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Yes, but look at the frontal area and mass of that machine versus a car.
yeah, i know its a streamlined motorbike, but that example is more about the fact he did it in 1967, and NOBODY believed that a guy living on the bones of his **** could drag that bike over from New Zealand back then and go that fast.
In 2019 anybody with a bit of money and dedication to the task could build and drive a C4 to 200 mph.
The big question is opportunities to safely drive to those speeds, what events allow that opportunity?
Why in the world would anyone trying to set a top speed goal not gear the car to make peak power at the car's projected top speed? More to the point, the question (16 years ago) was "How much power at the flywheel would it take to achieve 200mph?" That question can be answered without respect to gearing/rpm/blahblahblah: it takes however much power is required to overcome the total friction of the car. Total friction include mechanical friction, parasitic drag, and induced drag. However, much power (crank torque is irrelevant) it takes to equalize all the frictional forces at 200mph is how much power it takes. Let's say it takes 500rhp (it probably doesn't take quite that much): it doesn't matter if it's a diesel churning out 500hp at 2500rpm, or a 1000hp motor that is geared by moron so that it's only outputting 500hp at 200mph, or if it's an electric motor with no transmission, or a steam turbine geared to the wheels - whatever engine is powering the thing, at a steady speed of 200mph the wheels are seeing 500hp under those conditions. And yes, it does matter what altitude we run the car, because it takes less power to overcome the lower aero drag in less dense air. Same thing for ambient temps (hot air is less dense).
A C4 is not only a very slippery car (low drag coefficient), but also a small car in terms of frontal area. It is a pretty good car for top speed attempts. In a recent thread, we were just discussing that a stock L98 C4 could touch 150mph with 240hp or whatever it was rated at SAE net (crankshaft). And yes, this assumes that it was actually at its peak power at top speed (I don't know if it was geared that way, but let's just say it was to keep this simple). The math tells us that the same car would take 78% more power to increase speed to 200mph. So it should hit 200mph with 427 SAE net hp at the crank. This is not unreasonable. After all, a stock C6Z could get close to 200mph with 505hp (~440rwhp), and it is a draggier car than the C4 (it's way wider and has wider rear tires).
I have to respectfully disagree with the calculations. While aero drag goes up as the square of the speed, the horsepower required to overcome it goes up as the cube of the speed (at higher speeds, the engine is doing more work, and doing it in less time). Using the 240hp at 150mph baseline, the aero load at 200mph is increased by 77%. The horsepower required to overcome this 200 mph aero drag is 565 hp.
I have to respectfully disagree with the calculations. While aero drag goes up as the square of the speed, the horsepower required to overcome it goes up as the cube of the speed (at higher speeds, the engine is doing more work, and doing it in less time). Using the 240hp at 150mph baseline, the aero load at 200mph is increased by 77%. The horsepower required to overcome this 200 mph aero drag is 565 hp.
It's true that power to overcome drag increases with the cube of velocity: Pd = Fd*v (v=velocity, Fd is drag force) and drag force already contains v^2 so drag power = (1/2)*Cd*p*v^3*A
The units have to work out: Power is kg*m^2/s^3 which is why Fd needs to be multiplied by v: kg*m/s^2 * m/s
I don't agree with your 565hp number though. You have to convert everything to SI units (kW and m/s) then scale it by v^3 so …..186.4kW*(89.4m/s / 67m/s)^3 = 442kW which is 592hp
I have to respectfully disagree with the calculations. While aero drag goes up as the square of the speed, the horsepower required to overcome it goes up as the cube of the speed (at higher speeds, the engine is doing more work, and doing it in less time). Using the 240hp at 150mph baseline, the aero load at 200mph is increased by 77%. The horsepower required to overcome this 200 mph aero drag is 565 hp.
I disagree to your disagreement
If drag eats 240whp at 150mph then you need 241whp to have some acceleration left.
At 200mph drag increases by 78%. 240whp*1.78=427whp. If you have 428whp you will also have some acceleration. Acceleration will be very slow. You reach top speed when whp and drag hp meet.
To have constant acceleration you need the horsepower to go up with the cube of the speed.
I have to respectfully disagree with the calculations. While aero drag goes up as the square of the speed, the horsepower required to overcome it goes up as the cube of the speed (at higher speeds, the engine is doing more work, and doing it in less time). Using the 240hp at 150mph baseline, the aero load at 200mph is increased by 77%. The horsepower required to overcome this 200 mph aero drag is 565 hp.
I think you and Jay are exactly right. I don’t have much time to post right now, but a handy way to think of it is a car with a CVT. At 200mph, the gearing would be 33% taller to keep the engine at the same rpm. That means to keep the same tractive force as it has at 150mph, it needs 133% of the power output. BUT at 200mph the resistive force would have ALSO increased by the square of the proportional increase in speed from 150mph, or by 133%*133%. So in total we get 133% cubes for the total power required to go 200mph. That number seems more realistic based on empirical observation.
ETA: I can’t speak about the conversion to SI units.
Last edited by MatthewMiller; 01-17-2019 at 07:58 PM.
It's true that power to overcome drag increases with the cube of velocity: Pd = Fd*v (v=velocity, Fd is drag force) and drag force already contains v^2 so drag power = (1/2)*Cd*p*v^3*A
The units have to work out: Power is kg*m^2/s^3 which is why Fd needs to be multiplied by v: kg*m/s^2 * m/s
I don't agree with your 565hp number though. You have to convert everything to SI units (kW and m/s) then scale it by v^3 so …..186.4kW*(89.4m/s / 67m/s)^3 = 442kW which is 592hp
May I ask why you did your calculations using 250 hp (186.4kW) at 150mph, rather than the baseline/example number posted by MatthewMiller of 240hp to achieve 150mph?
If drag eats 240whp at 150mph then you need 241whp to have some acceleration left.
At 200mph drag increases by 78%. 240whp*1.78=427whp. If you have 428whp you will also have some acceleration. Acceleration will be very slow. You reach top speed when whp and drag hp meet.
To have constant acceleration you need the horsepower to go up with the cube of the speed.
As I said before, aero drag may go up by the square of the speed, but the horsepower required to deal with it goes up by the cube. Please look it up.
I don't agree with your 565hp number though. You have to convert everything to SI units (kW and m/s) then scale it by v^3 so …..186.4kW*(89.4m/s / 67m/s)^3 = 442kW which is 592hp
You do not have to convert to SI units. English units work just fine and in fact took us to the moon. SAE is a weight based system, SI is a mass based system as long as you keep that in mind and no the difference between mass and weight and account for it you are fine.
