How Much Does a 1000-Kg Sports Car Weigh?

A simple mistake, the SI unit of mass is kilograms, no further conversions are necessary. Multiplying kilograms by 2.2 is the correct conversion to pounds.

I flip back and forth between English and SI, but I generally solve this type of problem using the SI units. Pound in the English system is a unit of weight, and must be converted into a unit of mass. Slugs and poundals confused me more than I already am.

http://en.wikipedia.org/wiki/Poundal

I lost the last part of my post when I was editing. There is nothing wrong with expressing weight in kilograms, that is unless you're planning to leave the planet, in which case you will have to utilize a scale that either has springs or pressure sensors to get an accurate weight.

I could be wrong in my view of what constitutes a actual kilogram scale, however it should have a counterweighted assembly versus springs or pressure sensors. Your mass will be the same any place in the universe, your weight will vary according to the gravitational field that you happen to be in.

So in my opinion you are technically right, the kilogram is a unit of mass, however as long as you're here on earth the approximate conversion factor of 2.2 is perfectly acceptable when working these types of problems.

Thanks for your response. My post was so long that my main question got lost in it. When the website author calculated hp = 52 from 1000 kg mass accelerating to 100 km/hr in 10 seconds I converted everything to US units and couldn't get the answer to come out to 52 hp until I divided 2200 lbs by 32 to get mass. So far so good, but here's my problem. A mass of 1000 kg gives the same hp as a mass of 2200/32 lbs-m. If we use mass to calculate acceleration in US units, but weight is mass x 32, then why is the same answer obtained using SI units for a car that is going to weigh 9.8 x more? If there is a mistake I don't know what it is, but there has to be one and I can't locate it.

I know that people use kilograms to talk about weight, like my wife who says she weighs 64 kilos, but isn't that really her mass, with the term "weight" misused? As for the car in question, it can't be 1000 kg weight because 1000 is being used to get acceleration.

I thought maybe the conversion of kilogram mass to lb-f with the factor 2.2 was the problem since it causes some confusion too. Shouldn't mass in kilos go straight to mass in lbs-m or slugs? The factor is then 14.59 or 32.174/2.2.

But there I go raising too many questions again. My real concern for now is, why does a mass of 1000 kg and a mass of 68 lb-m both get 52 hp? The figures I used are:

1000 kg accelerated from 0-100 km/hr in 10 sec

68 lb-m accelerated from 0-62 mph in 10 sec

But 1000 kg is supposedly representing weight...so why does the hp come out the same either way?

Luther

1000 Kg car weighs 1000Kgf or 9800N - it is because when you mention Kg, it is generally assumed as Kgf (unit of force in MKS system and is used in SI also occassionally). If you like to avoid confusion, start using SI system in Toto then a car of 1000Kg weighs 9800N.

Kgf does not exist.

It is bridge for users of the imperial system to make them understand and be able to draw a comparison to lbf, which is needed in the Imperial system.

"lbf" is a weight or force measure on earth. In metric it is expressed in Newtons N. We need to convert the force to a mass to do mass calculations. Kg or slug.

2205 lbf / 32 = 69 slugs. A slug is the equivalent to kg in the metric system.

F = m * a, (or a = F / m), that is true in imperial and in the metric system.

Using gravity as acceleration and calculating F, we get:

Metric: F = 1000 kg * 9.8 m/s = 9800 N, that simple. N is a force like Lbf.

Metric: Force = Newtons, Mass = kg, (no "kgf" it will lead to a wrong answer by a factor of 9.8)

Imperial: Force = lbf, Mass = slugs,

In Imperial we first need to calculate the mass in slugs which is 2205 lb / 32 = 68.9 slugs. Then we calculate F = 68.9 slugs * 32 ft/s^2 = 2205 lbf. Conversion lbf to N is ~0.225 lbf / N. N = 2205/ .225 = 9800 N, the same as above.

I know it seems silly to do that in this example. But using F = m*a in general, e.g. for the acceleration of a car or such, we need to divide lbf by 32 m/s^2 to get slugs in the Imperial system. Slugs can be used the same as kg in the metric system. Problems solved.

