New shoes... 205/55/16 Michelin Hydroedge
#61
07-08-2011, 08:25 PM
68.5mph. Give or take.
#1 I don't buy that you're a physics teacher. Link to your class's web page? .edu email with an associate or prof status in a uni's directory?
#2
There's a massive error in this line of thinking which you still haven't found. We're talking driving force here, not engine torque. Great read:
Shift Points - Sport Rider
[DISCLAIMER: The following includes bad, mahout styled maths. Do not attempt or your brain may implode]
By extension of your own thinking, a fit would never be able to leave a red light, as it would not have enough torque. e.g. 17lbs wheels * 8" * 4 wheels + 18lbs tires * 12" * 4 tires = 117lbs ft. This is obviously not true.
#3
Sigh. Yes, it does. Energy (kWh, hp*sec, etc) is what gasoline turns into as its burned. The rate at which it is accumulated is called power. For instance, a large amount of KE over a little amount of time is a lot of power. Large KE / big time = less horsepower. Either way, it's burned dino. Reducing the amount you burn increases gas mileage.
#4
I don't think you have any legs to stand on here about physics... when you don't understand the difference between torque, power, and energy. I've stated, over, and over, and over again in this thread (which I'm certain you didn't bother to read before posting) that the increased mass and radius decreases mpg in stop / go only driving. At certain velocities, with no acceleration, it increases mpg by effectively reducing rpms and parasitic losses at a given car velocity. I've seen this work out first hand. Its similar, but much less apparent, to changing a final drive in a rwd car. 3.23s tend to get better mpgs than 4.11s. This is a well known fact.
Finally, I hope you really are full of it. If not it makes me grieve for our nation's future, our students, and also appreciate my PHY241 teacher.
#1 I don't buy that you're a physics teacher. Link to your class's web page? .edu email with an associate or prof status in a uni's directory?
#2
So if you want to accelerate the revs per mile you must apply a force of 48 inch pounds, 3 pound feet, i.e. 3 pounds more at a radius of 12" )which is 3 pound feet torque). For 4 wheels thats 12 pounds or 12 lb ft. Your Fit only has a maximum of 119 lb ft at max rpm and a lot less with lesser rpm.
Shift Points - Sport Rider
[DISCLAIMER: The following includes bad, mahout styled maths. Do not attempt or your brain may implode]
By extension of your own thinking, a fit would never be able to leave a red light, as it would not have enough torque. e.g. 17lbs wheels * 8" * 4 wheels + 18lbs tires * 12" * 4 tires = 117lbs ft. This is obviously not true.
#3
Kinetic energy has absolutely nothing to do with the work it takes to get there. And it is the work that consumes mpg. Can you state the relationship between horsepower and torque?
#4
The premise of this 'discussion' was to show why a heavier tire due to added width should have a smaller diameter seems to be lost in your misapplication of physics equations.
Finally, I hope you really are full of it. If not it makes me grieve for our nation's future, our students, and also appreciate my PHY241 teacher.
Last edited by z06dustin; 07-08-2011 at 08:29 PM.
#62
07-09-2011, 12:46 PM
68.5mph. Give or take. Show your math. you are incorrect but even a guess might be that close.
#1 I don't buy that you're a physics teacher. Link to your class's web page? .edu email with an associate or prof status in a uni's directory?
Sorry, but I don't use a college email address. You've critiicized my math and education enough show me yours. And I know you are not even a passable physics class student.
#2
There's a massive error in this line of thinking which you still haven't found. We're talking driving force here, not engine torque. Great read:
Shift Points - Sport Rider
The reference you use has an equationthat fails to account for the weight of the wheel/tire. It is merely a calculation of the torque applicable at the circumference of a tire based on the torque of the engine at the rpm turning. And that equation merely states again what I'm saying: except it does not address the effect of a heavier tire. The heavier the wheel the more torque required to turn it., same as the more weight the vehicle the more torque required to turn at a given rpm..