There was a thread on the C5 or C6 forums a week ago about this place: https://jbprovinggrounds.com/pages/about-us
Bonneville, as well.
I'm sure there are dozens, if not hundreds, of other options.
So that raises the next question (in amongst all the other arguing going on) how much horsepower you need also depends on the length of the circuit
Bonneville is usually around 10 miles long (depends on quality of the salt) so your minimum horsepower requirement is probably sufficient - youve got plenty of time to get there !
What happens when you run at the Texas mile ? Standing start and only 1 mile of acceleration time- how much horsepower does it then require ?
May I ask why you did your calculations using 250 hp (186.4kW) at 150mph, rather than the baseline/example number posted by MatthewMiller of 240hp to achieve 150mph?
Yes, if you convert 240hp to kw (178.96) and then do the math again and convert back, you get the same 565hp answer that 69427 got. So the units don't change anything.
BTW< it's worth pointing out again that the 565hp calculation is SAE rated at the crank, so not rwhp. 500rwhp would probably get you there (would have to calculate driveline losses, which is a an exercise fraught with imperfection). Also, I mentioned a C6 Z06 earlier because it can nearly make 200mph in stock form. It turns out to be more slippery than I realized. It has more front area than a C4 (22.something vs 19.4), but apparently a lower Cd (.31 measured independently vs .34 for the C4). So their total aero drag is very similar.
Yes, if you convert 240hp to kw (178.96) and then do the math again and convert back, you get the same 565hp answer that 69427 got. So the units don't change anything.
BTW< it's worth pointing out again that the 565hp calculation is SAE rated at the crank, so not rwhp. 500rwhp would probably get you there (would have to calculate driveline losses, which is a an exercise fraught with imperfection). Also, I mentioned a C6 Z06 earlier because it can nearly make 200mph in stock form. It turns out to be more slippery than I realized. It has more front area than a C4 (22.something vs 19.4), but apparently a lower Cd (.31 measured independently vs .34 for the C4). So their total aero drag is very similar.
I agree. We're just discussing the basic concepts of aero drag and the required horsepower to deal with that drag. If other posters want to get into the minutia of a particular vehicle's Cd, rho, driveline efficiency/losses, time/distance to get to terminal velocity, yada, yada, yada, I believe there's plenty of cerebral capability here to do that, but this present/recent conversation has just been on the basic overall concept of terminal speed issues.
As I said, from a physics standpoint I disagree. If aero drag goes up by the square of speed, then the horsepower required to deal with it also goes up by the square.It is the definition!Any extra horsepower will cause acceleration and thereby increase speed.
I think you're confusing power required to maintain a certain speed with force required (as I did when I typed the first erroneous calculation). Aero drag (and all restrictive forces like driveline friction, tire deformation, etc.) is a force. So we all agree that the force increases with the square of speed. But power is not the same as force. In order to exert the same force at the contact patch at 200mph as is produced at 150mph, you need 33% more power. This must be the case, because the basic equation for power is force*speed. Get way from thinking about power at the flywheel, and think about the power at the wheels: if the power level remains the same, the force and speed are inversely related. If the speed increases 33%, then the tractive force at the contact patch must fall by 33%.
So again, at 200mph you need 133% of the power you needed at 150mph, just to maintain the same tractive force as it had at 150mph. And you are also encountering an increased resistive force of 133%^2. So the total increase in force required is 133%^2, and therefore the total increase in power required to get that force is 133%^3.
Last edited by MatthewMiller; 01-18-2019 at 12:42 PM.
As I said, from a physics standpoint I disagree. If aero drag goes up by the square of speed, then the horsepower required to deal with it also goes up by the square. It is the definition!Any extra horsepower will cause acceleration and thereby increase speed.
I made some calculations and graphing in excel. If you want to download and play with it -> The Excel file
Again, I disagree. Below is where I disagree with you (I'm short on time, but I'll look at your graphs later today).We all agree that the areo drag goes up by the square of the speed. Let's just go with two simple speed numbers at the moment to simplify things (100mph and 200mph). At twice the velocity/speed, the aero drag will be four times as large.
Now, Work = Force times Distance (W=F x D). Horsepower, which is the rate that work gets done, is then Hp=W/t. (Work per second).
At twice the speed (200mph), the F (force to overcome the drag) in the work equation is 4x what it is at 100mph.
Also, at 200mph the distance traversed each second (D in the work equation) is 2x what it is at 100mph.
We then end up with an equation like this: Hp (@200mph) = (4xF@100mph)x(2xD@100mph)/(t=1sec) = 8x(FxD@100mph) = 8xW each second = 8 x Hp that was required at 100mph.
I'm late for an appointment. Will check back later. I welcome any correction to any mistakes made in this rushed post.
You do not have to convert to SI units. English units work just fine and in fact took us to the moon. SAE is a weight based system, SI is a mass based system as long as you keep that in mind and no the difference between mass and weight and account for it you are fine.
You're right, don't have to use SI for what I posted but I originally was going calculate a power based on an example density and cross sectional area. Since density consatins mass instead of weight it's typical to convert. Changed my mind to keep it simple but didn't remove the SI comment.
A 3000lb car has an engine that produces 500lb/ft of torque at the crank. How fast does it accelerate (assume it produces 500lb/ft of torque at all times, and you can ignore friction and pick a speed) or complete the quarter mile?
Two C4s have identical friction and drag. One car's engine outputs 250lb/ft at 8000rpm at top speed, and the other car's engine outputs 500lb/ft at 4000rpm at top speed. Which car has the higher top speed?
Neither example has enough information to solve. Many unknowns. I hope you know that, maybe you don't. But holy crap could you imagine an engine capable of producing 500ft-lbs at all times! The area under the curve would immense.
Originally Posted by MatthewMiller
Crankshaft torque alone tells you nothing about a car's performance. It's meaningless.
Saying Torque is meaningless makes you sound like you don't understand Force at all.
Originally Posted by MatthewMiller
This is wrong. Power is defined as the rate at which work is done, or force*distance/time. Horsepower is defined as 550lb per ft per second.
