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OK, this is going to look really F-ed up, but I think I'm getting somewhere.
I took my VE's and attempted to calculate engine power output strictly from cylinder fill, CFM, g-forces of angular acceleration on the pistons and crank, pumping losses due to compression, and RPM
***Frictionless horsepower based on airflow and perfect combustion = CFM * 1.44
Then, I used HP = TQ * rpm/5252 to get frictionless/resistanceless TQ
Used rotational speed, weight of pistons and connecting rods at 1.811 from centerline * .707 (average sine wave value) to get the acceleration of the pistons and rods at the top and bottom of stroke and then calculate force from that, which integrated to work. I ended up with:
*** piston/rod peak force = SqRt(Piston Acceleration in g's) * 1/2 stroke * 0.707 (avg sine wave value) * 0.8 (no clue, but I got it somehow)
Used 14.7 psi * decimal VE * DCR * pi * bore/2 squared * 1/2 stroke to estimate force exerted compressing air, then integrated to get work lost due to compression.
*** Compressive TQ = [14.7 * VE (as decimal) * (bore/2)^2 * DCR * 1.811 / 12 ]^2 all divided by 33 (must have ended up from unit conversion leftovers)
I'm not even sure how to type the whole equation out, but it accurately predicts my current HP and peak HP RPM between 6500 and 6800. Over estimates my peak TQ by 30, but accurately predicts peak TQ near 4800 rpm.
I was going to do an adder for friction, but I think I somehow have it included with the compression work.
So, net TQ = [Frictionless TQ] minus [TQ used for Piston Acceleration] minus [Compressive TQ]
Net HP = Net TQ * rpm / 5252
I have no idea how correct this all is, so feel free to harshly critique the maths. I'll try to plot this later on and show how it looks.
Most Pro Stock racers subscribe to the 8% past peak horsepower for a shift point to continue the acceleration from torque recovery (gear drop). Although they are not everyday normal engines they are one of the ultimate NA type racing machines
Until reading this thread, I was a firm believer in what carroll shelby said about horsepower selling cars and torque winning races. But...
The torque being static whilst horsepower is dynamic argument really has me thinking. The example of the two trucks really helped me visualize/conceptualize this (thank you, Tony).
I'm absolutely down for some mental masturbation, so bring on the math. I like the torque formula that was brought up earlier, but how would you derive your cylinder pressure to plug in to the equation? Is there a separate equation to find cylinder pressure based on volume of the cylinder, compression ratio, air/fuel ratio, and octane rating? It would be cool to have a spreadsheet program to be able to plug in different values and see the change in theoretical torque.
Given the same weight vehicles with the same amount of aerodynamic drag the vehicle with the most average power wins....period....doesn't matter if the other vehicle has a 500 ft/lb advantage in torque.....this assumes the driver is smart enough to shift where he utilizes the maximum average power which is typically 500-700 RPM past peak power.
If peak torque was what won races you would short shift the engine miserably and hurt your ET and trap speed by a significant amount
Modern day "Superbikes" (the likes of the ZX14 Kawasaki, Suzuki Hyabusa etc.) trap close to 150 MPH in the quarter mile right off the showroom floor (insane performance per dollar!!)....Is that because they are producing a whopping 80-100 ft/lbs of torque or close to 200 HP in a vehicle that weighs 700 lbs with driver.
Its all about the horsepower folks when discussing acceleration....or more specifically power to weight ratio.
And if someone has the time to Google Im sure there are other publications and articles that will back me on this.
Why would a sliderule who's sole purpose is to calculate ET and trapspeed only have HP and weight as inputs if torque was of any value??
And back to my diesel/gas truck analogy.....if peak torque wins races modded diesels would be the fastest vehicles on the planet.....their impressive considering their weight but their track performance only spits out exactly what their horsepower produced and weight would dictate.....nothing more.
-Tony
PS....Just for grins I inputted 700 lbs and 200 HP into the sliderule.....149.8 MPH trap speed!! If you had a 2800 pound race car that made 800 HP it would run almost the same trap due to the fact both are 3.5 lbs per horsepower (which is obviously a very impressive power to weight ratio).....the only difference here would be aerodynamics.....if the aerodynamic footprint was the same they would run exactly the same trap speed.....ET is dictated by traction and other factors and has more variables but trap speed is what tells you how hard the car is pulling and stuffing you in the seat!
