Marc '99T/A

iTrader: (22)

how do you report it?

GOaT Cheese

Just to **** you off

MrDude_1

iTrader: (4)

the button that looks like this: that is below everyones name in a thread (except your own name.. cant report yourself... or some admins..)

Marc '99T/A

iTrader: (22)

Ahhhhh...... never knew that. Thanks.

RomulusSmallBlock

ok i am a physics and engineering major so i will kill this. to do this i will use my bible of physics my fish yr physic book. power is the time rate of doing work. like this power = work / time or power = force * speed. personally i think the second equation is more important to us guys for use on cars. from this your see what your theoritical top speed in your car is and in for each gear. this is what i used to calculate my top speeds for each gear. if you know basic physics, it simple to do. i will walk you through it for top speed. for me it was very simple because my four banger in my s10 makes 120 hp (which explains why i want a V8 in it). i knew that in 5th gear my s10 couldn't hit redline at all. so i toped it out and saw at what speed and RPM. now for the physic part. in a straight and level road with no wind there are only to forces affecting you in your car: friction form the tirers and wind drag both forces apposing your forward motion. now for that one equation that is useful power = force*speed; because you are top out, the forces opposing your motion are in equalibrium with the force being made by your engine. like this force from engine = friction+drag. now you can do this speed * force from engine= speed*(friction+drag) or simplified HP from engine = HP of the friction +drag. you can guess pretty good what the friction of the tires are by looking up the average friction coefficent of the tires and multiply it by the weight of the car with you in it. so you can do this hp from engine - hp friction= hp drag. now you know your drag pretty well. ok now for your gears in the tranny, rear end and tire size. you have a physical limit to just how fast your car can go even if you had a engine that would produce unlimited hp i will call it potiential speed or P speed. P speed= (RPM *120 *PI* tirer height)/(rearend ratio*tranny gear*63360). this equation is a straight line on a graph and useful in a min. now you have what you need to calculate the hp of the opposing forces from any speed you are going. HP of opposing forces at any speed= (friction+(.5*drag constant)*P speed^2)*P speed. the graph of the hp of opposing forces is a curve that grows quickly. now for the beauty part. if your car is stock or you have dyno'd it you know your engine's hp curve. what you do is graph both your engine's HP curve and the HP of opposing forces curve together. they will cross each other in your higher gears for a T56 more than likely it will be 5th gear. they will meet at a certain RPM. now with this rpm punch it in to the P speed equation and that will give you your theoritical top speed. i did this for my s10, 4th gear for me is my fastest but because my truck has a cut off at 100 mph i had to test it for 5th gear. in 5th gear my theoritical top speed is 96 mph which mean i can test that. so i waited for a not to windy day and i tested it out. and of course at about 96 my truck topped out. if you are interested or bord and want to do this, it is best to put everything in SI units and convert them to US standard units because all of the coefficents are in SI units. all of the coefficents you can look up in a book or off the web i was just lucky to have all of them in my physics and engineering books.

PS i know there are typos and miss spellings probably everywhere

GOaT Cheese

I think you are missing my point. In terms of physics, It takes alot more than 380 lbs. of Rear wheel torque to accelerate a car beyond say more than 5 or 6 mph.,but that is what the Chassis dyno says the car is making. Most performance cars are making thousands of pounds of torque at the wheels! Why does the dyno not state this fact? My contention is that the dyno algorythms reverse engineer the work force measured at the wheels to a number that is more or less the output at the crankshaft. Furthermore, if that is the case, how can the horse power numbers measured at the rear wheels be accurate if the torque output is completely botched? I guess in short what I am asking is how can you have absolute faith in the authenticity of one measurement figure, when at the exact same time, at the exact same location (the rear wheels) another measurement figure is completely off? Not by a few lb. ft., but off by thousands of lb. ft.?

LS1IMPULSE

iTrader: (42)

I read this along time ago and the picture and all make it kinda easier to understand,

http://auto.howstuffworks.com/horsepower.htm

Quick1998Z28

iTrader: (1)

this is contradictory and wrong. Horsepower is a function of engine speed and torque, essentially it's output (torque) through distance (determined by engine speed), it isn't bull, it's dynamic and therefore only the integral, or area under the curve, is what matters, not peak numbers. Torque is dynamic too, but it isn't a function of engine speed, its source is the energy released from combustion and the stroke of the rod/piston it's moving as well as the other various factors like displacement, compression, energy per mole/fuel etc. Before you think it's contradictory realize that throttle, not engine speed is what determines the amount of air and consequently, fuel that is being delivered to the engine.