And please don't use Kgf, it simply does not exist. F in metric is N = Kg * 9.8.

Hello Lutherman

As long as you stick to a consistent set of units it's OK. There seem to be 3 systems being considered here.

1) foot - pound - second. Force unit is poundal = force to accelerate 1 lb at 1 ft/sec². Work/energy unit is ft.pndl. I believe this system is pretty much obsolete.

2) foot - slug - second. Slug = 32.2 lb. Force unit is the familiar pound = force to accelerate 1 slug at 1 ft/sec². 1 lbf is the same as force on 1 lb mass in Earth's gravity, 32.2 ft/sec². (or you could say definition of slug is that mass which 1 lbf accelerates at 1 ft/sec²). Work/energy unit is ft.lb.

In effect, you're working in the foot - slug - second system, that's why you had to divide mass in lb by 32.2 (to get slugs). You could have used the foot - pound - second system, that would give energy in ft.pndl, but you need to divide by 32.2 at that point because horsepower is defined as 550 ft.lb/sec, not ft.pndl/sec.

3) then there's the SI system, which I prefer, along with other posters if I'm not mistaken.

Don't know whether that adds anything to what Blink et al have said, but I've tried!

Cheers.....Codey

I don't know whether or nt I'm going to be able to explain any better than YWROADRUNNER (and narendra54) already did. I suppose neither of us will lose anything if I try, though:

for starters, when solving problems involving equations start by forgetting that overhere in europe we talk of our weight in terms of kilos - that won't help! Weight is always measured in Newton (N) which represents a force, as opposed to kg which represents a mass. Your 1000 kg sports car weighs 9800 N, not 9800 kg.

if you feel more comfortable thinking in terms of kg mass and kg force (or weight) then all you need to know is that the conversion factor between one and the other is 1, but in all equations when a force is expected it always comes in N. (in fact that is why we refer to kilos in general, regardless of whether it is a mass or a weight we're talking about, because they are numerically the same provided that we don't leave earth :) )

I hope this helps, or at least that it doesn't mess it up any further ...

P.S. I didn't review your calculations in US units, but I am assuming they are correct. I did review the SI calculations though and those are correct. I'm not positive because I'm utterly unable to think in US units, but I think your problem might be that you're performing one extra conversion that isn't coherent (like converting to mass something that already IS a mass ...)

Hugo

A site I always recommend for physics discussions is Hyperphysics. It is extensively hyperlinked, so you can find about as much or as little detail as you need on any subject. It also has many calculators built in, so you can play around with the numbers as see what happens.

The fact that you are confused by some of this indicates you are thinking. Even within systems of measurement there are many inconsistencies.

One of the few places in which the real English system unit of mass, the slug, is used routinely is in aerodynamics, as practiced in the US. More commonly, if the distinction between mass and weight is made (as it should be, given they are quite different things) the units end up being pounds mass and pounds force. This adds loads of confusion, as does the fact that the simple "pound" is used for each, indiscriminately.

My high school physics teacher would cut off students' fingers (usually only one at a time) if we used the English units of work and torque incorrectly. Torque is measured in lb-ft and work is measured in ft-lb. Of course, math tells you that ft x lb is the same as lb x ft, so given that math is integral to physics, it seems there should be better terms to make the distinction. Old torque wrenches were marked in kg-meters and ft-lbs, both incorrect from a physics perspective. You don't measure work (ft-lb) with a torque wrench, and clearly a mass (kg) times a distance (m) does not give you a torque. If you read road tests in car mags, you find that most older ones will quote torque in ft-lbs, although many more recent articles will use lb-ft.

Europeans should not say (to their physics teacher) "I weigh 100 kilos." They should say "I have a mass of 100 kilos, and on earth I weigh 980 newtons." In the US one should say "I have a mass of 6.83 slugs, and on earth I weigh 220 lbs." (In either case, unless this is a tall person, he or she should be thinking about reducing mass.)

