Stock spring rates

[Fred H.]

Just some more fodder to add to your data files on spring rates and suspension tuning.....

I've found some data that I think is correct on the OEM spring rates for the C14, and just wanted to post them for future reference, along with a link to where I found it.

Front fork springs - 1.2 kg/mm

Rear shock spring - 13.3 kg/mm (130 newton/meters or 742 lbs/in)

"Ideal rider weight" for the OEM rear shock spring is listed as 180lbs.

http://racetech.com/ProductSearch/2/Kawasaki/ZG1400%20Concours/2008-12

My actual weight measurements of my bike with a full tank of gas put it at 729 lbs. I found a fork spring calculator on line that also seems to confirm these are the correct rates.

http://www.sonicsprings.com/catalog/calculate_spring_rate.php#calculate

Interestingly, when I use the Race Tech spring calculator tool, I get the same number for the rear shock spring, but it actually recommends slightly lighter fork springs than stock.

 

[mcrider007]

Front fork springs - 1.2 kg/mm (11.7 newton/meters or 104 lbs/in)

Rear shock spring - 13.3 kg/mm (130 newton/meters or 1150 lbs/in)

Your conversion to lbs/in seems to be way off.

http://www.hraefn.net/projects/spring_rates.php

When I stalled a Penske shock with a 750 lbs/in spring, the sag was noticeably less than with the OEM spring so I suspect the OEM is a bit less than 13.3 kg/mm.....but since I don't have any way of knowing each spring's total preload its just a guess on my part.

 

[rcannon409]

Fred, I've found a lot of errors in the Race Tech spring guide calculator.  When errors were pointed out to Race Tech, they never bothered to correct them.  Also, no one was ever able to confirm that the recommended springs from them were designed to be used with or without the gold valve.  A person might think they could contact Race Tech and find out....good luck. 

I dont know if Kawasaki lists the rate in the official c14 manual, but it would be far more accurate. The figures COULD bea accurate from Race Tech, but its nto very reliable.

If YOU personally measured them, I'd be much more comfortable with the result.

 

[Fred H.]

Your conversion to lbs/in seems to be way off.

Maybe I'm doing something wrong, but I'm using a little converter program for torque that says 1 newton meter = 8.85075 inch lbs. Here is the conversion utility I'm using.
http://joshmadison.com/convert-for-windows/

This one also tells me 130 newton meters = 1150 in lbs
http://www.unitconversion.org/energy/newton-meters-to-inch-pounds-conversion.html

Fred, I've found a lot of errors in the Race Tech spring guide calculator.  When errors were pointed out to Race Tech, they never bothered to correct them.  Also, no one was ever able to confirm that the recommended springs from them were designed to be used with or without the gold valve.  A person might think they could contact Race Tech and find out....good luck. 

I dont know if Kawasaki lists the rate in the official c14 manual, but it would be far more accurate. The figures COULD bea accurate from Race Tech, but its nto very reliable.

If YOU personally measured them, I'd be much more comfortable with the result.

Thanks for that info, I assumed they were reliable. I'll see if I can recheck them somehow.

I do know that Ohlins told me for my weight (185), I was right between a 130 and 140 newton meter rear shock spring, so the 130 nm number for the OEM rear spring may indeed be correct. I'll have to double check, but I think my front Traxxion springs are 1.1 kg/mm.

 

[mcrider007]

Your conversion to lbs/in seems to be way off.

Maybe I'm doing something wrong, but I'm using a little converter program for torque that says 1 newton meter = 8.85075 inch lbs. Here is the conversion utility I'm using.
http://joshmadison.com/convert-for-windows/

This one also tells me 130 newton meters = 1150 in lbs
http://www.unitconversion.org/energy/newton-meters-to-inch-pounds-conversion.html

The problem is that the conversion rate for torque is different than the conversion rate for springs....which I didn't know until a few minutes ago.

http://www.motocd.com/spring_rate.htm

 

[Fred H.]

The problem is that the conversion rate for torque is different than the conversion rate for springs....which I didn't know until a few minutes ago.

http://www.motocd.com/spring_rate.htm

Thanks for pointing that out. I wasn't aware of that either. I edited the lb/in numbers in my first post to correct them.