I literally gave you a text book definition and you said the same thing. You are obviously not reading to comprehend. The definition is HERE
Originally Posted by MatthewMiller
Torque is defined as rotational force, yes. But you can make power in all kinds of ways that don't involve rotational force. The original concept for the HP unit was (a horse) lifting a 550lb block 1ft vertically in 1 second using a pulley and rope:
[img]file:///C:\Users\Nick\AppData\Local\Temp\msohtml clip1\01\clip_image001.jpg
No torque there!
You are kidding right? Do you know how to construct a free body diagram? The horse pulling the rope creates a torque around each pulley you see and rotates them. Changing the size of the pulleys here and connecting other ropes would also be an example of GEARING. At any rate, replace the horse pulling the rope over a pulley with an Engine or Motor that is rotating the pulley, and you would use the torque applied from the rotating source to move the rope. Either position would be fine. The pulley turns the linear force into a rotational once.... and a rotational force is... Torque. If you want to be technical about saying “No Torque here”, each one of the horses joints that allows him to move and pull the rope experiences torque. Besides we are discussing cars with engines, not a horse and buggy.
Originally Posted by MatthewMiller
But yeah, a piston engine must make torque to make any power. Of course, it also has to make rotational velocity (distance/time). A steam piston engine (external combustion) can make tons of torque on a crank without moving it, and no power is made at all. Same thing happens when you have a small electric motor and turn it on but hold with your fingers so it is stalled (can't turn): torque is produced, but zero power. So again, torque by itself tells us nothing about how fast work gets done. Nothing.
You are right in a way, but you are confusing Torque with Stall torque. Even in your example it is not telling you "nothing". It is telling you the potential you have to move mass. Because a Force is equal to mass times acceleration (F=M*a). To accelerate mass you need a force. Power is not a force.
Originally Posted by MatthewMiller
ORLY? Then tell me how you can put a car on a Dynojet, leave the rpm connection (ignition sensor) off completely, and still get an accurate power number? You'll get power over mph instead of rpm because the dyno doesn't know rpm, and you won't get torque; but the power curve will be as accurate as ever. How can it know power without knowing torque or rpm? It's because it doesn't measure torque and then calculate power from rpm: it directly measures power and then derives torque from that...if you tell it the rpm of the engine. If you don't tell it the rpm, it still gives you power, but it can't calculate torque.
It's not defying physics. There are lots of ways to measure power without knowing force. Power is the change in energy measured over time. The inertial dyno knows the inertia of its drum and it can sense the drum's angular velocity, so it knows the kinetic energy (K) in the drum at any given time: K = 1/2 * (inertia) * (angular velocity)^2. From there, it just measures the change in kinetic energy over time and it then knows the power that was applied to it: P=(K1-K2)/(T1-T2). It gives zero ***** how fast the engine is spinning or how much torque it has at the crank.
The Dynojet relies on a couple laws of physics, a fixed weight and diameter set of rollers (the big round things your tires sit on) and the idea that Force = Mass x Acceleration. Understanding that the drum has a fixed mass (M) the Dynojet software measures acceleration (A) of the drums, multiples the two (MxA) and outputs FORCE. If these look familiar, that's because we are working with Newton's 2nd Law. Knowing the FORCE applied to the drum, the Dynojet software then multiplies Force by Distance (or, how many revolutions the rollers made during the run) to calculate Work, which it then divides by Time to get Horsepower. So... IT RELIES ON MEASURING FORCE. A pure inertia-only dyno can only calculate power by measuring the rate of change in acceleration (that’s why it’s called an inertia dyno). Since the wheels and drums are round, that force is rotational, and a ROTATIONAL FORCE is TORQUE. While your are kind of right, P can always be substituted because it is defined by F (Force).
Originally Posted by MatthewMiller
To be fair, an old-school brake bench dyno does directly measure torque at the crank and then calculate power based on rpm. But...what about a brake dyno that is a chassis dyno: what torque is it measuring, and what happens when you don't give it an rpm signal? Yep, you still get a power reading and no torque reading. It's measuring torque at the tire's contact patch, and again it gives zero ***** how much torque the engine has or at what rpm it's spinning.
With no torque from the engine, you have no torque at the wheels, no torque means no force to get anything moving or accelerating. I suppose if your argument is where it gets measured and not what is measured. Sure, I will concede a chassis dyno doesn’t care what an engine is making because by design it is to measure what gets to the outside of the tire and include any losses the chassis causes before it gets there. Or hub if directly connecting the hub to the dyno which takes wheel/tire diameter and mass out of the equation.
While discussing dynos and RPM triggers... well it needs to know the RPMs to plot the graph. The X axis, to be accurate. Otherwise it would need to know tire diameter and gearing to get engine RPM from Drum RPM.
While we are discussing dynos, I had to wrap my head around how gearing doesn’t affect the dynos reading when we know going from a 2.73 to a 3.73 gear allows a car to accelerate faster through mechanical advantage, but at the sacrifice of top end velocity. This one’s tricky. First, there are potential discrepancies because different gears have different inertia values, generate more friction, and change the amount of tire slip. Higher numerical gears tend to be more inefficient, so as gear ratios increase numerically, power levels tend to slightly drop, particularly on an inertia dyno. When torque is multiplied by steeper gears, tire slippage also tends to increase. However, there’s another, often overlooked, factor in the brew: rpm and torque are inversely related to calculating horsepower, so changing the rear axle ratio or testing in other than a 1:1 transmission gear seemingly shouldn’t change the horsepower numbers. But this doesn’t take into consideration the fact that changing gear ratios changes the engine’s rate of acceleration. For example: We know that on an engine dyno, if you change a sweep test’s acceleration rate from, say, 300 rpm/second to 600 rpm/second, the flywheel power number (bhp) drops due to the faster rate of acceleration. As an engine accelerates at a higher rate, the power required to accelerate the engine increases, and a greater portion is consumed before it gets to the flywheel. Going to numerically higher gear ratios-whether in the trans (testing in a lower gear) or in the rearend-is like increasing the rate of acceleration in a sweep test. Whether this actually changes a given chassis dyno’s reported results depends on how the specific dyno manufacturer does its math. For the most consistent results, always test in the same trans gear (generally 1:1) and re-baseline the vehicle after a rear-axle ratio change.
You chose the wrong Dyno… if you understood what you were trying to argue, you would have picked a fluid power dyno. It does measure Power directly, fluid power… and theoretically the fluid power should equal the vehicle power. Though, in backing out how that happens, it’s about fluid flow through a rotor that requires a Force to rotate it… which is torque applied to the shaft from the moving fluid, but here the force (Torque) is a known constant in the dyno.
