large holes are what make up the balance specification. Please double check my math. Barring a math error I'm calling the riddle solved. |
#54
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(I added a photo of the drill bit angle. I get 30 degrees.)
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#55
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I don't know how much difference it makes but the flywheel isn't cast iron, it's cast steel.
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#56
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Thanks for pointing this out John. The flywheel I have is described as "cast" by Mopar Performance. The word "steel" is used for a different part number and they refer to it a a "race" part. (I had to leave out the www etc in the links as my post count does not allow links).
Quote:
Quote:
I used the site below to find the weight of cast steel vs cast iron. Cast steel weighs very close to steel, (typical cast steel is 0.280 lb/in3) with little difference between alloys. Cast iron is more variable depending on the alloy ranges from 0.254 to 0.264 lb/in3 (6 to 9% lighter). The cast iron weight I used (0.265) is 5% lighter than steel. If the flywheel is cast steel, the math changes as follows. 0.280 x 16 = 4.48 oz/in3 1.098 x 4.48 = 4.92 4.92 x 3.9 = 19.19 in oz I'm guessing its not enough to matter but it would be nice to know if its cast steel or cast iron with certainty. Do you have more information? matbase.com/material-categories/metals/ferrous-metals/cast-steel/[/url] |
#57
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I am really impressed by your scientific approach. Please keep us informed. Did you ever read beleneagle's (David) experiences with balancing?
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#58
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Thanks for the complement. It's just me being stubborn and plugging away with google searches.
I had not found that thread. Interesting, they came up with 14 in oz. The problem of there being so many numbers out there is the reason I put all my documentation up for folks to evaluate. The question of cast steel vs cast iron will eventually be solved as well. Based on that thread, selling the weights without instructions could lead one think there's an effort to preserve sales of overpriced flywheels. |
#59
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__________________
https://t.me/pump_upp |
#60
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Someone on FABO pointed out an error in my calculation of total balance weight. Two of the holes are at different degrees on the flywheel so they contribute less to total balance weight. I found a method to calculate the vectors by drawing a parallelogram. Below are the corrected results.
I drew a parallelogram with 3 vectors, one for each hole in the flywheel. I set the center vector at 100 mm. The two other vectors came out to 63 mm and 64 mm so I used .635. Here's the calculation. Total weight = 4.65 oz. 4.65/3 = 1.55 oz per hole. The distance from flywheel to center is 3.84" (1 x 1.55) + 2(.635 x 1.55) = 3.52 oz at the center vector 3.52 oz x 3.84" = 13.5 in oz Feel free to critique the method. Note that its quite close to the value successfully used by 340 Duster (2.28 oz x 5.625" = 12.825 in oz) |
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