 

[gfinca]

Spring rate is newton per meter (N/m) or lb per inch (lb/in) and torque is newton-meter (N x m) or lb-in (lb x in)

 

[Fred H.]

Spring rate is newton per meter (N/m) or lb per inch (lb/in) and torque is newton-meter (N x m) or lb-in (lb x in)

Ah ha, now it all makes more sense.

 

[Fred H.]

FYI, I found a good spring rate calculator that lets you convert between newton/millimeter and lbs/in and kg/millimeter.


http://hraefn.net/projects/spring_rates.php

130 newton/millimeter = 13.256 kg/millimeter = 742.3 lbs/in

 

[dunlop]

All the Kawasaki have rock front suspention, my ZX6R also have realy stiff front suspention who don't match the rear. Their point is a more sport ride with a stiff front, but still confortable with a rear more soft. In real world, this is excellent for the road if you set it correctly. For the track, this is an another question.

 

[Fred H.]

All the Kawasaki have rock front suspention, my ZX6R also have realy stiff front suspention who don't match the rear. Their point is a more sport ride with a stiff front, but still confortable with a rear more soft. In real world, this is excellent for the road if you set it correctly. For the track, this is an another question.

I was talking to Dan Kyle the other day, and he indicated that the spring rates on the C14 replacement springs are some of the highest ones Ohlins makes. I don't know what the OEM spring rate is, but I get the feeling it is pretty stiff as well (I believe both the front and rear springs are progressive in nature). When Traxxion sent me replacement springs, they gave me a 1.0 kg/mm for the left leg and a 1.1 kg/mm spring for the right leg, since my weight put me in between the two spring rates. I believe Dan Kyle said Ohlins uses a 1.3 kg/mm which is significantly stiffer than what the Traxxion uses.

I'm real happy with the Traxxion set up. It feels very well planted and in control at all times, but not too harsh, so I think they got the right spring rate for me.

I've also recently returned my Ohlins shock to Dan to switch it back from a 130 n/mm spring to a 140 n/mm. The 130 felt good, but was a bit on the soft side, and I'm worried that it's going to be a problem when I load up the bike and travel on it. Dan is going to remove about 3mm of the initial pre-load applied to the 140 n/mm spring, to soften it up a bit for me. I'll let you know how this works when I get it back.

 

[timmyrodriguez8968]

Just some more fodder to add to your data files on spring rates and suspension tuning.....

I've found some data that I think is correct on the OEM spring rates for the C14, and just wanted to post them for future reference, along with a link to where I found it.

Front fork springs - 1.2 kg/mm

Rear shock spring - 13.3 kg/mm (130 newton/meters or 742 lbs/in)

"Ideal rider weight" for the OEM rear shock spring is listed as 180lbs.

http://racetech.com/ProductSearch/2/Kawasaki/ZG1400 Concours/2008-12

My actual weight measurements of my bike with a full tank of gas put it at 729 lbs. I found a fork spring calculator on line that also seems to confirm these are the correct rates.

http://www.sonicsprings.com/catalog/calculate_spring_rate.php#calculate

Interestingly, when I use the Race Tech spring calculator tool, I get the same number for the rear shock spring, but it actually recommends slightly lighter fork springs than stock.

Hey Fred, try to convert to LBS here: https://oneconvert.com/unit-converters
Couse your numbers are bit misteken.

 

[Fred H.]

My conversion rates are correct. You need to use a proper spring rate conversion utility.

13.3 kg/mm = (approx) 744 lbs/in

.: Spring Rate Conversion Calculator :.

Enter the value you want to convert Conversion results After you have entered in a value please select a conversion. Convert lbs/inch kg/mm Convert lbs/inch N/mm Convert kg/mm lbs/inch Convert kg/mm N/mm Convert N/mm lbs/inch Convert N/mm kg/mm Convert inches mm Convert mm inches

norwestsuspension.com

Spring Rate Calculator - Springrates.com

How it Works Our Spring Rate Conversion Calculator Tool is designed for reference purposes only in order to help determine the approximate rate of existing unmarked helical round wire compression coil springs. There are many variables which are not represented in these calculations, therefore it...

www.springrates.com

Spring Rate Conversion The Spring Store - Over 70 Trillion Custom & Stock springs