The truth is any shift with sufficient power on hand ( and power means torque) as happens with a good sport bike will lunge when the clutch is re-engaged. For decades the best upshifts are ones that occur at the maximum power rpm such that the rpm does not drop below maximum torque range where the rpm does not exceed the upper torque limits. It is torque that turns the wheel/tire.
[DISCLAIMER: The following includes bad, mahout styled maths. Do not attempt or your brain may implode]
By extension of your own thinking, a fit would never be able to leave a red light, as it would not have enough torque. e.g. 17lbs wheels * 8" * 4 wheels + 18lbs tires * 12" * 4 tires = 117lbs ft. This is obviously not true.
In case you've never seen a dyno curve (which measures the torque and uses rpm to convert to hp) even at 1000 rpm a Fit engine develops some 25 lb-ft which is more than enough to rotate the wheels. Rotation rate (angular velocity) increases as the rpm i and torque increase, up to maximum. In all out drag racing thats why you drop the clutch at 3000 rpm or higher; to get the maximum torque that doesn't spin the wheels.
#3
Sigh. Yes, it does. Energy (kWh, hp*sec, etc) is what gasoline turns into as its burned. The rate at which it is accumulated is called power. For instance, a large amount of KE over a little amount of time is a lot of power. Large KE / big time = less horsepower. Either way, it's burned dino. Reducing the amount you burn increases gas mileage.
Nonsense. The combustion of fuel such as gasoline creates energy.
A lot of change in kinetic energy in a short time is power. And that power is supplied by torque at a given rpm.
#4
I don't think you have any legs to stand on here about physics... when you don't understand the difference between torque, power, and energy. I've stated, over, and over, and over again in this thread (which I'm certain you didn't bother to read before posting) that the increased mass and radius decreases mpg in stop / go only driving. At certain velocities, with no acceleration, it increases mpg by effectively reducing rpms and parasitic losses at a given car velocity. I've seen this work out first hand. Its similar, but much less apparent, to changing a final drive in a rwd car. 3.23s tend to get better mpgs than 4.11s. This is a well known fact.
Finally, I hope you really are full of it. If not it makes me grieve for our nation's future, our students, and also appreciate my PHY241 teacher.
#1 I don't buy that you're a physics teacher. Link to your class's web page? .edu email with an associate or prof status in a uni's directory?
Sorry, but I don't use a college email address. You've critiicized my math and education enough show me yours. And I know you are not even a passable physics class student.
#2
There's a massive error in this line of thinking which you still haven't found. We're talking driving force here, not engine torque. Great read:
Shift Points - Sport Rider
The reference you use has an equationthat fails to account for the weight of the wheel/tire. It is merely a calculation of the torque applicable at the circumference of a tire based on the torque of the engine at the rpm turning. And that equation merely states again what I'm saying: except it does not address the effect of a heavier tire. The heavier the wheel the more torque required to turn it., same as the more weight the vehicle the more torque required to turn at a given rpm..
The truth is any shift with sufficient power on hand ( and power means torque) as happens with a good sport bike will lunge when the clutch is re-engaged. For decades the best upshifts are ones that occur at the maximum power rpm such that the rpm does not drop below maximum torque range where the rpm does not exceed the upper torque limits. It is torque that turns the wheel/tire.
[DISCLAIMER: The following includes bad, mahout styled maths. Do not attempt or your brain may implode]
By extension of your own thinking, a fit would never be able to leave a red light, as it would not have enough torque. e.g. 17lbs wheels * 8" * 4 wheels + 18lbs tires * 12" * 4 tires = 117lbs ft. This is obviously not true.
In case you've never seen a dyno curve (which measures the torque and uses rpm to convert to hp) even at 1000 rpm a Fit engine develops some 25 lb-ft which is more than enough to rotate the wheels. Rotation rate (angular velocity) increases as the rpm i and torque increase, up to maximum. In all out drag racing thats why you drop the clutch at 3000 rpm or higher; to get the maximum torque that doesn't spin the wheels.