Originally Posted by MatthewMiller
Gearing has nothing to do with the power an engine produces at the crankshaft or at the tires. Except for the subtraction of drive line friction and inertia, a car has the same power at the wheels as it has at the crank.
True, but it does set what wheel rpm you get at for a given engine RPM, which we can know what the engine produce at that point. Whether that is making sure you are not redlining before hitting the traps in a 1/4 mile or you are not RPM limited trying to achieve a top speed. If you have read anything that makes you think I am saying a gear increases Engine HP, you are comprehending it wrong. Let me just be straight right here. It doesn’t add Power or Torque output from the engine, but it does increase Torque from Mechanical Advantage. 1st gear produces more torque to the real wheels than 2nd, 2nd more than 3rd and so on.
Originally Posted by MatthewMiller
Torque at the what? Wheels or crank? Because one is very different than the other.
What gets to the wheels is the point of everything. I am not sure what is so confusing to you. Without the engine, you have no torque to the rear wheels. More engine torque - more wheel torque. More gearing, more wheel torque.
Originally Posted by MatthewMiller
And actually, it would be more accurate to say the force you need to calculate is the linear force at the contact patch of the drive tires, which is called "tractive effort." But here's the drop-the-mike moment: if you know the tractive effort (or wheel torque) being produced at the contact patches, and you know the vehicle speed, then you know the power being applied to the wheels. You don't have to know gearing and torque and tire diameter and rpm of the engine. You already know power, and again you don't need to give any ***** about how fast the engine is spinning or how much crank torque it is making at that speed.
You would drop the mic on this? While you are correct if you know resistance and the force needed to overcome it then you have solved for the force needed. But that doesn’t tell you what you need to know to build a machine to do it. To build the car to achieve this given mph as we are discussing about land speed cars, once you know the force, you will have to produce that at the wheels.
So, now that we know the force, you will have to take the tire diameter, differential gear, and transmission gear all into account to make sure your engine produces said amount of force at the RPM you will be running to get the force needed tot he wheels. Are you really trying to say if you are trying to achieve 200mph with a 400hp engine, it doesn’t matter if it is spinning at peak hp or 3000 RPMs below it? That is nonsense, engine Torque and power output are not equal at all RPMs and you will need to know what power is being outputted so you know what can get to the wheels. The produced force must be equal to or greater than required force. You would be solving two simultaneous equations. 1 Equation to define what is needed, then 1 equation that defines the result of what is available at the tires based off engine production.
Originally Posted by MatthewMiller
All of which means that the tractive effort to overcome all the resistance at any given speed is created by power, not crank torque. The link I provided on tractive even states that explicitly: "Tractive effort available: maximum tractive effort of a vehicle as limited by the available power." That's why literally every chart you'll ever see on attainable road speeds is given in terms of required power, not required torque. Because power is the measure of how fast work gets done, such as how many hours it takes to travel a certain number of miles against a known amount of resistance.
That’s fine, without torque, you have no power. Just to reiterate Horsepower is defined as 550lb per ft per second. In a car, that 550ft-lbs comes from crank torque, most commonly transferred through a transmission, connecting to a drive shaft that connects to a differential, that then turns the dears that rotates the axles. These axles have a hub at the end that the wheel/tire mounts to, and is TWISTED by the hub. All this rotation and twisting force is TORQUE!
Just because you have the Power available, also doesn't mean you can achieve speed without other factors known. An engine making 400hp at 5000RPM will not go 200mph if the gearing puts 5000RPM at velocity of 140mph. The Same 400hp at 5000 RPM will also not make 200mph if the gearing is too long and 200mph is occurring at 2000Rpms and the engine is not in its "power band" and making the required force (Torque) at that speed (Power)to overcome resistance. So gearing means a lot. A lot of factory cars, the final over drive gear will not achieve higher velocity than a lower gear. The engine falls out of its "power band" and changing to a lower rear gear can change the multiplication factor and turn high gear into a pulling gear.
Originally Posted by MatthewMiller
You're confusing yourself. If you know all the gearing and the road speed and tire diameter (or conversely, the engine's rpm), then you know the power being produced. If you don't know power, you can't know the road speed attainable. Period.
I am not confusing myself. You are confused. Power is calculated from a Force and. When doing a Free Body Diagram to define the motion and forces of any component, you lay out the forces applied to it. Even if you know it’s a POWER source, your concern will be what Force it applies. Conversely, if you know the Power needed, you will need to find the FORCE to generate that power. PERIOD!
Originally Posted by MatthewMiller
The only tweaking you do for top speed is to set up the total gearing so the engine is making peak power at the terminal velocity. Only an idiot would tune for peak torque.
I never said anything about tuning for peak torque. You tune for peak Horsepower. Basically, Peak Horsepower occurs when then engine can no longer effectively produce torque as engine speed increases. Horsepower defines the performance of an engine because gearing allows you to use mechanical advantage to multiply torque. 1st gear accelerates faster due to a larger force. But you run out of ability to produce torque at speed and must reduce the gearing, which thus reduces the output torque but increases speed. The more RPM you have, the more multiplication you can do. The reason why a TPI doesn’t run better ET’s with a 4.56 ring gear over a 3.42. It doesn’t have enough RPM band, too many shifts, and runs out of breath. This wouldn’t happen in a S2000.
I guess I need to get involved with Land Speed cars again. It would be nice to just build something and it be right out of the gate and never have to tweak it for top speed. I need to tell the guys in the pits to stop adjusting spoiler angles, air duct openings, suspension geometry, and engine tuning based on acquired data from runs.
That’s ridiculous. They make tweaks to their set up as much as drag and circle track racers do. Just with different goals in mind. Including engine adjustments.
Originally Posted by MatthewMiller
That's why truck commercials talk about torque so much: to sell trucks to ignorant customers even though their truck doesn't make as much power as the competition. Because the only thing that determines how fast you can accelerate a given load is power. Same with locomotives and ships: they're rated in terms of power, not torque. The only reason anyone needs to know peak torque of an engine is to spec a transmission and clutch that won't break under the load.