Spring Rate is the amount of force it’ll take your spring to travel one inch or 1 millimeter. The unit used to measure this load can be pounds or newtons and, in some rare cases, kilograms. This page is to show you how you can convert these values.

www.thespringstore.com

 

[Pete_COG_TN]

Actually (from the nitpicking engineer) Kg/mm doesn't really make sense, though that's what the industry seems to use. Kg is a measure of mass not force. It should be either N/mm (or N/m) or Lbf/inch. That's always bugged me but I've just bit my tongue (til now LOL). To relate Kg to a force, the gravitational constant has to be included. N is the measure of force in metric or SI units. IIRC using F=ma, 9.8N = 1Kg x 9.8m/sec2 . So a kg of mass will produce a different force on the moon vs. the earth. Of course the spring rate of a spring won't change regardless of what gravity is. Yeah the bike spring industry won't change just because I posted this (LOL). Never seen automotive springs rates in Kg/mm. Maybe some are. This always seemed odd to me.

As far as C14 spring rates, they've always seems stiff to me, like they're designed to work well for two-up loaded down.

 

[Fred H.]

Spring rates are expressed in how much MASS makes the spring move a DISTANCE. There is no acceleration component applied in the measurement.

Thus a 742 lb/in spring will sit one inch lower when 742 lbs of mass is placed on it. It's a static measurement, not a dynamic one.

We're all on earth so the gravitational component is assumed. But I guess if you're riding on the moon, then you might need different springs, plus a space suit and some extra O2 for the engine to run on.

 

[Fred H.]

And by the way, it you did want to measure the actual acceleration component involved, you'd need to use Hookes law equation to calculate it, but that's way above the intended topic of discussion here.

 

[Pete_COG_TN]

Hi Fred, I wasn't being critical of you, but the industry as a whole for using Kg/mm. You and everyone here are just using the units as in practice by much of the spring industry.

But I have to disagree with your first statement. Mass won't move anything without any acceleration. No forces are in play, F=ma. If a=0 then F=0. A spring will only be compressed or extended from a force. Either an applied force or a force resulting from accelerating a mass. The 742 lbs of mass in your example is a mass that produces 742 lbf. within earths gravity, g. That's the a. It's the 742 pounds of force from ma that is compressing the spring. We're all always subject to 1g of acceleration standing still on the ground. And typically when we describe mass in terms of lbs we use pound-mass or lbm.

Yes I agree Hooke's law is the correct relationship for force of a spring vs it's movement F=kx, k being the spring rate F/x. And I believe that is the correct way the spring rate should be described. The spring can be compressed by a force regardless of if there's any mass involved, like if it were compressed in a hydraulic machine. Which is why I think using Kg/mm is erroneous for describing a spring rate. I believe it's Newton's law that brings the acceleration into it, which I stated in my post above and this one.

Again I wasn't intending to be critical of any of you but the fact that those units are being used improperly by the industry. There's plenty of wrongly used nomenclature out there. But we can agree to disagree if we have too.

 

[Fred H.]

What I need to know to select a spring, is how much it compresses when I place a known amount of weight on it and if that compression rate is linear or progressive over the length of the spring, and the springs overall length. Expressing its rate in lb/in (or kg/mm) gives me the information I need for my purposes.

 

[Steve in sunny Fla]

Hi Fred, I wasn't being critical of you, but the industry as a whole for using Kg/mm. You and everyone here are just using the units as in practice by much of the spring industry.

But I have to disagree with your first statement. Mass won't move anything without any acceleration. No forces are in play, F=ma. If a=0 then F=0. A spring will only be compressed or extended from a force. Either an applied force or a force resulting from accelerating a mass. The 742 lbs of mass in your example is a mass that produces 742 lbf. within earths gravity, g. That's the a. It's the 742 pounds of force from ma that is compressing the spring. We're all always subject to 1g of acceleration standing still on the ground. And typically when we describe mass in terms of lbs we use pound-mass or lbm.

Yes I agree Hooke's law is the correct relationship for force of a spring vs it's movement F=kx, k being the spring rate F/x. And I believe that is the correct way the spring rate should be described. The spring can be compressed by a force regardless of if there's any mass involved, like if it were compressed in a hydraulic machine. Which is why I think using Kg/mm is erroneous for describing a spring rate. I believe it's Newton's law that brings the acceleration into it, which I stated in my post above and this one.