So, I think your confusion is all in linguistics. The physics is much easier and more logical to deal with if you work in SI units, where the distinction in units of mass vs force is clear, and where the terms for each are commonly used.

And my wife, who is Asian, comes in kilos, while I come in pounds. Is she really bigger than me?

What is gray and comes in quarts?

But other resistances to a car's motion, such as rolling resistance and grade resistance, use weight, not mass, since the whole point is friction against the road or a road that climbs, so gravity is part of the equation. Right?

Right, sort of. Each of these calculations should use both mass and force. The intrinsic quality of a car is it mass. So to get friction you need to calculate the normal force first (from its mass and acceleration due to gravity) and apply that to the coefficient of friction. To make the car go up a grade, you will need a tractive force (which again means a mass times an acceleration). For a steady state speed up the grade, the acceleration is again (a part of that) due to gravity (depending on grade angle -- for most grades of interest, the grade percentage is close enough to use as a factor).

I say "should use both mass and force." In practice in the US, because of the strange way we use pounds indiscriminately, we may be unaware that we have already done the calculations to go from our 2200 lb mass car to a force of 2200 lb force against the road. So we can, in the blink of an eye, say that if the rolling resistance coefficient is .010 (about the average for modern radials) the tractive force required to overcome rolling resistance in this 2200 lb car is 22 lbs. We can also, in the blink of an eye, say that driving up a 10% grade will require a tractive force of 220 lb. (We can also say that acceleration at a rate of .1 g will also require 220 lb tractive force.) By happy coincidence, wheels on smallish cars (with 165 x 13 tires, for instance) have a radius of one foot, so the 220 tractive force also takes 220 lb-ft of torque at the wheels. (This, by the way, is far more than a 53 hp engine produces, thus the need for a gear reduction to enable grade climbing).

My solution if anyone would care to comment would be to choose an engine powerful enough to make the car, when loaded down, go up a pretty steep grade like a 10% grade faster than necessary at high constant speed like 70 mph. Then at low speeds on normal grades there will be plenty of extra power for acceleration. Right? I'm not talking about precision here but general ballpark guessing that isn't way off.

This could get you reasonably close, depending upon your needs. I work with electric cars for which a certain number of myths exist, such as there being no need for any more than a single speed gearbox. (With AC motors, this can be fairly close to being true, but even then, is not really true. The Tesla as a good demonstration. Despite having loads of horsepower, its top speed is very low by performance car standards, at 120 mph. [In a recent Car and Driver test of fast econobox (Civic class) cars, a couple had top speeds over 150 mph, box stock.] So depending on how much you are investing in designing a car, you typically need to at least calculate the grade climbing ability at low speeds (where 40-50% ability is typical of a standard ICE car) and also at high speeds, taking into account the amount of hp consumed in overcoming aero drag and rolling friction. For acceleration, you need to at least calculate tractive force in each gear, and pick a speed at perhaps the middle of each gear for calculating aero drag. (In practice, with a spread sheet, you can make the increments much smaller, for only a very small amount of extra work.)

But if your car is a typical ICE with a typical gearbox, then your 10% grade ability at 70 would translate to ample grade climbing ability at low speeds, and also adequate acceleration in traffic.

A 53 hp 1000 kg car will not, by the way, accelerate to 62 mph (100 km/h) in 10 seconds. As the car approaches 60, aero drag increasing absorbs horsepower. The old 1300 cc VW bug was 53 hp, and about 1000 kg with a driver, and would not accelerate to 60 in 10 seconds. I don't have a test in front of me (I'm sure you can find one on the web) but I'd guess the figure was close to 20 seconds. The 40 hp VW had a top speed of 72 mph, so the later ones would have had a top speed of maybe 77 or so -- the speed at which all of the developed hp is overcoming (mainly) aero drag, leaving nothing for acceleration. At 62, you are fairly close to that speed, so acceleration slows to a crawl.

Has the author of the website made the mistake of using 1000 kg as a mass when sports cars don't weigh 9800 kilos?