#3
Sigh. Yes, it does. Energy (kWh, hp*sec, etc) is what gasoline turns into as its burned. The rate at which it is accumulated is called power. For instance, a large amount of KE over a little amount of time is a lot of power. Large KE / big time = less horsepower. Either way, it's burned dino. Reducing the amount you burn increases gas mileage.
Nonsense. The combustion of fuel such as gasoline creates energy.
A lot of change in kinetic energy in a short time is power. And that power is supplied by torque at a given rpm.
#4
I don't think you have any legs to stand on here about physics... when you don't understand the difference between torque, power, and energy. I've stated, over, and over, and over again in this thread (which I'm certain you didn't bother to read before posting) that the increased mass and radius decreases mpg in stop / go only driving. At certain velocities, with no acceleration, it increases mpg by effectively reducing rpms and parasitic losses at a given car velocity. I've seen this work out first hand. Its similar, but much less apparent, to changing a final drive in a rwd car. 3.23s tend to get better mpgs than 4.11s. This is a well known fact.
Finally, I hope you really are full of it. If not it makes me grieve for our nation's future, our students, and also appreciate my PHY241 teacher.
The point remains that a heavier tire at a larger diameter will always have less acceleration than a lighter tire at less diameter for the same power aplication and vehicle weight and traction.
Last edited by mahout; 07-09-2011 at 12:53 PM.
#63
07-09-2011, 02:29 PM
I'm not doing your sons physics homework for him, sorry. Here's the scrap I had, the rest was done in my calculator, you'll have to have him do that part himself. https://lh3.googleusercontent.com/-T...I/8d5wbyUu6yI/
Regardless, you're sidetracking.
I also 'critiicize' your spelling and honesty. I know enough physics to have a P.E. registration.
Right. I know that. And apparently now so do you, in direct contrast to your original statement here: https://www.fitfreak.net/forums/2nd-...ml#post1010078. But hey, teachable moment, you're improving I'm proud.
Right. There's a break even point, at some point a larger tire would certainly hurt things I completely agree. We've increased OD by 3.5% here though. It's not a major change.
Here's the difference: I've seen the results first hand, an increase in MPG. All you have is a pontification that decreasing rpm at 80mph for all sets of driving will result in decreased MPG. I have results, you have a guess. End of discussion.
Completely agree. All I'm saying, is you're neglecting the decreased rotational velocity that a larger tire requires at a given actual vehicle speed. Which is all I care about. This is a squared relationship, which offsets some (definitely not all) of the increase in the moment of inertia. The moment of inertia is only a factor when accelerating the system. When you're cruising down the highway, it has no effect.
Here are my original calculations about the em of the system if you care. I'm done, your idiocy has finally exhausted me.
https://www.fitfreak.net/forums/2nd-...tml#post979894
Regardless, you're sidetracking.
You've critiicized my math and education enough show me yours. And I know you are not even a passable physics class student.
In case you've never seen a dyno curve (which measures the torque and uses rpm to convert to hp) even at 1000 rpm a Fit engine develops some 25 lb-ft which is more than enough to rotate the wheels.
You need to get an education in efficiency. Simply reducing rpm does not necessarily increase mpg. Thats why manufacturers test a wide range of tires and sizes. All engines have a minimum specific fuel consumption point. An increase or decrease from that point results in less mpg. Unless your rpm is too low to start with if you change the rpm by 22% you will get a reduction in mpg but that is not indicative of the changes in rpm by changing a tire size on a Fit.
Here's the difference: I've seen the results first hand, an increase in MPG. All you have is a pontification that decreasing rpm at 80mph for all sets of driving will result in decreased MPG. I have results, you have a guess. End of discussion.
The point remains that a heavier tire at a larger diameter will always have less acceleration than a lighter tire at less diameter for the same power aplication and vehicle weight and traction.
Here are my original calculations about the em of the system if you care. I'm done, your idiocy has finally exhausted me.
https://www.fitfreak.net/forums/2nd-...tml#post979894
#64
07-09-2011, 08:04 PM
I'm not doing your sons physics homework for him, sorry. Here's the scrap I had, the rest was done in my calculator, you'll have to have him do that part himself. https://lh3.googleusercontent.com/-T...I/8d5wbyUu6yI/
Regardless, you're sidetracking.