Again I wasn't intending to be critical of any of you but the fact that those units are being used improperly by the industry. There's plenty of wrongly used nomenclature out there. But we can agree to disagree if we have too.

Pete, I’m not an engineer so please respond like you’re teaching a 3rd grader.

Q: a mass weighing 742# is compressing a spring 1”. There is no movement , no acceleration. Yet the spring is compressed and I assume storing potential energy. Since there is no acceleration in the system, does that mean no force is being applied in the system?

Steve

 

[x01660]

Pete, I’m not an engineer so please respond like you’re teaching a 3rd grader.

Q: a mass weighing 742# is compressing a spring 1”. There is no movement , no acceleration. Yet the spring is compressed and I assume storing potential energy. Since there is no acceleration in the system, does that mean no force is being applied in the system?

Steve

In this case, what would the "system" entail? The spring is being compressed in relationship to what? Wouldn't the "accerleration" be a constant 9.8m/s^2 with a mass of 742 lbs, compressing the spring 1 inch against Earth's gravitational constant?

Edit: the mass of 742 lbs (assuming we're talking lbs to kg as mass, and not lbs as newtons) with a constant acceleration due to gravity (9.8m/s2) is compressing the spring 1 inch?

Edit 2: isn't the way we normally refer to "weight" in kg and lbs actually a measure of force? That's why you "weigh" different of different planets? Because your g is different?

So in the case of lb-in, the acceleration is included in the "lbs" unit because we're talking about it as a force as opposed to a mass?

Or am I completely mistaken?


-Z

 

[Pete_COG_TN]

Pete, I’m not an engineer so please respond like you’re teaching a 3rd grader.

Q: a mass weighing 742# is compressing a spring 1”. There is no movement , no acceleration. Yet the spring is compressed and I assume storing potential energy. Since there is no acceleration in the system, does that mean no force is being applied in the system?

Steve

Hi Steve,

The force F=ma is being applied to the spring, in this case a = 1g. And that's really a mathematical device to say it that way; F is being calculated using mg, knowing what m and g are. So maybe it's kind of backward to say with regard to gravity (for an object at rest) that for F=mg the acceleration of gravity is producing the force, which my explanation to Fred above might sound like, and isn't quite correct. I believe it's more proper to say that the force of the earth's gravity will produce 1g acceleration if there are no opposing forces (the ground, or your spring the mass is sitting on, or air resistance). That acceleration being 9.8m/sec/sec as 1660 said. If the ground or a spring does provide an opposing force that prevents any acceleration, then that force will be equal and opposite the gravity force (and calculate that using mg). So my explanation above maybe sounded a little backwards or unclear and probably prompted your question. So therefore the ever present force gravity will accelerate a mass free-falling toward earth (if no air resistance) at an acceleration of 1g or if the mass is resting on an object, the object has to produce an equal and opposite force mg, on the mass if the mass is to be at rest. In the case of your spring, it will compress 1 inch before providing the equal opposite force, then come to rest with no movement, ignoring oscillations that damp out.

Hopefully that's a more correct clearer explanation.

 

[Pete_COG_TN]

What I need to know to select a spring, is how much it compresses when I place a known amount of weight on it and if that compression rate is linear or progressive over the length of the spring, and the springs overall length. Expressing its rate in lb/in (or kg/mm) gives me the information I need for my purposes.

Yes Fred it does work since gravity is consistent and the spring industry seems to use Kg/mm frequently. But in 40+ years I've never read a text book or met another engineer that defined Kg as a unit to measure force, always mass. So sometimes it's hard to "bite my tongue" when I see Kg/mm. But it does work for choosing springs and I've noticed lots of springs are rated that way when I browse for them.

 

[x01660]

Yes Fred it does work since gravity is consistent and the spring industry seems to use Kg/mm frequently. But in 40+ years I've never read a text book or met another engineer that defined Kg as a unit to measure force, always mass. So sometimes it's hard to "bite my tongue" when I see Kg/mm. But it does work for choosing springs and I've noticed lots of springs are rated that way when I browse for them.

I know nothing about the engineering field, but:

Do you think that the original measurement used in the US was lb-in (using lbs as a force; mg) because SAE, and as we started using metric, they just made the conversion over? So that's why they're using kg/mm (a "direct" conversion of units) ?