A 1000 kg sports car weighs (exerts a force on the ground of) 9800 Newtons. It has a mass of 1000 kg. Unfortunately, in different settings, these terms mean different things. In a physics class, your teacher might break your arms if you say a car weighs 1000 kg. A 1000 kg car weighs different amounts on the moon vs here, but possesses the same mass in either place. You weigh more for the first second of so of an upward elevator ride, and much more for the first minutes of a rocket ride.

What is gray and comes in quarts?

You just couldn't resist could you?

I couldn't. I tried. But I couldn't. Lurking within is the juvenile.

I think we should all be glad and give thanks to Ken that he is laying it out so nicely. He spared no effort and took the time. We should be glad he couldn't resist.

Thanks Ken.

Regards

Rolf

Thanks, Rolf. I've spent some time teaching, and have always enjoyed it, so these discussions on CR4 are their own reward, particularly so when the person with the question takes the time to clearly state it, and particularly when he is so clearly engaged in thinking about the issues. If the world over worked only in SI units, understanding these basic physics issues would be much easier.

There are several conversions that almost everyone in the US, technically aware or not, is fairly comfortable with: most people know that a mile and a kilometer are quite similar in size (and in the US, you can easily get a sense for the the conversion just by looking at most speedometers). We know, in the US, how big a two liter bottle of Coke is (because soda is one thing that is sold that way -- but oddly, soda is sold in 12 oz cans). Many people "know" that 2.2 lbs is equal to a kilogram. But if you "know" that and then come across what should be a basic, fundamental truth, F=MA you can become very quickly confused. Then, as if there are not already enough units and enough confusion, we get the proposal to use kg force!! It makes you wonder if distance should be measured in mph-seconds.

Fortunately, in physics classes in the US, SI units are used all the time. But otherwise, we are surrounded in things measured in feet and pounds, and we rarely distinguish between pounds mass and pounds force. In engineering we've developed some shortcuts that work easily and quickly: for example, for acceleration of a car, it is quick and easy to think in "g's": if you want to go from 0-60mpg (0-88 fps) in ten seconds, then you need to accelerate at 8.8 fps/s. Therefore, you need to accelerate at an average of .275g (i.e., 8.8/32). If the car weighs 2000 lbs, you know that you would need to push with a force of 550 lb. You can do those calculations without ever using a slug or mentioning "pounds mass" or "pounds force". (Of course if we did the math carefully, crossing out units appropriately, we'd run into some stumbling blocks.)

Fortunately, I do not have a large sphere of influence, so I don't have to feel too guilty for helping to perpetuate a system that makes little sense*. (I imagine that if it were not for the differences in measuring systems, the question starting this thread might not have been raised at all.)

*(although I did advocate for using liters per 100 km as a measure of fuel consumption for the Automotive X Prize... nevertheless, they are sticking with MPG.)

Ken,

Thanks so much for the information you provided. I am sure there is plenty there for me to study and come up with the answers I need. I have been burning the midnight oil on this one because I need to become familiar with the concepts so that when I grow up maybe I can be an inventor. At 52 I am hoping to grow up within the next few decades. As for your answer, the quote below is where I get lost, and I'll study your post in detail when I get home.

In practice in the US, because of the strange way we use pounds indiscriminately, we may be unaware that we have already done the calculations to go from our 2200 lb mass car to a force of 2200 lb force against the road.

I have seen glimmerings of hope as of last night after midnight that I was going to understand this soon, and then in references copied from wikipedia and other sources I find information that seems conflicting. Nobody tells the whole story simply in one place with examples in both units so putting that together is the hardest part. I downloaded a physics book and will look at the Hyperphysics site also.

Your statement about the same car having a force of 2200 lb and a mass of 2200 lb is going to see me up late again tonight. Of course you are right and now I'll have to figure out why! You probably gave me plenty of info to intuit it out of the ozone, if I can just locate an intuition about this thing.

I have decided that the figures used on that website were all correct and like you say it's a matter of linguistics from here to the finish line, at a slow crawl no doubt.