My son is 52 and also an engineer. We thank you for your entertainment. Read on for my use of your equation to properly show you why there just isn't any practical reason for your assumption.
I also 'critiicize' your spelling and honesty. I know enough physics to have a P.E. registration.
We can check my list of PE's. Couldn't find one in your neighborhood.
Right. There's a break even point, at some point a larger tire would certainly hurt things I completely agree. We've increased OD by 3.5% here though. It's not a major change.
Here's the difference: I've seen the results first hand, an increase in MPG. All you have is a pontification that decreasing rpm at 80mph for all sets of driving will result in decreased MPG. I have results, you have a guess. End of discussion.
I bet we have more manufacturer test logbooks than you do. Keep reading.
Completely agree. All I'm saying, is you're neglecting the decreased rotational velocity that a larger tire requires at a given actual vehicle speed. Which is all I care about. This is a squared relationship, which offsets some (definitely not all) of the increase in the moment of inertia. The moment of inertia is only a factor when accelerating the system. When you're cruising down the highway, it has no effect.
Here are my original calculations about the em of the system if you care. I'm done, your idiocy has finally exhausted me.
https://www.fitfreak.net/forums/2nd-...tml#post979894
Regardless, you're sidetracking.
My son is 52 and also an engineer. We thank you for your entertainment. Read on for my use of your equation to properly show you why there just isn't any practical reason for your assumption.
I also 'critiicize' your spelling and honesty. I know enough physics to have a P.E. registration.
We can check my list of PE's. Couldn't find one in your neighborhood.
Right. There's a break even point, at some point a larger tire would certainly hurt things I completely agree. We've increased OD by 3.5% here though. It's not a major change.
Here's the difference: I've seen the results first hand, an increase in MPG. All you have is a pontification that decreasing rpm at 80mph for all sets of driving will result in decreased MPG. I have results, you have a guess. End of discussion.
I bet we have more manufacturer test logbooks than you do. Keep reading.
Completely agree. All I'm saying, is you're neglecting the decreased rotational velocity that a larger tire requires at a given actual vehicle speed. Which is all I care about. This is a squared relationship, which offsets some (definitely not all) of the increase in the moment of inertia. The moment of inertia is only a factor when accelerating the system. When you're cruising down the highway, it has no effect.
Here are my original calculations about the em of the system if you care. I'm done, your idiocy has finally exhausted me.
https://www.fitfreak.net/forums/2nd-...tml#post979894
Ok lets do some proper math. Hang onto your hats, we may not come this way again.
First, rotational kinetic energy has a fixed axle so no linear motion occurs. not applicable.
Second, K = mvv/2 applies to the kinetic energy of a rotating wheel whose axle is in linear motion. Thats the application.
No matter which diameter tire the difference in mass is the only change in kinetic energy between tires as the axle linear velocity is the same in both cases.
The work difference between the two is then power x time. Power = torque x rpm / 5250. The heavier tire always has the burden of requiring more work to achieve the same linear velocity. More work = less mpg.
So the work is the necessary force at the radius applied at the rpm divided by 5250. A heavier tire cannot succeed at the original diameter so the radius must decrease.
The resistance to turning a wheel and tire must be overcome to increase rotation so the greater the torque resistance the less excess torque the engine develops. Therefore, if you do not want to handicap your Fit's engine do not saddle it with more torque to rotate a wheel/tire than you can get away with. That always means if you go heavier the radius must reduce.
So you go with a wider and heavier tire, always reduce the diameter of the tire. many times you cannot completely make up for the loss in torque resistance but you can minimize at the benefit of increased cornering.
Now, having shown the inapplicability of your premise by your own equation, I'll not respond further to your posts.
Tell me ,do you really expect to intimidate techinal people with insults?