What units do they use in Europe?

-Z

 

[Steve in sunny Fla]

Hi Steve,

The force F=ma is being applied to the spring, in this case a = 1g. And that's really a mathematical device to say it that way; F is being calculated using mg, knowing what m and g are. So maybe it's kind of backward to say with regard to gravity (for an object at rest) that for F=mg the acceleration of gravity is producing the force, which my explanation to Fred above might sound like, and isn't quite correct. I believe it's more proper to say that the force of the earth's gravity will produce 1g acceleration if there are no opposing forces (the ground, or your spring the mass is sitting on, or air resistance). That acceleration being 9.8m/sec/sec as 1660 said. If the ground or a spring does provide an opposing force that prevents any acceleration, then that force will be equal and opposite the gravity force (and calculate that using mg). So my explanation above maybe sounded a little backwards or unclear and probably prompted your question. So therefore the ever present force gravity will accelerate a mass free-falling toward earth (if no air resistance) at an acceleration of 1g or if the mass is resting on an object, the object has to produce an equal and opposite force mg, on the mass if the mass is to be at rest. In the case of your spring, it will compress 1 inch before providing the equal opposite force, then come to rest with no movement, ignoring oscillations that damp out.

Hopefully that's a more correct clearer explanation.

Pete, let me regale you with a stunning display of ignorance.
Since I live in our world and have more gravity than alot of people, I naturally understand that there is 742# being applied by the mass and being resisted by the spring, once it has compressed 1”. But the devil is in the details… the formula F=MA requires acceleration. In my mind acceleration is movement - sideway, downward, upward. In the subject system there is no movement. I understand there are resisting forces, but no acceleration. Gravity, in my mind is the same for this system as if the 742# mass was sitting on the ground. Yes there’s stored energy in the spring. But nothing is moving, so there’s no acceleration.
So ther you go. I naturally know the spring is applying an equal and countering force to the mass, but in my mind there is no “a” left on the formula.
Please make sense of this for me. I understand changing it to f=mg, but isn’t the gravitational force the same on the ground as it is on top of the spring? Make the formula make sense for me, please.
Thanks. Steve

 

[x01660]

Pete, let me regale you with a stunning display of ignorance.
Since I live in our world and have more gravity than alot of people, I naturally understand that there is 742# being applied by the mass and being resisted by the spring, once it has compressed 1”. But the devil is in the details… the formula F=MA requires acceleration. In my mind acceleration is movement - sideway, downward, upward. In the subject system there is no movement. I understand there are resisting forces, but no acceleration. Gravity, in my mind is the same for this system as if the 742# mass was sitting on the ground. Yes there’s stored energy in the spring. But nothing is moving, so there’s no acceleration.
So ther you go. I naturally know the spring is applying an equal and countering force to the mass, but in my mind there is no “a” left on the formula.
Please make sense of this for me. I understand changing it to f=mg, but isn’t the gravitational force the same on the ground as it is on top of the spring? Make the formula make sense for me, please.
Thanks. Steve

I think when you're looking at your "subject system", you have to take into account Earth's gravity. Remember that for every action, there is an equal and opposite reaction (Newton's 3rd Law of Motion- refer to the section titled "Pushback").

Gravity is a constant force. And Force=Mass x Acceleration. In the case of Earth, we can figure out WHAT that force due to gravity (Fg: i.e. your weight) is, using the following equation: Fg=G(big G, gravitational constant)m1m2/r2. So to figure out your weight (which is your Fg, which we generally measure in lbs or kg) you take the gravitational constant, multiply that by your MASS, then by the mass of the Earth, and divide the whole thing by the distance between your center of mass and Earth's center of mass, squared.

That's why in my earlier comment, I mentioned that the way we generally refer to weight (in lbs and kg) is actually as a FORCE (F=MA) and not as a mass. And that's why we "weigh" different on different planets (because m2 and r are different, so Fg is different).