Well, this is just a different animal all together. Unless you are talking about someone who wants to achieve the fastest truck. Trucks are made to pull and haul. The speed at which it is done is rarely as important as the ability to just get it done. This is why trucks have “granny” gear transmissions, Low Gear transfer cases, and 18 wheelers have 16 speed transmissions. To multiply that torque and provide enough force to get extremely high mass loads moving. Diesels are notorious for having high torque low power ratings due to not spinning at 5000RPMs and above. I sure see a lot of fast Diesels out there in the last 15 years though that are getting moving with Torque ratings higher than HP ratings.
I guess I should just ask you this….
Car A: C4 with a gas engine that makes 400Hp @ 6000 RPMs weight is 3000lbs with stock body.
Car B: C4 with a diesel engine that makes 400Hp @ 4000 RPMs weight is 3000lbs with stock body.
So, you say only power matters. They have the same power, nothing else matters, they both achieve the same top speed?
I say yes, but the gearing has to be different. Car B must be geared taller to achieve the same speed at the same RPM. Since Car A has more RPM to utilize TORQUE multiplication and it will accelerate faster to speed.
Car A is producing 350ft-lbs of crank torque, Car B is producing 525ft-lbs of crank torque. But based on gearing, for simplicity sake, say the result is they both make the same torque at the wheels. So either engine could be used in the chassis. Just have to have the same output torque at the wheels.
Now go back to your dyno without tracking engine RPM. I have never seen this done, usually they plug into the OBD port to pull engine data, but I would hypothesize these two car’s power curves at the tires would look very similar against wheel RPM, but would look vastly different against engine RPM. When tuning the ECM/PCM to maximize output, you must reference Engine RPM. You wouldn't be doing timing and fuel curves based off wheel speed or drum RPM. The files in the engine management system are RPM based.
When I say crank torque means everything, I mean just that, but I think you are not absorbing my main point in that statement - the fact that you can’t have Power without Torque. You could use any engine to produce the torque required and adjust your gearing to get the resulting force at speed you need (Power). The difference in how it is made will show up in acceleration due to the ability to multiply the torque, but Torque is the basis of any rotational power output by definition.
So, in a performance car you want the RPM to take advantage of torque multiplication to accelerate. In a truck, you want torque to get moving down low and don’t want to have to rev up to get going, you use gearing to multiply torque for "Brute" force. In a quest for ultimate top speed, some just use a push car to overcome initial inertia forces and use gearing to accelerate at speed.
I think you're confusing power required to maintain a certain speed with force required (as I did when I typed the first erroneous calculation). Aero drag (and all restrictive forces like driveline friction, tire deformation, etc.) is a force. So we all agree that the force increases with the square of speed. But power is not the same as force. In order to exert the same force at the contact patch at 200mph as is produced at 150mph, you need 33% more power. This must be the case, because the basic equation for power is force*speed. Get way from thinking about power at the flywheel, and think about the power at the wheels: if the power level remains the same, the force and speed are inversely related. If the speed increases 33%, then the tractive force at the contact patch must fall by 33%.
So again, at 200mph you need 133% of the power you needed at 150mph, just to maintain the same tractive force as it had at 150mph. And you are also encountering an increased resistive force of 133%^2. So the total increase in force required is 133%^2, and therefore the total increase in power required to get that force is 133%^3.
Originally Posted by 69427
Again, I disagree. Below is where I disagree with you (I'm short on time, but I'll look at your graphs later today).We all agree that the areo drag goes up by the square of the speed. Let's just go with two simple speed numbers at the moment to simplify things (100mph and 200mph). At twice the velocity/speed, the aero drag will be four times as large.
Now, Work = Force times Distance (W=F x D). Horsepower, which is the rate that work gets done, is then Hp=W/t. (Work per second).
At twice the speed (200mph), the F (force to overcome the drag) in the work equation is 4x what it is at 100mph.
Also, at 200mph the distance traversed each second (D in the work equation) is 2x what it is at 100mph.
We then end up with an equation like this: Hp (@200mph) = (4xF@100mph)x(2xD@100mph)/(t=1sec) = 8x(FxD@100mph) = 8xW each second = 8 x Hp that was required at 100mph.
I'm late for an appointment. Will check back later. I welcome any correction to any mistakes made in this rushed post.
I stand corrected and I have corrected my post with the graphs and the Excel file.
I got fooled by the statement that you need cube of hp for keeping up with square of hp drag, not noticing that drag had changed from hp to force.
With 275 whp the top speed calculated to 183mph after 110 seconds with gearing to match max hp at top speed.
To get 200 mph you would need 350 whp with optimal gearing.
That is exactly right, and it's exactly my point. Thank you for making it for me in such a short sentence! The fact is, though, that horsepower is the parameter of engine performance that tells you everything you need to know about how fast the car can go. Crank torque literally doesn't mean everything. You just said it yourself: just knowing torque doesn't provide enough information. Crank torque by itself tells us nothing.
Many unknowns. I hope you know that, maybe you don't.
Nope, there's just two unknowns: the distance the crankshaft covers and the time it takes to cover it. Remember, power is just force*distance/time. But with crank torque, you only know the force. Or you could say one thing is unknown: rpm (which is just distance/time). The point is, if you know the engine's power, you know exactly how much tractive force the car can put to the ground at any speed, and therefore you know how fast it can go against its total resistive force, or how fast it can accelerate its total mass.
But holy crap could you imagine an engine capable of producing 500ft-lbs at all times! The area under the curve would immense.
Ummm, yeah...it would be infinite. But back in the real world...
Saying Torque is meaningless makes you sound like you don't understand Force at all.
I didn't say all torque is meaningless for all things. I said crankshaft torque is meaningless in predicting vehicle performance. And it is, as you agreed above.
I literally gave you a text book definition and you said the same thing. You are obviously not reading to comprehend. The definition is HERE
You're absolutely right. I overlooked that you wrote ft-lbs instead of just lbs. My mistake there.
You are kidding right? Do you know how to construct a free body diagram? The horse pulling the rope creates a torque around each pulley you see and rotates them. Changing the size of the pulleys here and connecting other ropes would also be an example of GEARING.
Are you kidding? In the diagram I posted, the pulleys only serve to change the direction of the linear force that moves the block. The pulleys aren't doing any work (it obviously assumes the pulleys have frictionless bearings). The block isn't moving in rotation. It's moving straight up, in response to the linear force that is pulling it that way. Changing the size of the pulleys here would change nothing about the magnitude or direction of the force moving the block. But when you start talking about connecting other ropes and examples of gearing, I see how you tend to get yourself caught up in extraneous aspects that don't actually matter to the example at hand.
