I think your calc's are fine but you used a different method from me to calculate the Torque. Your lever arm is the length of a line from the crank centre to the rod at 90 deg I take it? I just considered that F is the only force acting on the system, and since the crank stroke and angle (lever) are the same for both, F is the sole factor of Torque. Which is why the ratio of your 6.008/5.64 is the same as the ratio of your instantanious loads 0.158/0.147.
Of course they cover the separate parts of one rev in the same time. The engine has accelerated the Inertia of the system to get to a given rotational speed. Now, at that speed, the system has built up an Angular Momentum due to the product of Moment of Inertia and the Rotational Speed. L= I x Omega. I is fixed for the system, so for Omega to change within part of a rev, the Angular Momentum must change. But, from Conservation of Moment, the only way to change the Angular Momentum is to apply an external force. But you can't do that in one rev, we only have the one detonation force starting the rev off. OK, the force does reduce during one rev, to zero at bottom dead centre, but the whole Angular Momentum of the system is keeping the speed, section by section constant. You havn't put another forse into the either piston part way through a rev! If you know some new Physics which changes this, I believe you should explain how the crankshaft can change speed during part of one rev! And post a full explanation!
A retired engine designer told me that you decide your block height, your crank throw, your small end height within the piston, and the the distance then between the small end and big end you join up with a conrod... My thinking on it all without blinding the place with maths....a short rod puts more of a load on the bores. A short rod also causes less piston dwell time at tdc. This can be a good thing as it lessens the time for the chance of detonation to take place meaning you may be able to run more timing. A short rod is also lighter, and the engine also ends up being shorter, meaning lower centre of gravity. Theres as many pros as cons for both long and short.
Here's how to calculate torque. If new physics dictates a different definition of torque, please post calcs to explain it!
The crank speed omega is proportional to the crank torque. The crank torque is proportional to the piston force (same for both) and the piston to rod angle. At TDC the angle is the same for both and therefore the torque is equal. Past TDC the piston to rod angle for the two will differ, and hence the torque too, and hence the angular velocity. If we take it that the system has no Inertia or friction, the piston acceleration will be the same for both engines which will lead to differing crank accelerations. If we include inertia then the difference in crank acceleration will reduce and we'd see a difference in piston accelerations. The larger the inertia the more it tends toward constant omega, which is closer to the real life situation. But you did say in your initial calculations that you were ignoring Inertia, when in fact you were considering it as infinite.
Correct, my first calc did ignore all outside effects for an instantaneous torque. Not infinite but zero, just leaving points in space. Torque is not proportional to speed. Torque is purely force multipled by lever arm. Rotational speed does not come into the equation. For power it does! I was explaining, in my separate post, why the rotational speed, from angle to angle, cannot, as you implied, vary during one revolution. That's why I brought in Inertia and Momentum! Of course the pistons accelarate differently, that's what I said before. For constant RPM, the shorter rod will accelarate away from TDC more quickly than the longer one.
Your physics is fine and your calculations look fine. I was just saying you didn't need to go through all the hassle of those working outs when what you ended up with was the same as your F2 ratios. ie if the crank angle is the same Torque = F2 * constant.
Take your point. I was trying to show the numbers on a diagramatic representation, so people could see the instant at 45 deg.
I don't understand this. Angular accel = Torque / Inertia So why can't a change in Torque produce a change in acceleration?
OK, torque is proportional to a change of speed then. Torque will affect speed. Still, an increase in Torque leads to an acceleration which increases speed.
Good write up as always mate, do you think that the gap between the m/s 9A and ABF is down to mapping and the late cam swap? I also noice the 9A is using the KR style inlet manifold and two stage T/B do you think that could of played a part in the torque dropping off from 4700 like it did?
Hi there, We have an ABF on K-Jet that would be as you know similar, yet much more aggressive than this car in terms of drive. The throttles are not the issue, in fact the crossectional area of the twin thottle is very similar to the unported throttle found in both the Digifant 3.x and ABF MS vehicle so that is transparent. I would have thought the use of ABF cam sets, even I knew this after, would have resulted in a different response, but no. Indeed dyno development time will give the oppertunity to keep the torque higher thus making urgency more noticable at 4500-6500rpm. But my predictions based on unpublished data is it will just about match the unmapped digifant vehicle for torque + a bit more torque post 5 to 6K.
ok so the 9A would sit between the oem ecu'd ABF and m/s ABF. so the cam swap and mapping messed it up, shame! A good test would be a 9A(with kr cams or abf ones) on k-jet Vs ABF on k-jet that way both engines are playing on a level field.
Pardon? Rather than me retype was already said, I suggest very politely, you read post 1 and understand the purpose of this thread.
maybe I am missing something here. 9A vs ABF 139bhp vs 150bhp in stock form. why is this? electronic management and fuel injection? the extra compression ratio? add different ecu's and maps to the mix... and.... different results, making ABF look "better", which in stock form was and is... so no huge shock ABF makes more power/torque, and optimised MS version can make more again... I think I know the answer but invite comment
9A is 136 in stock form, the reason for this is the lame inlet cam fitted as standard. the way I understand it the reason the inlet cam is so lame is due to the k-jet system not being good enough to prevent damage to the cat otherwise. for the same reason the cat equipped PL engine power is bad compared to the full fat KR lump. (129? vs 139bhp)
Stock 9A vs Stock ABF. I think Rubjonny covered that. But the 9A on this comparison has ABF cams and some mapped management. Now I, for one, expected it to perform inline with the mapped ABF. The only differences I can see between the hardware now are: - A different type of plenum & injectors - A difference in Rod/Stroke ratio and thus possibly different dwell at TDC? - Valve seats with better blending on the ABF vs 9A from factory - Less shrouded exhaust valves/seats on the ABF vs 9A I believe the compression is the same. So could these factors influence the way the torque is being developed? Or is the engine just a little tired in comparision? Gurds
Didn't late 9a's use the same head as the abf? This is the point I try'd to make on post No58 but got 'shoot down' I only asked why did the torque drop off at 4.7k as it did? cos in the past both engines on k-jet (9a's with in kr cam) are very close.
I am not 100% sure about the 9A engines, but the late Audi 6A engines do use the 051 103 373D head. It is fitted with a slightly different inlet cam than the ABF 051 103 373D head though. Possibly to do with one running k jet and the other running efi. But they still both have the better combustion chamber shape and 27mm exhaust valves.