I think that slide ruler sums it up very well. Horsepower is derived from torque and rpms... So if you use horsepower, you are factoring in torque and rpms. That's what makes it dynamic, that's what makes it accurate as a tool of predictions, and it's what makes horsepower so valuable in the real world.
In order for the horsepower value to increase, or even exist at all, there has to be some measurable amount of torque, AS WELL AS a rpm value... Metering that aforementioned amount of torque against some measure of time.
If the slide ruler asked for torque as an input, it would also have to have rpms as an input... which would just give you horsepower, so why complicate things more than necessary? The same thing absolutely applies to discussions, debates, and arguments everywhere.
Infinite torque, literally all of it, wins not a single race if there's no movement. Torque is a rotating force, not to be confused with actual rotating movement. You can apply 100lb/ft of torque to something and not move an inch if you need 101lb/ft to go... But you still absolutely have 100lb/ft of rotating force being applied.
There has to be movement for there to be horsepower, and in most races, whoever is moving the fastest wins.
Until reading this thread, I was a firm believer in what carroll shelby said about horsepower selling cars and torque winning races. But...
The torque being static whilst horsepower is dynamic argument really has me thinking. The example of the two trucks really helped me visualize/conceptualize this (thank you, Tony).
I'm absolutely down for some mental masturbation, so bring on the math. I like the torque formula that was brought up earlier, but how would you derive your cylinder pressure to plug in to the equation? Is there a separate equation to find cylinder pressure based on volume of the cylinder, compression ratio, air/fuel ratio, and octane rating? It would be cool to have a spreadsheet program to be able to plug in different values and see the change in theoretical torque.
The two-truck example helped me a lot also. And it makes sense, you see these 400 HP 800 ft-lb pick-up truck diesels all the time. great torque, but can't rev, so net/net can't accomplish any more work.
Another visual that worked for me is that you can gear torque down torque but not horsepower. If I make 470 HP and 380 pounds of torque at 6500 RPM, and then reduce the output with a 2:1 gear, At the other end of the gear, it sees 760 pounds of torque and 470 horsepower at 3250 gear RPM.
On the cylinder pressure, the way I'm seeing it, there are two cylinder pressures to be concerned with. One generates power and the other is parasitic. The cylinder pressure created during the compression stroke is parasitic - stealing from the engine's output to perform work. It's pretty easy to calculate at 14.7 x VE x DCR. The cylinder pressure at the start of the power stroke is the hard one. And I didn't bother. I just used air to calculate horsepower potential and then calculated TQ from there, assuming loss-less. What makes it worse is that at no point in the cycle is the pressure even constant.
If the equations below are even close, then the thing I'm seeing robbing the motor of power at the very top end is the increased acceleration on the rotating assembly. As the engine moves faster, it takes more force to reverse the direction (negative acceleration) of piston travel, and it is a parabolic function:
g-forces at the piston at 800 rpm = 32.9196 g
g-forces at the piston at 8000 rpm = 3291.96 g
so, RPM went up 10x, g-force went up 100x. Mass was constant, so force went up 10x. Nothing else accounts for the increased torque drop off at higher RPM
Now, to Tony's point about increasing airflow at higher RPM -
At 7200 rpm, the VE is 95%. If you change the VE in the model at 7200 to 105% (10% increase), the HP increases from 470 to 530, for a 13% gain. Above 5252, you get greater than 1:1 for the increases, and below 5252, you get less than 1:1.
Part of this is that I'm trying to understand why the motor doesn't continue to put out more power as RPM increases. So far, the only thing that is robbing that kind of power internal to the engine is the acceleration of the pistons and connecting rods at TDC and BDC
OK, this is going to look really F-ed up, but I think I'm getting somewhere.