Last edited by mahout; 07-09-2011 at 09:36 PM.
#66
07-12-2011, 04:09 PM
And be glad that nerds or technical designed the car you're driving, and most other things you use.
#67
07-12-2011, 05:45 PM
and
in the same post... uh huh.
I don't know about maturity... but, if you're old enough to drive, I would think you'd be old enough to stop calling anyone a nerd.
Behave and be nice to each other.
I don't know about maturity... but, if you're old enough to drive, I would think you'd be old enough to stop calling anyone a nerd.
#68
07-12-2011, 06:25 PM
Didn't read the entire thread, but it seems to me that OP thinks a taller, heavier and wider tire is going to net him more MPG just because he is cruising at ~3% less RPMs on the highway. Is this accurate?
OP...forget physics for a while. Have you ever pedaled a bicycle with wider (slick) tires vs one with narrow (slick) tires? You do realize what contact patch does to rolling resistance, right? And you realize what weight (especially weight so far out on the "wheel") does to acceleration, right?
This is, perhaps, a case of OP trying to "out-think" the obvious.
What's even funnier is that I've known folks that have went from a 4.4 final drive to a 4.7 and reported exactly the same MPG before/after. Internal combustion engines have a lot of factors when it comes to the MPG they acheive and those aren't "bettered" by simply upping OD of a tire.
OP...forget physics for a while. Have you ever pedaled a bicycle with wider (slick) tires vs one with narrow (slick) tires? You do realize what contact patch does to rolling resistance, right? And you realize what weight (especially weight so far out on the "wheel") does to acceleration, right?
This is, perhaps, a case of OP trying to "out-think" the obvious.
What's even funnier is that I've known folks that have went from a 4.4 final drive to a 4.7 and reported exactly the same MPG before/after. Internal combustion engines have a lot of factors when it comes to the MPG they acheive and those aren't "bettered" by simply upping OD of a tire.
#69
07-13-2011, 12:25 PM
I still haven't seen that anyone has denied Dustin's premise. He has measured better MPG and attributes that to lower cruising RPM under his own driving conditions. Has anyone rufuted that? I don't think so.
Maybe his brakes wear out a bit sooner, but his warranty lasts 3.5% longer.
#73
11-02-2011, 05:09 PM
Second, K = mvv/2 applies to the kinetic energy of a rotating wheel whose axle is in linear motion. Thats the application.
No matter which diameter tire the difference in mass is the only change in kinetic energy between tires as the axle linear velocity is the same in both cases.
No matter which diameter tire the difference in mass is the only change in kinetic energy between tires as the axle linear velocity is the same in both cases.
The linear translation (motion) of the wheels has energy, but they also have angular momentum. Energy is required to spin up the wheel and tire assemblies to 800-odd RPM. The axle linear velocity is not the only factor in the energy analysis.
Consider a fifth wheel and tire assembly, identical to four grounded wheels (i.e., a full size spare with an identical wheel). Accelerate up to 60 MPH. The non-rotating fifth wheel has the same linear kinetic energy as the four grounded wheels, but the grounded wheels are also spinning at about 850 RPM. Those four wheels have angular momentum; the fifth wheel in trunk (or sitting in the passenger seat, or mounted on the tailgate, or sitting stationary anywhere in the car) has only linear energy. Momentum is energy, whether it's linear or angular.
Now consider your tire and wheel assembly spun up to measuring speed on the balancing machine. There's energy in that stationary spinning wheel. You can illustrate this by trying to spin down that assembly with your hand--you would burn off the skin trying. (Better to use the balancing machine's brake).
THe angular momentum of a larger diameter tire with have greater angular momentum than a smaller tire of identical mass. The only way to change angular momentum is to do work through engine torque (acceleration) or braking and friction (deceleration). Angular momentum follows the same rules as linear momentum (think of billiard balls, which are really good at transferring momentum via elastic collisions, or less desirably, your car crumpling against an immovable pylon, where the momentum is consumed by deformation of the vehicle and contents--a plastic collision).