You are right in a way, but you are confusing Torque with Stall torque. Even in your example it is not telling you "nothing". It is telling you the potential you have to move mass. Because a Force is equal to mass times acceleration (F=M*a). To accelerate mass you need a force.
It is telling you nothing about how fast the motor can do work, which again was the question at hand. Knowing the force(torque) the engine can apply when stalled does sort of tell us about its potential to move mass, but it tells us nothing about how fast it can move it (again, please see the original question we were trying to answer in this thread). I said "sort of" in the last sentence, because the stall torque of the engine itself doesn't really tell us the potential to just move mass, because we can multiply torque. I can gear any engine of any torque output down to move any mass - it just will move it awfully slowly if I have to gear it down by very much to multiply its force. Which again is the point: if I only know the torque of an engine, I have no idea the speed at which it can apply that torque, and therefore I have no idea how fast it can move a car.
So do you want to stick with this? Because that's very, very wrong. Power isn't just a force, but it most definitely includes force as one of the three key pieces of info you need to find out how fast an engine can move a car. If you think power is not a force, it's no wonder you would think that "torque is everything," and that you wouldn't understand why power is everything.
Knowing the FORCE applied to the drum, the Dynojet software then multiplies Force by Distance (or, how many revolutions the rollers made during the run) to calculate Work, which it then divides by Time to get Horsepower. So... IT RELIES ON MEASURING FORCE. A pure inertia-only dyno can only calculate power by measuring the rate of change in acceleration (that’s why it’s called an inertia dyno). Since the wheels and drums are round, that force is rotational, and a ROTATIONAL FORCE is TORQUE.
Well, it's actually not measuring torque - it's measuring tractive force at the contact patch, which is a linear tangent to the drum's circumference. It doesn't know how tall the tire is, so it has no idea how much torque the axle/tire is exerting; nor does it know the rpm at which the axle is spinning. Again, I never said no force matters. I spent a lot of time talking about tractive force, and even linked a definition for it. Force at the contact patch is the Force in F=MA for a car! But the measure that determines the tractive force of the drive tires at any given speed is power. You don't have to know the gearing, the tire diameter, or the actual torque output at the crank to know how much Force you have available to Accelerate your Mass. You just need to know the power and the speed at which the car is traveling. Remember, the power at the crank is also the power at the wheels, so you if you know the speed of the car then you also know the tractive force available to accelerate it or to overcome resistive force.
What I said was that torque at the crankshaft doesn't tell us anything about a car's performance. And one way you know that's true is that an chassis dyno can measure power without knowing the engine's rpm, and therefore without having any idea what the engine's torque actually is. You can put an engine that makes 100hp and 200lb/ft at 2626rpm, another that makes 100hp and 100lb/ft at 5252rpm, and another that makes 100hp and 50lb/ft at 10504rpm. They will all get exactly the same power readout, and the dyno doesn't need the tach sensor connected to determine it. And they all power a vehicle to the exact same terminal velocity. The crank torque doesn't make a bit of difference to the dyno or to the performance of the car.
With no torque from the engine, you have no torque at the wheels, no torque means no force to get anything moving or accelerating.
Seriously, you're reading lots of stuff into my post that I never wrote. I never said an engine doesn't produce torque or that it doesn't need to. Of course it produces force, starting as cylinder pressure which is then translation to rotational force at the crankshaft. You can't produce power without force! When I say power is all you need to know, that includes that force! Because (unlike what you wrote above), power most definitely does include force. My saying torque tells us nothing about vehicle performance isn't me saying an engine doesn't make torque. Come on!
While discussing dynos and RPM triggers... well it needs to know the RPMs to plot the graph. The X axis, to be accurate. Otherwise it would need to know tire diameter and gearing to get engine RPM from Drum RPM.
Wrong. Have you been around chassis dynos? If you don't connect the tach sensor, you get your X axis plotted on mph instead of rpm, and you only get a power curve and no torque curve. The dyno doesn't care about engine rpm at all. The tach sensor is an add-on accessory to give torque fanbois something to talk about (okay, I'm being facetious, but only a little bit). If you have the tach sensor connected, the dyno calculates the engine torque from the change in kinetic energy of the drum and the rpm of the engine. It doesn't measure crank torque at all, and it doesn't calculate power by multiplying crank torque*rpm/5252 as you claimed.
While we are discussing dynos, I had to wrap my head around how gearing doesn’t affect the dynos reading when we know going from a 2.73 to a 3.73 gear allows a car to accelerate faster through mechanical advantage, but at the sacrifice of top end velocity. This one’s tricky.
It's not tricky at all. It's because the engine is putting out the same power regardless of whatever gearing you install after it. You're tricking yourself because you're comparing acceleration to engine rpm, which isn't the correct measure of performance. The correct measure is acceleration vs mph. That's really, really important to understand here. The car with 3.73s in 4th gear can pull 37% more Gs of acceleration at a given rpm than it can with 2.73s at the same rpm, but it's also going 37% slower when it does it. It doesn't really change how fast it accelerates at the same speed though. The only thing that determines how fast a car can accelerate at any given speed is how much power the engine is sending to the contact patches at that speed. And gearing doesn't change that. People often get confused when the try to imagine which gear they'd be in at what speed, etc. It's much easier to understand this if you imagine a car with a CVT.
First, there are potential discrepancies because different gears have different inertia values, generate more friction, and change the amount of tire slip. Higher numerical gears tend to be more inefficient, so as gear ratios increase numerically, power levels tend to slightly drop, particularly on an inertia dyno. When torque is multiplied by steeper gears, tire slippage also tends to increase. However, there’s another, often overlooked, factor in the brew: rpm and torque are inversely related to calculating horsepower, so changing the rear axle ratio or testing in other than a 1:1 transmission gear seemingly shouldn’t change the horsepower numbers. But this doesn’t take into consideration the fact that changing gear ratios changes the engine’s rate of acceleration. For example: We know that on an engine dyno, if you change a sweep test’s acceleration rate from, say, 300 rpm/second to 600 rpm/second, the flywheel power number (bhp) drops due to the faster rate of acceleration. As an engine accelerates at a higher rate, the power required to accelerate the engine increases, and a greater portion is consumed before it gets to the flywheel. Going to numerically higher gear ratios-whether in the trans (testing in a lower gear) or in the rearend-is like increasing the rate of acceleration in a sweep test. Whether this actually changes a given chassis dyno’s reported results depends on how the specific dyno manufacturer does its math. For the most consistent results, always test in the same trans gear (generally 1:1) and re-baseline the vehicle after a rear-axle ratio change.