And if you take into account that the equation for force is: F=MA, you can see that there's an acceleration component to it (if the acceleration was 0, the force would be 0.) That acceleration is 9.8 meters/sec/sec on the surface of the Earth, "pulling" towards the center of Earth's gravity (which is why it can be slightly different depending on where you are on Earth due to the height differential. Its also the reason why satellites have to have an adjustment so they stay in sync with atomic clocks on earth; the distance from Earth causes a slight time dilation that can be measured. Clocks in space run faster than those on Earth (or rather a clock closer to the center of gravity of an object will run slower RELATIVE to a clock that's further from the object's center of gravity, but that's a discussion for another time)

You can go here to see how its derived: https://byjus.com/jee/acceleration-due-to-gravity/

And also see here: https://www.vcalc.com/wiki/force-of-gravity








You'll notice that the unit for weight is pound-FORCE (lbf). That's my weight. I put my mass in kg in the equation, the mass of the Earth, and the equatorial radius of Earth, and it spat out my actual weight in lbs. But its lbf, because its a force. And see here:









Notice that the unit for kilograms is actually kilogram FORCE. This is what Pete was talking about when it bothered him that he never sees kg referred to as a force. Because kg is a unit of mass. Techincally, the spring unit should be kgf/mm. NOT kg/mm.

So with all that said, you have to take into account that Earth's mass that is curving spacetime in 3 dimensions all around it (i.e. gravity) is permeating your entire system, and therefore HAS to be a part of the calculation.

Let me put it to you this way: if we were at Sun-Earth L2 (Lagrange point 2- gravitational balance point between 2 bodies. In this case, on the other side of the Earth, where the James Webb Space Telescope is sitting now), and you had your spring and you put a 742 lb mass "on top of it" (even top/bottom, up/down are determined by gravity), would it compress the spring? Why or why not?

In addition, if we were to take that 742 lb "mass" and the spring, and go to the Moon, would that mass compress the spring 1 inch? Why or why not?

Also, an astronaut on the ISS can easily lift/push/pull an object that would be impossible to move on Earth. Why? Your muscles exert a force (MA) and since the acceleration due to gravity (A) is MUCH lower, the force needed to move the object is much lower. Same mass, but lower acceleration due to gravity (due to the distance from Earth's center of gravity).

Hope that helps.

-Z

Edit: Don't feel too bad, Steve. I probably curve spacetime a LOT more than you, lol! My Fg is positively MASSIVE!

Edit 2: I've been thinking of a way to make this more concise; a TL;DR , if you will.... so here it is:

The reason that a mass can exert a force on something is because of acceleration. In other words, if there was no acceleration due to gravity (9.8m/s^2), then the mass would have 0 effect on the spring. In any Earth-based "system" in which you're calculating a force, Earth's gravity is an inherent part of that calculation.

 

[Pete_COG_TN]

Pete, let me regale you with a stunning display of ignorance.
Since I live in our world and have more gravity than alot of people, I naturally understand that there is 742# being applied by the mass and being resisted by the spring, once it has compressed 1”. But the devil is in the details… the formula F=MA requires acceleration. In my mind acceleration is movement - sideway, downward, upward. In the subject system there is no movement. I understand there are resisting forces, but no acceleration. Gravity, in my mind is the same for this system as if the 742# mass was sitting on the ground. Yes there’s stored energy in the spring. But nothing is moving, so there’s no acceleration.
So ther you go. I naturally know the spring is applying an equal and countering force to the mass, but in my mind there is no “a” left on the formula.
Please make sense of this for me. I understand changing it to f=mg, but isn’t the gravitational force the same on the ground as it is on top of the spring? Make the formula make sense for me, please.
Thanks. Steve

Steve, yes the gravitational force on the ground is the same as it is on top of the spring. So maybe a better analogy to your 'mass on the spring' example is the compression of the spring by a hydraulic machine. The force is applied to (let's say slowly so we can ignore dynamics) the spring and compresses it without any acceleration. So we can have forces without acceleration and maybe my reply to Fred that there has to be acceleration to have a non-zero force was confusing. We can have mechanical forces. And so then it may be more accurate to describe gravity as a "mechanical" force on any mass and can be manifested as applying that force to an object it is resting on or causing acceleration of a mass that is above the ground and falling, accelerating toward the ground at 1g. So 1g is the acceleration of a mass falling toward the ground. Then I think a somewhat man-made definition of weight was created as W=F=mg (in several different systems of units), with the force varying directly with the mass since gravity is constant. We're essentially defining the amount force gravity creates on an object as the mass of an object times the accel. of the free-falling object, maybe somewhat arbitrarily. So maybe we should think of F=mg as a man-made equation to describe what the force gravity does. But F=ma holds for say a mass sitting on a friction-less surface and pushed by a lateral force. It will accelerate laterally at accel=F/m, like a C14 on friction-less bearings and tires and no wind resistance! So my pet-peeve on kg/mm is that in the man-made system of SI units, F is in Newtons, m is in kg, and a is in m/sec/sec.