What seems complicated (and mathematically, it is) is how to define that rotating mass. A one ounce piece of wheel nearest the axle will contribute much less to angular momentum than an ounce of tread at the perimeter. For the purposes of calculating angular momentum, the mass of a rotating wheel is expressed as its "moment of inertia".
THink of ice dancers: when they go into a spin and pull their arms close in, their mass does not change. But the moment of inertia does, and their rotational speed must increase to conserve angular momentum. (Momentum must be conserved in the angular world no differently than in the linear world).
The question that goes beyond my scope here is whether the larger diameter tire's slower rotation reduces angular momentum sufficiently to make a difference in acceleration or fuel consumption in stop-and-go conditions. A determination would require knowledge of the tires' moments of inertia in some detail, particularly with respect to the distribution of mass as measured from the tire bead to the outer edge of the tread.
On a 2500 pound car, 80 pounds of moving AND rotating mass comprises about 3 percent of the vehicle's mass. That affects efficiency, but so do rolling resistance, air resistance of the different profiles (which varies with speed), and drivetrain efficiency at a particular ratio of gearing and tires.
A quantitative overall answer (i.e., what really matters in terms of diameter, mass, efficiency, etc. would require extensive, well controlled experimentation. To understand which individual factors made the most difference would need detailed, explicit models and many more measurements.
Besides being a 2009 Fit owner in search for replacement tires, I have an MS in mechanical engineering. I now work with databases instead of equations, but I'm qualified to discuss the qualitative aspects of the physics.
#74
11-03-2011, 01:38 PM
Sorry to bump such an old thread, but I don't understand the first statement.
The linear translation (motion) of the wheels has energy, but they also have angular momentum. Energy is required to spin up the wheel and tire assemblies to 800-odd RPM. The axle linear velocity is not the only factor in the energy analysis.
Consider a fifth wheel and tire assembly, identical to four grounded wheels (i.e., a full size spare with an identical wheel). Accelerate up to 60 MPH. The non-rotating fifth wheel has the same linear kinetic energy as the four grounded wheels, but the grounded wheels are also spinning at about 850 RPM. Those four wheels have angular momentum; the fifth wheel in trunk (or sitting in the passenger seat, or mounted on the tailgate, or sitting stationary anywhere in the car) has only linear energy. Momentum is energy, whether it's linear or angular.
Now consider your tire and wheel assembly spun up to measuring speed on the balancing machine. There's energy in that stationary spinning wheel. You can illustrate this by trying to spin down that assembly with your hand--you would burn off the skin trying. (Better to use the balancing machine's brake).
THe angular momentum of a larger diameter tire with have greater angular momentum than a smaller tire of identical mass. The only way to change angular momentum is to do work through engine torque (acceleration) or braking and friction (deceleration). Angular momentum follows the same rules as linear momentum (think of billiard balls, which are really good at transferring momentum via elastic collisions, or less desirably, your car crumpling against an immovable pylon, where the momentum is consumed by deformation of the vehicle and contents--a plastic collision).
What seems complicated (and mathematically, it is) is how to define that rotating mass. A one ounce piece of wheel nearest the axle will contribute much less to angular momentum than an ounce of tread at the perimeter. For the purposes of calculating angular momentum, the mass of a rotating wheel is expressed as its "moment of inertia".
THink of ice dancers: when they go into a spin and pull their arms close in, their mass does not change. But the moment of inertia does, and their rotational speed must increase to conserve angular momentum. (Momentum must be conserved in the angular world no differently than in the linear world).
The question that goes beyond my scope here is whether the larger diameter tire's slower rotation reduces angular momentum sufficiently to make a difference in acceleration or fuel consumption in stop-and-go conditions. A determination would require knowledge of the tires' moments of inertia in some detail, particularly with respect to the distribution of mass as measured from the tire bead to the outer edge of the tread.
On a 2500 pound car, 80 pounds of moving AND rotating mass comprises about 3 percent of the vehicle's mass. That affects efficiency, but so do rolling resistance, air resistance of the different profiles (which varies with speed), and drivetrain efficiency at a particular ratio of gearing and tires.