This is all a good example of how you're confusing yourself by making this more complicated than it is.
You chose the wrong Dyno… if you understood what you were trying to argue, you would have picked a fluid power dyno. It does measure Power directly, fluid power… and theoretically the fluid power should equal the vehicle power. Though, in backing out how that happens, it’s about fluid flow through a rotor that requires a Force to rotate it… which is torque applied to the shaft from the moving fluid, but here the force (Torque) is a known constant in the dyno.
Again, my point was that any chassis dyno of any design measures power at the tires, not the crank. And any of those can tell us the engine's power without knowing its rpm, and therefore without knowing its torque.
True, but it does set what wheel rpm you get at for a given engine RPM, which we can know what the engine produce at that point. Whether that is making sure you are not redlining before hitting the traps in a 1/4 mile or you are not RPM limited trying to achieve a top speed. If you have read anything that makes you think I am saying a gear increases Engine HP, you are comprehending it wrong. Let me just be straight right here. It doesn’t add Power or Torque output from the engine, but it does increase Torque from Mechanical Advantage. 1st gear produces more torque to the real wheels than 2nd, 2nd more than 3rd and so on.
All of which tells us nothing about how much power is required to push a given car with given resistive force to a certain top speed...which again was the original question being asked. You do understand that the only reason we have multi-gear or CVT transmissions in racing cars at all is to maximize the power the engine can apply to the tires at any given speed, right? If you want your car to accelerate the hardest at any particular speed, you gear it so that the engine is at peak power rpm at that speed. You don't gear it for peak torque. Not ever. Same for trying to reach a desired terminal velocity: you will gear the car so the engine is making peak power at the desired speed, not for peak torque.
What gets to the wheels is the point of everything.
I'm glad you agree with me on that. If only I could convince you that it's engine power that determines the torque at the wheels (or more correctly, the tractive force at the contact patch) then we'd be in total agreement.
I am not sure what is so confusing to you. Without the engine, you have no torque to the rear wheels.
This is true.
More engine torque - more wheel torque.
Oh yeah? Answer this: A car has an engine that produces 500lb/ft of torque at the crank; so how much torque does it have at the wheels, or how much tractive effort at the contact patch, at 100mph? You can't answer that, because without knowing the engine's power, you can't know the gearing required. Since you don't know if the engine is creating 500lb/ft at 1000rpm or 8000rpm, you can't know the torque at the wheels.
OTOH, if I know the power the engine is supplying to the contact patches, then I know exactly how much tractive force is being generated at any road speed, irrespective of engine speed and gearing. I won't know the torque at the wheels unless I also know the tire diameter, but since it's the tractive effort that moves the car (which again is a linear force and not a torque), then it doesn't matter.
Let's modify that: More gearing, more wheel torque.
But less road speed. At the same road speed, the wheel torque doesn't change (again, think CVT here). Which again is why power is what matters here: it's the only thing that determines how much tractive effort the tires generate at a given road speed.
You would drop the mic on this? While you are correct if you know resistance and the force needed to overcome it then you have solved for the force needed. But that doesn’t tell you what you need to know to build a machine to do it. To build the car to achieve this given mph as we are discussing about land speed cars, once you know the force, you will have to produce that at the wheels.
The question wasn't how to build a car to achieve a certain amount of power or torque. The question was how much power it takes to move a C4 at 200mph. Power is the only thing that matters here. Earlier, we calculated that it should require 565hp to do that. All I need to know is that any engine that can produce 565hp will push the C4 to 200mph, as long as I make sure it's actually outputting 565hp at 200mph.
Just because you have the Power available, also doesn't mean you can achieve speed without other factors known. An engine making 400hp at 5000RPM will not go 200mph if the gearing puts 5000RPM at velocity of 140mph.
Yes, of course, which is why I wrote: "Why in the world would anyone trying to set a top speed goal not gear the car to make peak power at the car's projected top speed?" If you're trying to go 200mph and you gear your car to make peak power at 140mph, then you're an idiot. There's no other way to put that. Again, the question was how much power does it take to go 200mph in a C4. The answer to that doesn't depend on rpm or crank torque. It takes however much power it takes, no matter how you get the power.
I never said anything about tuning for peak torque. You tune for peak Horsepower.
Which is why power is the only thing that matters in determining how fast the car goes! Why is that? Because power is what determines the tractive force at the contact patches of the driven tires. The more power you have, the more tractive force you have at a given speed. The crankshaft torque you have...doesn't tell us anything about the tractive force at any speed.
Basically, Peak Horsepower occurs when then engine can no longer effectively produce torque as engine speed increases.
Yeah...that's not true. Peak horsepower always occurs at an rpm higher than peak torque - it always occurs after torque output is already dropping.
Horsepower defines the performance of an engine because gearing allows you to use mechanical advantage to multiply torque.
No, horsepower defines the performance of the engine in a car because it is what determines the tractive force available to accelerate the car at a given speed. You gear the car to allow it to produce the most power possible, not to multiply torque. If multiplying torque were all-important, then you'd gear the car infinitely low (numerically high) to have the most torque possible. The car could pull an infinitely high load with infinitely high acceleration, but it would only achieve an infinitely low speed. There's an obvious reason we don't do that: because speed is actually also important to vehicle performance! Ergo, power is what's important and not crankshaft torque.
I guess I need to get involved with Land Speed cars again. It would be nice to just build something and it be right out of the gate and never have to tweak it for top speed. I need to tell the guys in the pits to stop adjusting spoiler angles, air duct openings, suspension geometry, and engine tuning based on acquired data from runs.
That’s ridiculous. They make tweaks to their set up as much as drag and circle track racers do. Just with different goals in mind. Including engine adjustments.
Dude, you were talking about tweaking the drivetrain to get torque to the wheels: "Once you have the torque needed then you look at how to get that torque there. Rear gear, tire diameter, RPM range, and transmission gearing all come into play. You build a combination that is delivering enough torque to the rear wheels." Obviously you work on aero trim, suspension, etc. But your goal in setting up the engine is to deliver enough power to the rear wheels so they can create enough tractive force at the road. You are trying to maximize power from the engine, not its torque.