My big problem right now is that the 2200 lbs is considered weight or mass, "depending". I have concluded that in the US when we say "I weigh 170 pounds," we are talking about our weight and not our mass, and I concluded the same thing about what we call weight of cars--it's weight. Yesterday I had the opposite conclusions based on what I've seen on the internet. So it's back to the jungle for more days of head scratching and thanks again to everyone for your responses.

Luther

Hello Lutherman

Your statement about the same car having a force of 2200 lb and a mass of 2200 lb is going to see me up late again tonight.

2200 lb mass = 2200/32.2 = 68.32 slug, and if you work in the foot - slug - second system everything's OK.

It's possible to use lb for force and mass, but then you need to put in g at the appropriate places to make formulas come out right. Early versions of the Chemical Engineers' Handbook (Perry) did this, g being called a "dimensional constant" i.e. a fudge factor to make the dimensions work out! I believe the later issues use SI so it isn't needed.

Cheers........Codey

I am glad this is starting to make sense to you mow. Probably a big weight mass slug concern removed from you.

Protons have mass? I didn't even know they were Catholic.

I'll respond again later, but for the time being, it may be interesting to think about "weighing" someone with two different scales, in two different places. With a balance scale, likes doctors sometimes still use, will your reading be the same or different here vs on the moon? With a spring scale (typical old bathroom scale) will the reading be the same in both locations, or different?

The spring balance has a spring which has a K value which will remain constant. At moon the spring will be compressed less so it will indicate a lower weight than earth.

( this is in response to Ken Blink's doubt.)

Actually I was not in doubt, but was offering the question for the original poster to think about. Now you've spoiled it! (I'm kidding -- he's had loads of time to think about it.) Although the scales serve roughly the same function on earth, one is in fact measuring mass and the other weight.

Hi Ken, thanks for the challenge question. I was not ignoring you and I had plenty of time to think about it but I live in Asia so I have to travel each week to an internet cafe.

From my research I have concluded that a balance scale "weighs" mass, even though the hunk of metal on one side for comparison is called a "weight". The spring scale is affected by gravity so it weighs weight. I'd guess that the balance scale is also affected by gravity but equally on both sides so it cancels out? Anyway they're proportional, so a scale that weighs slugs can be marked to read out pounds.

Also in the US apparently we refer to our weight as weight and the weight of our cars also in weight. Not so in the metric world, my wife says she weighs 64 kilos but that's really her mass. It was easy to figure out why they do that...first, they learned English from us so misuse the word "weight" the same way we do, but more importantly, what woman will admit she weight 627 Newtons?

But things that are sold by weight--which in most countries doesn't include people and cars--are listed by their mass in lbs mass or ounces mass which is still 1/16 of a lb whether its weight or mass, but this "net weight" of canned beans or whatever is really mass. There is even a law saying that the word "weight" is to be used in reference to mass when weighing things in commerce that are sold by weight--which is really mass.

Thanks to everyone for straightening me out...assuming I'm not still confused.

Luther

Its a problem with the imperial system and not knowing why and where the conversion come from.

In SI system kg is mass, N-Newton is weight or force. 1 N = 1 kg * m/s^2.

On Earth gravity is 9.81 m/s^2. So your 1000kg car will weight: 1000 kg * 9.81 m/s^2 = 9810 N.

Problem with imperial system is people usually use lb-pound for both mass and weight which works only on earth.

Its better to use lbmass/lb for mass and lbf for weight/force which bring in more confusion. Since 1 lbf = 1 lb * earth gravity = 1 lb * 32.2 ft/s^2 which make it "easy" when the mass is only affected by earth gravity and damn difficult everywhere else.

This is why its so confusing when you try to calculate HP. The car is not accelerating at 32.3 ft/s^2 all the time, so you can't assume 1 lbf = 1lb hence you need to keep multiplying and dividing 32.2.

To get it right you'll have to know if you need to use mass or force. Its a lot easier to calculate everything in SI then convert the result to Imperial.

NASA failed one Mars landing due to miscalculate of lbf and lb. The craft was landing about 32.2 times too fast.