A quantitative overall answer (i.e., what really matters in terms of diameter, mass, efficiency, etc. would require extensive, well controlled experimentation. To understand which individual factors made the most difference would need detailed, explicit models and many more measurements.
Besides being a 2009 Fit owner in search for replacement tires, I have an MS in mechanical engineering. I now work with databases instead of equations, but I'm qualified to discuss the qualitative aspects of the physics.
The linear translation (motion) of the wheels has energy, but they also have angular momentum. Energy is required to spin up the wheel and tire assemblies to 800-odd RPM. The axle linear velocity is not the only factor in the energy analysis.
Consider a fifth wheel and tire assembly, identical to four grounded wheels (i.e., a full size spare with an identical wheel). Accelerate up to 60 MPH. The non-rotating fifth wheel has the same linear kinetic energy as the four grounded wheels, but the grounded wheels are also spinning at about 850 RPM. Those four wheels have angular momentum; the fifth wheel in trunk (or sitting in the passenger seat, or mounted on the tailgate, or sitting stationary anywhere in the car) has only linear energy. Momentum is energy, whether it's linear or angular.
Now consider your tire and wheel assembly spun up to measuring speed on the balancing machine. There's energy in that stationary spinning wheel. You can illustrate this by trying to spin down that assembly with your hand--you would burn off the skin trying. (Better to use the balancing machine's brake).
THe angular momentum of a larger diameter tire with have greater angular momentum than a smaller tire of identical mass. The only way to change angular momentum is to do work through engine torque (acceleration) or braking and friction (deceleration). Angular momentum follows the same rules as linear momentum (think of billiard balls, which are really good at transferring momentum via elastic collisions, or less desirably, your car crumpling against an immovable pylon, where the momentum is consumed by deformation of the vehicle and contents--a plastic collision).
What seems complicated (and mathematically, it is) is how to define that rotating mass. A one ounce piece of wheel nearest the axle will contribute much less to angular momentum than an ounce of tread at the perimeter. For the purposes of calculating angular momentum, the mass of a rotating wheel is expressed as its "moment of inertia".
THink of ice dancers: when they go into a spin and pull their arms close in, their mass does not change. But the moment of inertia does, and their rotational speed must increase to conserve angular momentum. (Momentum must be conserved in the angular world no differently than in the linear world).
The question that goes beyond my scope here is whether the larger diameter tire's slower rotation reduces angular momentum sufficiently to make a difference in acceleration or fuel consumption in stop-and-go conditions. A determination would require knowledge of the tires' moments of inertia in some detail, particularly with respect to the distribution of mass as measured from the tire bead to the outer edge of the tread.
On a 2500 pound car, 80 pounds of moving AND rotating mass comprises about 3 percent of the vehicle's mass. That affects efficiency, but so do rolling resistance, air resistance of the different profiles (which varies with speed), and drivetrain efficiency at a particular ratio of gearing and tires.
A quantitative overall answer (i.e., what really matters in terms of diameter, mass, efficiency, etc. would require extensive, well controlled experimentation. To understand which individual factors made the most difference would need detailed, explicit models and many more measurements.
Besides being a 2009 Fit owner in search for replacement tires, I have an MS in mechanical engineering. I now work with databases instead of equations, but I'm qualified to discuss the qualitative aspects of the physics.
The problem is its two different applications. One is linear, the other rotational. The energy to change velocity of a vehicle is linear; the energy to slow a rotating wheel is rotational. The slowing of a vehicle is much greater than the effort to slow a rotating tire/wheel.
The mass of a vehicle includes all four wheels/tires as well as the chassis while the mass of a wheel/tire includes only the wheel/tire. Multiply by 4 to get all 4 wheels/tires.
Perhaps the easiest way to observe is to jack the vehicle off the ground and calculate the work involved to slow or speed rpm of the wheels/tires compared to the work to slow - or speed up - the entire vehicle. It takes a lot more work to change velocity of the entire vehicle than just the four tires.