Z yes SI units came from Europe I think (somewhere outside the US). I suspect Kg was more familiar to the general population and Newtons never caught on as a measure of force, such as ones weight, so using kg/mm felt more familiar when they started marketing springs. Any calculations engineers or physicists make with forces and masses in SI units, Newtons is the measure of force.

Hope that is clearer

 

[Steve in sunny Fla]

Steve, yes the gravitational force on the ground is the same as it is on top of the spring. So maybe a better analogy to your 'mass on the spring' example is the compression of the spring by a hydraulic machine. The force is applied to (let's say slowly so we can ignore dynamics) the spring and compresses it without any acceleration. So we can have forces without acceleration and maybe my reply to Fred that there has to be acceleration to have a non-zero force was confusing. We can have mechanical forces. And so then it may be more accurate to describe gravity as a "mechanical" force on any mass and can be manifested as applying that force to an object it is resting on or causing acceleration of a mass that is above the ground and falling, accelerating toward the ground at 1g. So 1g is the acceleration of a mass falling toward the ground. Then I think a somewhat man-made definition of weight was created as W=F=mg (in several different systems of units), with the force varying directly with the mass since gravity is constant. We're essentially defining the amount force gravity creates on an object as the mass of an object times the accel. of the free-falling object, maybe somewhat arbitrarily. So maybe we should think of F=mg as a man-made equation to describe what the force gravity does. But F=ma holds for say a mass sitting on a friction-less surface and pushed by a lateral force. It will accelerate laterally at accel=F/m, like a C14 on friction-less bearings and tires and no wind resistance! So my pet-peeve on kg/mm is that in the man-made system of SI units, F is in Newtons, m is in kg, and a is in m/sec/sec.

Z yes SI units came from Europe I think (somewhere outside the US). I suspect Kg was more familiar to the general population and Newtons never caught on as a measure of force, such as ones weight, so using kg/mm felt more familiar when they started marketing springs. Any calculations engineers or physicists make with forces and masses in SI units, Newtons is the measure of force.

Hope that is clearer

That is clearer. I mean obviously I understand how all this works in the physical world but when a formula describes acceleration it seemed a little “off” to me, which is why I felt compelled to ask.
X, I appreciate your effort, but I wonder how many 3rd graders would get your response. My ignorance is legendary, so maybe I should have asked for a response for a kindergartner. Regardless, thank you both, Pete and x for the free education.
Steve

 

[x01660]

That is clearer. I mean obviously I understand how all this works in the physical world but when a formula describes acceleration it seemed a little “off” to me, which is why I felt compelled to ask.
X, I appreciate your effort, but I wonder how many 3rd graders would get your response. My ignorance is legendary, so maybe I should have asked for a response for a kindergartner. Regardless, thank you both, Pete and x for the free education.
Steve

I wouldn't say you're ignorant, Steve. If you were, you wouldn't be on here trying to learn in the first place. That demonstrates a recognition that you have a deficit, a capacity to learn, and an innate desire to improve. Those three things are contradictory to an "ignorant" mindset. So actually, quite the opposite, unless you're saying that you don't think you have a capacity to learn, or don't have a desire to improve! But I don't think that's the case.

That said, I used to teach this stuff (at a VERY basic level) at a museum, and I HAD to do it at a grade appropriate level. And I got to researching what I wrote above, and guess what? You're right. That description was at a high school level, not a 3rd grade one. And since you asked for it at a 3rd grade level, I can understand how my previous explanation may have been a bit incomprehensible. Sorry about that.

So that got me to thinking, how WOULD I go about explaining the concept of gravity, then explain the acceleration, at a level that a 3rd grader WOULD understand (which is a good challenge)...