Quick1998Z28

iTrader: (1)

It's because there is no load on a dyno. Basically what happens is your wheels are moving another wheel, well a cylinder in this case. It isn't measuring the multiplied output at the axle, it is measuring how fast the tires turn the cylinder on the dyno. The rotation of the cylinder is where the numbers come from, the algorhythms probably take into account efficiency and friction to a certain extent. This here is also why dynos are useless for estimating E/Ts and real world performance. Tracks aren't freely spinning metal cylinders, they are sticky, grippy strips of concrete. There is a lot more friction on a track than on a dyno, and that friction is only overcome by one thing, torque. Because of the increased resistance from the concrete through friction (as well as air) the engine speeds don't increase as fast as they do on a dyno, making HP harder to attain. With tiny import motors that spin like crazy but lack torque to overcome friction and drag, you get the all to common dyno queen putting down gobs of HP, but as soon as it encounters a load, it's lack of torque becomes noticable and they really stink it up at the track, getting bested by torque monsters with much lower HP peaks. However, in the real world, those tiny import motors are NOT putting down the HP numbers they make at the dyno. Neither are the larger engines, but the error is far greater in the import engine. Simple as that.

MrDude_1

iTrader: (4)

because... you're measuring the engine RPM vs how fast it accelerates a known mass (the large dyno roller).

sure you make more torque with thoes 4.60 gears... but your engine RPMs also climb faster.... so while the sample time is smaller, the data is the same as when you have 2.23s...

if you were to say, hook it up to your wheel speed sensor instead of your tach, then it would measure that rear wheel torque. or you could calculate it based on time to acclerate the drum to a speed.

its all just math.

BJM

iTrader: (1)

For anyone who is interested, about a year ago I opened up a cheap Gtech and recorded acceleration data myself. I posted about it here. Since we are talking about inertia, the lower the gear you are in, the more torque you lose to accelerating the drivetrain, less is available for accelerating the car. I measured it and show it in the various figures in that posting.

The math is nearly the exact same as used on a Dynojet. It uses just the one sensor, an angular encoder, to determine how fast the drum speeds up and then backs out an effective thrust applied by the tires to the drum. An accelerometer does the same thing but in the moving car.

J-Rod

I moved the thread, but here is one thing. You don't dyno in low gear. Yes, there is a multiplication effect from using a lower gear ratio. But, if you'll look, you'll see how inertia dyno's derive HP by calulating the acceleration of a rather large and heavy drum. It measures rates of rotational speed change.

BTW, if memory serves me it derives torque from HP and RPM. If you pull the tach lead off on the dyno, you read no TQ, as it can't back it out of the equation. So, on a interia dyno, Tq isn't measured, its calculated.

Hp = Absorber Torque (in foot pounds) x RPM / 5252

In reality It's not measuring SPEED of the rotation ...it's measuring the force applied to the drum. What FORCE is exerted to roll that drum.

Horsepower cannot BE measured.......it's derived BY measuring torque and converting. Horsepower is in a sense, is unimportant. Horsepower doesn't move a vehicle. It is really a measurement of how capable the engine is of maintaining the force that really is moving the vehicle


F=ma
Where F is Force, m is mass, and a is acceleration
For our purposes, we'll rearrange it to be:

a = F / m
That is, the more force or less mass, the faster an object (such as a bike) will accelerate.

The force pushing a bike forward is the wheel spinning against the ground. The wheel has a sort of "rotational force," known as Torque. Following this torque up the line through the spinning axle, gears, the clutch, and the flywheel. Torque is related to force through the equation:

Q = F * d
Where Q is Torque, F is Force, and d is distance.

Rearranged for Force (since that's what moves the bike):

F = Q / d
That is, the higher the torque and the smaller the distance between the center of the spinning object and the place it is applying force, the more force is pushing the bike.

Bringing our two equations together:

a = (Q / d) / m
OR

a = Q / (d * m)
With a car, the mass will not change nor will the distance between the center of a wheel and the outside of the tire Since this is the case, the only thing that changes a car's acceleration is the torque at the wheels, which is the result of torque from the engine. In an ideal world, that is. In reality friction has its affects, which is where horsepower comes in.

So what's important is not only how much torque an engine makes, but also for how long it makes its good torque (referred to as its torque band). A dynometer plots out the torque over the entire range of an engine's operation. The best indicator of a bikes's acceleration is this plotted curve. Or more accurately, the area under the curve (ie, the torque times the length of the band).

Some light reading....

http://www.factorypro.com/dyno/true1.html

dynocar

Let's try this one - at any given MPH, you will produce the most axle TQ (acceleration) by selecting a gear that puts your engine closest to your peak HP (work being done) RPM. This is why we select our shift points so we average the most HP, not TQ, whether we are drag racing or pulling a heavy load up a hill with a truck.

dynocar

Back to the original question, how does a dyno calculate engine TQ? Easy, by manipulating this formula, HP = TQ X RPM / 5252. So if it's an enertia dyno that basically measures HP, or a load dyno that basically measures TQ, if it also knows the engine RPM we can compute HP or TQ from some variation of that formula very accurately.