I took my VE's and attempted to calculate engine power output strictly from cylinder fill, CFM, g-forces of angular acceleration on the pistons and crank, pumping losses due to compression, and RPM
***Frictionless horsepower based on airflow and perfect combustion = CFM * 1.44
Then, I used HP = TQ * rpm/5252 to get frictionless/resistanceless TQ
Used rotational speed, weight of pistons and connecting rods at 1.811 from centerline * .707 (average sine wave value) to get the acceleration of the pistons and rods at the top and bottom of stroke and then calculate force from that, which integrated to work. I ended up with:
*** piston/rod peak force = SqRt(Piston Acceleration in g's) * 1/2 stroke * 0.707 (avg sine wave value) * 0.8 (no clue, but I got it somehow)
Used 14.7 psi * decimal VE * DCR * pi * bore/2 squared * 1/2 stroke to estimate force exerted compressing air, then integrated to get work lost due to compression.
*** Compressive TQ = [14.7 * VE (as decimal) * (bore/2)^2 * DCR * 1.811 / 12 ]^2 all divided by 33 (must have ended up from unit conversion leftovers)
I'm not even sure how to type the whole equation out, but it accurately predicts my current HP and peak HP RPM between 6500 and 6800. Over estimates my peak TQ by 30, but accurately predicts peak TQ near 4800 rpm.
I was going to do an adder for friction, but I think I somehow have it included with the compression work.
So, net TQ = [Frictionless TQ] minus [TQ used for Piston Acceleration] minus [Compressive TQ]
Net HP = Net TQ * rpm / 5252
I have no idea how correct this all is, so feel free to harshly critique the maths. I'll try to plot this later on and show how it looks.
First off, I apologize for quoting the long post, but I have some specific questions.
The .707 number, derived from average Sin waves... What does this mean? What is this a measure of? Are you finding angular distance at TDC and BDC to use for acceleration? Please explain. I said I was down for some math, I never implied I was good at it. I guess it's not the math that is confusing me, just the application in this regard.
Is that 386.1 value in your piston acceleration formula the piston/rod weight?
And in the name of keeping this somewhat scientific, this being your peer review process, the 0.8 multiplier in the piston/rod peak force equation... where and how was this number derived? In order for this equation to be repeatable, we cannot have arbitrary values that happen to work. Same thing with the "33" in your compressive torque formula.
I'm not trying to bust your ***** about it, I'm actually trying to understand this and possibly make a spreadsheet program using the formulas produced in this thread.
First off, I apologize for quoting the long post, but I have some specific questions.
The .707 number, derived from average Sin waves... What does this mean? What is this a measure of? Are you finding angular distance at TDC and BDC to use for acceleration? Please explain. I said I was down for some math, I never implied I was good at it. I guess it's not the math that is confusing me, just the application in this regard.
I plotted a sine wave over 1000 points and then took the average absolute value. Since position is a sine wave, so is acceleration. So instead of having to calculate thousands of accelerations, I found an average number so I could treat it as a linear approximation.
Is that 386.1 value in your piston acceleration formula the piston/rod weight?
386.1 is the gravitational constant (g) in the unit "inches per second per second". The weight of the pistons and rods I ended up using was 8.4-KG or 18.52 pounds total (1050 g for each piston + rod assembly)
And in the name of keeping this somewhat scientific, this being your peer review process, the 0.8 multiplier in the piston/rod peak force equation... where and how was this number derived? In order for this equation to be repeatable, we cannot have arbitrary values that happen to work. Same thing with the "33" in your compressive torque formula.
I'm trying to duplicate the work, because it doesn't make sense to me, either. I think they are artifacts of unit conversion, but I need to re-do the equations. For example, the 33 I'm pretty sure was from an integration over time combining with other factors.
I'm not trying to bust your ***** about it, I'm actually trying to understand this and possibly make a spreadsheet program using the formulas produced in this thread.
No, I appreciate the questioning. It's constructive, IMO. Edit - And I just realized I haven't included anything for the weight of the crank counterweights. Oy!
Last edited by Darth_V8r; 10-26-2015 at 05:10 PM.
Reason: I just realized I haven't included anything for the weight of the crank counterweights.
I think we are nuancing about a peak TQ number. I'd happily take a lesser TQ number but one that carried a long way. When I look at dyno charts I like to see a big flat TQ curve. A high HP number then has to be there due to the math so in that sense it is all about the torque.