Well, this is just a different animal all together. Unless you are talking about someone who wants to achieve the fastest truck. Trucks are made to pull and haul. The speed at which it is done is rarely as important as the ability to just get it done. This is why trucks have “granny” gear transmissions, Low Gear transfer cases, and 18 wheelers have 16 speed transmissions. To multiply that torque and provide enough force to get extremely high mass loads moving. Diesels are notorious for having high torque low power ratings due to not spinning at 5000RPMs and above. I sure see a lot of fast Diesels out there in the last 15 years though that are getting moving with Torque ratings higher than HP ratings.
It's not a different animal at all. The truck that can accelerate the heaviest load (or accelerate a heavy load the fastest) is the truck that can apply the most power to the wheels (obviously assuming the rest of the truck remains constant, like chassis and suspension) over the total acceleration run. It's crank torque has nothing to do with that. There are good reasons that low-rpm diesels are used for heavy hauling, but the ability to accelerate the heaviest load isn't one of them. Trucks have granny gears and lots of ratios so that they keep their engines running around peak power rpm for a lot more of their speed ranges. Even more importantly, it's so they don't destroy clutches or torque converters trying to get heavy loads moving from a stop.
I guess I should just ask you this….
Car A: C4 with a gas engine that makes 400Hp @ 6000 RPMs weight is 3000lbs with stock body.
Car B: C4 with a diesel engine that makes 400Hp @ 4000 RPMs weight is 3000lbs with stock body.
So, you say only power matters. They have the same power, nothing else matters, they both achieve the same top speed?
I say yes, but the gearing has to be different. Car B must be geared taller to achieve the same speed at the same RPM.
This much is correct. Of course they will both achieve the same top speed as long as they are both geared to output peak power at that speed.
Since Car A has more RPM to utilize TORQUE multiplication and it will accelerate faster to speed.
This is incorrect. If they both are geared to achieve the same top speed at peak power, then they will both accelerate to that top speed in the same amount of time and distance. Car A has the same horsepower and 33% shorter gearing, but its engine also has 33% less torque at the crank. So with 33% shorter gearing multiplying 33% less crank torque, you get...wait for it...the exact same torque at the wheels as Car B, at all road speeds.
When I say crank torque means everything, I mean just that, but I think you are not absorbing my main point in that statement - the fact that you can’t have Power without Torque. You could use any engine to produce the torque required and adjust your gearing to get the resulting force at speed you need (Power).
That's just not true. You simply aren't understanding how this works. Your wrong statement above about Car A accelerating to the same top speed faster is proof that you don't understand it. Like said before, the force at the tires is determined by the power being produced, not the crank torque. An engine produces the power required, not just the torque. In other words, it's not just the torque at the crank that matters, but also the rpm at which it produces that torque that also matters. In your example above Car A has only 66% of the torque that Car B has, but they both produce exactly the same tractive force at a given road speed.
OTOH, say you have two engines: Engine A makes 400lb/ft of torque at 3000rpm, and Engine B makes 400lb/ft of torque at 6000rpm. You can't just "adjust your gearing to get the resulting force at speed you need (Power)." If, for example Engine A produces 1000lbs of tractive force at 100mph, then Engine B will produce 2000lbs of tractive force at 100mph. There is no way to gear yourself out of the fact that Engine A only has half the power of Engine B. You can't create a gearset that will allow Engine A to produce 2000lbs of tractive force at 100mph like Engine B does.
The difference in how it is made will show up in acceleration due to the ability to multiply the torque,
Nope. The power is what will determine how fast a vehicle accelerates at a given speed. Period. How it makes the power or what its crankshaft torque peak is won't matter.
but Torque is the basis of any rotational power output by definition.
No, torque is one of three equally important components of power: force, distance, and time. If either of the first two are zero, you have zero power. If the third is zero, you have undefined power. They all matter just as much as one another.
So, in a performance car you want the RPM to take advantage of torque multiplication to accelerate. In a truck, you want torque to get moving down low and don’t want to have to rev up to get going, you use gearing to multiply torque for "Brute" force. In a quest for ultimate top speed, some just use a push car to overcome initial inertia forces and use gearing to accelerate at speed.
Nope, you're buying into all those dumb truck commercials. If you have a 20,000lb trailer, the truck that will accelerate it the fastest will be the one that outputs the most power to the tires. If you have a 400hp diesel engine that makes 800lb/ft in your truck, and I have a 500hp engine that only makes 250lb/ft in my otherwise-identical truck (and obviously at a way higher rpm), then I will accelerate that trailer faster than you. Crankshaft torque peaks have nothing to do with how hard you accelerate at a low (or any other) speed, and the mass of the load doesn't change that either. Again, only power determines that.
I'm going to go back and requote my original statement, which you said "couldn't be more backwards and wrong":
Originally Posted by MatthewMiller
...the question (16 years ago) was "How much power at the flywheel would it take to achieve 200mph?" That question can be answered without respect to gearing/rpm/blahblahblah: it takes however much power is required to overcome the total friction of the car.
This is neither backwards nor wrong. It doesn't matter how the car is producing the required power: rpm and torque (and required gearing) don't matter. If it takes 565hp to push a C4 to 200mph, then the only thing that matters is that the engine produces 565hp at 200mph. The car doesn't care if I'm making 565hp at 1000rpm with 2967lb/ft of torque, or if I'm making 565hp at 20,000rpm with 148lb/ft of torque. It's going to reach 200mph either way.
Last edited by MatthewMiller; 01-18-2019 at 11:29 PM.
This is a good example of what a dyno needs.
1) Something to measure the rpm of the roller.
2) Something to measure breaking torque. In this case a load cell measuring force on a lever arm with a known length.
If you measure 100 pound force and the lever arm is 2 feet then the torque is 200 lb/ft.
Knowing rpm you can calculate hp. Nothing else matters, not geraing of the car or diameter of the rollers.
Plotting hp as a function of roller rpm does not make much sense but you still get a hp curve with the correct shape.
The diameter of the roller is known so it makes more sense to calculate mph from roller rpm and use that as x-axis in the graph.
If you know the engine rpm you can calculate engine torque from hp and plot both as a function of engine rpm.
1) Use an extra sensor to directly measure engine rpm,
2) Calculate engine rpm from wheel diameter, rear end gear ratio and gear ratio of used gear in the gearbox. A tourqe converter with unknown slip will result in a lower calculated rpm and higher calculated torque.
01-17-2019, 03:19 AM