RomulusSmallBlock

sorry man your engine doesn't make HP. HP is a derived quanity. like you just said torque and RPMs. engines don't produce rates. basically HP is the gauge we use to see the rate of transfer of energy of the engine. now for the answer to the guys original posting... i thought about it for like an hour or so and came up with alot and it should answer your question because technically you asked two questions. how does my engine overcome the forces acting on your car and how do dynos work. i put it all in my next post

RomulusSmallBlock

ok we will use my s10 as a model because i have all the numbers for it. in this example i will exclude both friction and drag, so the only forces acting on your car is the force needed to move forward, normal force, and weight. because the question only requires forces in the horizontal or x plane, we can ignore weight and normal forces. with me in my truck the total weight is 3172 lbs which is 14109 N (for ease i will convert to the SI system and when i am done i will convert the answer back into US standard system). we do need the mass of me combined with my truck to do that we use this equation: M=Fw/Ag (M=mass; Fw= weight force; and Ag=gravitational acceleration) 14109 N/ 9.81 m/s^2=M=1438 kg. now lets say i want to accelerate me and my truck from 0 mph to 5 mph in one second; we will use this equation to get the acceleration in the x direction: Ax=(Vf-Vi)/T (Ax= acceleration in the x direction; Vf= final velocity; and Vi= initial velocity) 5mph=2.2352 m/s... (2.2352 m/s-0 m/s)/1 s=Ax= 2.2352 m/s^2. now that we have the mass and acceleration in the x direction lets see how much force is needed to accelerate me and my truck. because Fx=ma (Fx=the force in the x direction) 1438 kg * 2.2352 m/s^2= Fx= 3214 N or about 723 lbs. this is where your confusion and mine too last night comes in. how can an engine that makes 140 ft-lbs of torque move me and my truck at this force and what about the gears as torque multipliers? I will do the torque multipliers first. your tirer and wheels, tranny gears, and rear end aren't torque multilpliers at all. this is a ignoracne. i thought the same until i did the math on the subject. think of these things as lever multipliers. don't think of a wheel as a wheel..think of it as the perfect lever. Torque=Force*sine of the angle made by the lever to the force*distance from where the force is applied to the lever to the axis of rotation: T=F*sin(Q)*R. you get the max torque when the lever is 90 degrees to the force being applied to it. Now for a wheel, a wheel is in constant contact with the ground so its radius is always at a 90 degree angle to the force and why I said to call it a perfect lever. Now saying that, my wheel’s radius or lever is 13.55 in or 1.1294 ft long. I for this part US standard system will be easier to use and sin (90)=1. So T=Fx*R, 723 lbs * 1.1294 ft = T = 817 lbs-ft. now that is a number that is why bigger than what the engine makes what’s the deal on that. Simple, the gears of the tranny and rear end in affect me my lever smaller and reducing the torque needed to move the truck. Here are my gears for 1st (3.94 to 1) and rear end (3.73 to 1). So the equation with the gears involved is this T=Fx*R/(3.94*3.73); 723 lbs*1.1294 ft /(3.94*3.73) =Torque needed to be made by my engine =55.6 lbs-ft now that is a realistic number. Now for the check. HP =(Torque*RPM)/5252. we need to know what rpm that is at 5mph to do that use this: 5mph*(3.73*3.94*63360)/(13.55 in* 120*PI)=RPM= 911 rpm. HP=(55.6 lbs-ft*911)/5252= 9.64 hp. This to can be used to get what the HP is; Force*V=power ( I will use the SI system for ease on this part) Power=Fx*Vf; 3214 N * 2.2352 m/s = 7184 Watts. To cover it in to HP do this: 7184 W / 746 W/hp = 9.63 hp. Notice that the two are the same 9.64 hp and 9.63 hp; the .01 hp difference is because of rounding due to when I converted from SI to US systems and vice versa.

RomulusSmallBlock

An dyno works on the theory of, um...inertia, which is a property of rotating objects determined by their shape and mass. A large-diameter object has more inertia than a smaller object of similar weight, and that extra inertia requires more torque to accelerate. On a dyno, your car's rear wheel is spinning a heavy drum that has a known inertia value, and the engine is used to turn and accelerate the drum. Accurately measuring the acceleration over a small amount of time gives a value for torque (torque = inertia x acceleration), and horsepower is determined from the rpm of the engine and torque value. bam you like

RomulusSmallBlock

and most importantly the generation 3 and 4 chevy small blocks should be called Romulus Small blocks and the old SB chevys need to be called Classic small blocks

DynoDR

iTrader: (1)

On a load cell dyno, I WOULD say a real dyno! J/K, You measure torque applied on the roller with a load cell or strain gauge. The dyno does read the actual force applied to the roller and shows that if desired along with calculated engine torque and hp. Of course like it was said in an earlier post the dyno uses the engine rpm input to calculate the numbers from roll force.

Also if it is not an inertia dyno the drum is not free spinning. We can stop and hold most vehicles at a chosen speed or rpm on the dyno with the absorber. Then you are getting measured TQ at that speed without any acceleration. This has very little value for most testing but it may help someone understand better.

6speeder

iTrader: (25)

It all comes back to the fact that the dyno MEASURES rear wheel horsepower, but it CALCULATES engine torque based on the MEASURED rear wheel horsepower using the engine rpm from the pickup placed on the plug wire. So it's not actually measuring rear wheel torque it is simply calculating what it would be at the engine based on the data acquired to accelerate the drum.