This PDF shows the NGSS (Next-Generation Science Standards) for 3rd grade. We can see what physical science standards there are in Bundle 4:



So with that in mind, lets see if I can provide an explanation that will fit:

Assuming we are on Earth, Earth is pushing on us with the same amount of "force" that we are pushing on it. That "force" is called gravity. Gravity is the reason why we have a weight. Without gravity, we would have no weight, and there would be no way for us to "push" on something, because it would just float away.

So going back to the spring and the "weight" (which is actually a mass), the reason why the mass can "push" the spring down is because Earth is pushing back up on the spring from the other direction. And since Earth is SO much bigger, it has enough "force" to push back with not only the force of the spring, but the mass on top of it. And Earth will always push with the same force that it's being pushed with. This is what we call (Issac) Newton's 3rd Law of Motion; for every action (force), there is an equal and opposite action (force).

Imagine the spring is pushing on Earth with a force of 1 (units don't matter here). And the mass is pushing on the Earth with a force of 4. The mass on top of the spring push on Earth with a force of 5. And since Earth is so big, it pushes back on the spring and the mass with a force of 5. So now you have a spring in the middle of two forces. That's what causes the spring to compress. That "force" that is pushing back up on the spring is what we call gravity. And mathematically, gravity is measured as an acceleration, which is calculated (on the surface of the Earth) to be 9.8m/s^2.

This is how the equation F=ma works; the spring, the earth, and the mass on the spring all have mass (m). The "acceleration" here is gravity. If there's no gravity, acceleration goes to 0. And any number (in this case, force) multiplied by 0 is 0.

This is why Pete and others mentioned that F=W=ma=mg. Weight is an inherent calculation of our mass (kg) x an acceleration. Since we all live on Earth (though I'll admit my head is often in the clouds....), the (a) is automatically assumed to be 1g (9.8m/s^2). So your weight is a force. And that force is calulated by multiplying your mass times the acceleration due to gravity. And the use of kg to measure weight is incorrect. It should either be kgf (kilogram force) or newtons. But we still call it an ATM machine and say: "RSVP, please!" So take it for what it is.

Let me know if that's any better!!

Edit: Here's a nifty little exchange that discusses the use of kg as a weight as opposed to a mass, and why: https://physics.stackexchange.com/q...of-newtons-to-measure-weight-in-everyday-life
Also, this: https://en.wikipedia.org/wiki/Mass_versus_weight

-Z

 

[Pete_COG_TN]

Z thanks for reinforcing my pet-peeve! Yeah it's likely most people that use SI units will state their weight in Kg even though it's really their mass. Newtons is probably really just familiar to us nerdy types.

 

[Freddy]

Join for the bike - stay for the education.

 

[x01660]

And I just thought about your previous comment, Steve, where you said:

Pete, let me regale you with a stunning display of ignorance.
Since I live in our world and have more gravity than alot of people, I naturally understand that there is 742# being applied by the mass and being resisted by the spring, once it has compressed 1”. But the devil is in the details… the formula F=MA requires acceleration. In my mind acceleration is movement - sideway, downward, upward. In the subject system there is no movement.

What you are describing is a CHANGE in acceleration. But that would require getting into vectors, and that IS above a 3rd grade level, so I'll hold off. But if you're (or anyone else on here is) interested, I can do a quick writeup of that. See if it fits.

In any case, let me know if I did better with my description above!

And again with that ignorance stuff... you're not ignorant, Steve! Though I understand that its probably just a literary device you're using. Faux humility, I think? You're a pretty smart guy. Wouldn't be out improving Connies if you weren't. I mean, I HAVE a set of your overflow tube carbs in my bike. And your carb tuning is EXCELLENT! Voyager runs great!!! But, I digress.....

Z thanks for reinforcing my pet-peeve! Yeah it's likely most people that use SI units will state their weight in Kg even though it's really their mass. Newtons is probably really just familiar to us nerdy types.

Yup! There's lots of little things like that that get on my nerves, too! Its like, "no, the Sun is never gonna explode; its not massive enough to overcome the Chandrasekhar limit and start fusing iron.... Ah well, that's why we ride!

Join for the bike - stay for the education.

“Wisdom is not a product of schooling but of the lifelong attempt to acquire it.” — Albert Einstein



-Zak