However, as the original posts were about it is the affect of the greater mass or diameter of 4 tires on mpg, even if the vehicle weight is identical.. Well, OK, only as long as the deformation energy of the different tires is the same. Thats why added tire pressures increase mpg.
And why narrower treads help mpg too. Add mass or radius (diameter aka gearing) pretty much requires more work and more work translates into poorer mpg.
Help?
#75
11-03-2011, 02:37 PM
I'll pass on commenting about the weight or size of the tires 
What I will say is that my Dad had a set of Hydroedge's on his last car and they went 90,000+/- miles and are now on display at his local Tire Kingdom store to show what the tires are capable of. They also handle surprisingly well for that type of tire.
What I will say is that my Dad had a set of Hydroedge's on his last car and they went 90,000+/- miles and are now on display at his local Tire Kingdom store to show what the tires are capable of. They also handle surprisingly well for that type of tire.
#77
09-06-2012, 03:36 PM
The problem is its two different applications. One is linear, the other rotational. The energy to change velocity of a vehicle is linear; the energy to slow a rotating wheel is rotational. The slowing of a vehicle is much greater than the effort to slow a rotating tire/wheel.
The mass of a vehicle includes all four wheels/tires as well as the chassis while the mass of a wheel/tire includes only the wheel/tire. Multiply by 4 to get all 4 wheels/tires.
Perhaps the easiest way to observe is to jack the vehicle off the ground and calculate the work involved to slow or speed rpm of the wheels/tires compared to the work to slow - or speed up - the entire vehicle. It takes a lot more work to change velocity of the entire vehicle than just the four tires.
However, as the original posts were about it is the affect of the greater mass or diameter of 4 tires on mpg, even if the vehicle weight is identical.. Well, OK, only as long as the deformation energy of the different tires is the same. Thats why added tire pressures increase mpg.
And why narrower treads help mpg too. Add mass or radius (diameter aka gearing) pretty much requires more work and more work translates into poorer mpg.
Help?
The mass of a vehicle includes all four wheels/tires as well as the chassis while the mass of a wheel/tire includes only the wheel/tire. Multiply by 4 to get all 4 wheels/tires.
Perhaps the easiest way to observe is to jack the vehicle off the ground and calculate the work involved to slow or speed rpm of the wheels/tires compared to the work to slow - or speed up - the entire vehicle. It takes a lot more work to change velocity of the entire vehicle than just the four tires.
However, as the original posts were about it is the affect of the greater mass or diameter of 4 tires on mpg, even if the vehicle weight is identical.. Well, OK, only as long as the deformation energy of the different tires is the same. Thats why added tire pressures increase mpg.
And why narrower treads help mpg too. Add mass or radius (diameter aka gearing) pretty much requires more work and more work translates into poorer mpg.
Help?
The easiest way to understand is leave the vehicle on the ground and measure the work it takes to move the vehicle by turning the axles by hand. Its easier to turn the axle when the radius of the wheel/tire is less. You're the engine. And of course, if the wheel/tire is heavier its harder yet to increase the rpm of the wheel/tire.
So if you were turning the wheel at the circumference of the wheel/tire its easier to turn the wheel but the engine isn't working at the tire circumference. Having the wheel carry all that extra weight of the vehicle just makes turning at increasing rpm all the more work. More work, less mpg.
Now at a constant rotation on level ground the bigger tires have a good chance of increasing mpg due to less effort to 'rotate' the tires because the ground offers less resistance to rotation.
In my case they aren't rotating at constant speed very much, hence less mpg, less acceleration, but longer life. adding weight makes the penalty worse. Work doesn't care whether rotational or linear, only about force times distance.
Last edited by mahout; 09-06-2012 at 03:44 PM.
#78
02-05-2013, 12:26 PM
I just put 205/55/16 Michelin Hydroedge on my2009 fit.
It rides so much better and theyare so quite.
Thanks for posting about them.
How manymiles did you end up getting out of them?