OK, well something is off. I'm overcalculating torque on the low end. The thick blue line is calculated HP, and the thick green line is calculated TQ. Thin red line is air flow in CFM, thin gray line is HP based off of pure air and gas converted to energy in a loss-less system, thin yellow line is TQ calculated off the gray "perfect" HP line. Once it hits 3600 RPM, it actually looks reasonable
Something is wrong. You know the formula. If torque is constant (and in your graph it is close) and the rpm goes up the HP has to follow. On your graph the blue line drops.
So here is an older head/cam LS6 combo dyno graph from my C5. Here are the facts.....and you tell me where you shift and why ? 2995lb race ready w/me 25.7" rollout 4.10 gear T56 (m6 ; 2.66/1.78/1.30/1.0) You show me how hp wins....and I'll show you how it's all about rear wheel torque .
I say shift at 7300.
I'll use second and third as my math comparison. The attempt would be to straddle 6800 as much as possible. Here is why:
6800 rpm is 400 pounds of torque, but in second gear, that's 712 at the wheels vs 559 at the wheels in third (430 x 1.30) at 5000 rpm, close enough to speed matching.
Basically, I would ride the lower gear as hard as possible to take advantage of the geared up torque.
But I will also say that if you ride it too long, torque falls off too far, so at some point, you're better off up shifting. So, I say straddle peak power to get the best possible mechanical advantage to the wheels.
Lsoholic, I would shift your car at 6500-ish. Still above 400lb/ft and close enough to peak hp to not feel like I am missing out on anything. I always thought that people shift at or around peak hp just to make sure you are still above peak tq after the rpm drop at shifting.
I'll use second and third as my math comparison. The attempt would be to straddle 6800 as much as possible. Here is why:
6800 rpm is 400 pounds of torque, but in second gear, that's 712 at the wheels vs 559 at the wheels in third (430 x 1.30) at 5000 rpm, close enough to speed matching.
Basically, I would ride the lower gear as hard as possible to take advantage of the geared up torque.
But I will also say that if you ride it too long, torque falls off too far, so at some point, you're better off up shifting. So, I say straddle peak power to get the best possible mechanical advantage to the wheels.
Are you taking rpm drop into consideration (per gear) ?? Which affects average power on a larger scale than some understand.
Also...rear wheel torque is trans gear + rear gear x torque at that specific rpm...hint...hint.
So, do you still stick by your 7300 ??
What do you think the accelerometer would say about a 7300rpm shift ??
Hmmm, you can see that the power curve is steepest when torque is close to its peak (or within say in the range 90%-100% of peak)...
the slope of the power curve is related to vehicle's acceleration...
P = m * v * dv/dt
dv/dt is maximized when the power curve is the steepest which occurs when torque is sufficiently close to peak (which with a wide torque curve, a wide usable RPM range is available).
Are you taking rpm drop into consideration (per gear) ?? Which affects average power on a larger scale than some understand. Also...rear wheel torque is trans gear + rear gear x torque at that specific rpm...hint...hint. So, do you still stick by your 7300 ?? What do you think the accelerometer would say about a 7300rpm shift ?? .
Right, left out the rear ratio, because I assumed your dyno was done in fourth with the rear already figured in. So, I figured everything relative to fourth.
I'm actually working with a friend to set up an accelerometer experiment. We are trying to figure out how to tie in rpm vs acceleration readings.
So now we need to see about rpm drop on the 1 - 2 shift. 2.66 to 1.78 is roughly 33% drop/ loss (example; shifting at 7500 drops to 5019......and 7700 drops to 5153.
So...look at the power at the wheels (rear wheel torque) and where the rpm drop back puts it.
I'll say this....you do not what to shift to the next gear...until the next gear is putting more power to the tire than the gear your leaving.
I need to fill in the calc's using every 100 rpm....but do you still stand by your 7300 comment ??
Chassis dyno accounts for the gear (one of the inputs of the parameters)...thus it's reading engine torque. So the rear gears needs to be calculated for true rear wheel torque.
As you can see, the duration is on the small side. That's why the tq falls quickly.
No need to fill in the hundreds. I see your point. But, at 7000, your torque is still higher in first than it will be in second. If you fill in the hundreds, you'll probably hit break even at 7100, which is probably where you'll say to shift? And I'll buy that. But if that's the case I was only off by 200 rpm. Not like you're telling me you should shift at 6200 and I'm 1000 off. Or are you?
10-26-2015, 02:57 PM
