Shorter Stopping Distance for ultralight vehicles?

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  • #1
mheslep
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The concept of ultralight vehicles is intended to allow greater fuel efficiencies in part by the use of composite structures to reduce mass by 2 or 3x. In several discussions of these vehicles I have seen and heard mention of the supposed additional safety benefit of shorter stopping distances, but I have not found any elaboration on why this is so, implying I fear that I missing something obvious.

Of course I reached for the standard stopping distance derivation: the kinetic energy of the vehicle and the work done by friction are both linearly related to mass, so that stopping distance is independent of mass as shown here:
http://hyperphysics.phy-astr.gsu.edu/HBASE/crstp.html
giving the familiar distance = velocity^2/(2*Cf*gravity)

So is there some other mass related factor here that is, say, a practical result of chassis, suspension, tires, or brake design? Reduced sway?

The ultralight vehicle article is here:
http://www.rmi.org/images/PDFs/Transportation/T95-27_VehicleDsnStategies.pdf [Broken]
Is lengthy covering several disciplines and I do not mean to introduce it all here. I am referring to the safety section on pg 14:
Design and Materials for Safety
Lightweight vehicle design, while presenting new challenges,does not preclude crashworthiness and could even improve it under some conditions. Lightweight design also improves maneuverability and stopping distance, allowing the driver to avoid many potential collisions. Using proven technologies for energy absorption, force-limiting occupant
restraints, and rigid passenger compartment design, even ultralight vehicles can surpass the safety of today’s cars in many types of collisions. The possible exceptions to this are
high-speed head-on collisions with, and side impacts from, a significantly heavier collision partner, though these might be effectively dealt with through innovative and careful design.

Thanks for any comments
 
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  • #2
mheslep
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Bump.

Am I missing the obvious? :confused:
 
  • #3
Mech_Engineer
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While lighter vehicles are easier to stop, technically stopping time has more to do with the capacity of the brakes, and the contact patch of the tires. You can make a relatively heavy vehicle stop very fast with big enough brakes; but big brakes are expensive, require more clearance (larger wheels on the car), and more maintinence.

Really the largest benefit of lightning a vehicle is the kinetic energy required to get it moving, reducing fuel consumption when accelerating. In the case of race cars, reducing weight increases acceleration with a set amount of power generation.
 
  • #4
NateTG
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In racing you can also use softer tires with a lighter car which improves the coefficient of friction. Lighter cars don't need the same amount brake ventilation (also power related) and lighter cars benefit more from aerodynamic downforce.

For suspension, lighter cars will have less unsprung weight, but heavier cars may have proportionally less, and I'm not sure which end comes out favored there.
 
  • #5
mheslep
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While lighter vehicles are easier to stop, technically stopping time has more to do with the capacity of the brakes, and the contact patch of the tires...
Yes. Modern brakes have ability to lock up the wheels immediately anything up to super large SUVs AFAIK, so this should give no advantage to ultralights.
You can make a relatively heavy vehicle stop very fast with big enough brakes; but big brakes are expensive, require more clearance (larger wheels on the car), and more maintenance.

Really the largest benefit of lightning a vehicle is the kinetic energy required to get it moving, reducing fuel consumption when accelerating. In the case of race cars, reducing weight increases acceleration with a set amount of power generation.
Yes of course. One of the oft cited reasons for not building cars that obtain these advantages is safety - low mass losing to high mass in collisions. Now there are claims that low mass vehicles have the advantage in stopping distance which helps the safety case and could allow the efficiency savings to go forward. Unfortunately I don't see how this (stopping dist) is accomplished.
 
  • #6
mheslep
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In racing you can also use softer tires with a lighter car which improves the coefficient of friction.
And also increases rolling resistance, which is antithetical to the concept of ultralights. I don't think that is how they get there?

Lighter cars don't need the same amount brake ventilation (also power related)
Yes I can see the reduced structural mass allows many other things like brakes to also grow smaller, but I don't see how that helps with safety and stopping distance?
 
  • #7
Mech_Engineer
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Now there are claims that low mass vehicles have the advantage in stopping distance which helps the safety case and could allow the efficiency savings to go forward. Unfortunately I don't see how this (stopping dist) is accomplished.

Yes I can see the reduced structural mass allows many other things like brakes to also grow smaller, but I don't see how that helps with safety and stopping distance?

The point is that there isn't any fundamental reason an ultra-light car can stop faster than a heavy one, as long as the brakes and tires on each car are sized appropriately. Theoretically, if you are using the same size brakes and tires on two different weight cars, the lighter one will stop faster; but, this argument is not really applicable to a vehicle that is being designed from scratch and can have brakes designed accordingly.

The advertising claim that super-light cars stop faster than heavy ones is really just trying to sell them; it isn't necessarily based in fact. It could be argued that it is easier and cheaper to make a light car stop quickly, but that's about it (and it's easier and cheaper to do most anything performance-based in a lightweight car).
 
  • #8
montoyas7940
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An ultralight aircraft can fly much slower and land at a steeper angle. The slow speed means less energy to dissipate with brakes. The steep approach angle means a more precise touchdown and better obstacle clearance at the end of the runway.

This is not the same subject but I have always thought it was interesting. A lightly loaded airplane will not glide as far as the same airplane heavily loaded.
 
  • #9
mheslep
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Did a bit of surfing and collected stopping distance specs as tested by Edmunds

Stopping distance from 60mph
BMW M3: 3726 lbs, 19" tires, 100 ft (best any vehicle Edmunds tested)
Jaguar XF: 4200lbs, 20" wheels: 108 ft
Pontiac G8 GT: 4000lbs, 109 ft
Audi A6: 114ft
Lexus LS 400: 4500lbs, 120 ft
VW Golf GTI (1998): 2800lbs, 139 ft
Jeep Wrangler Rubicon (year?): 165 ft
1997 Wrangler: 184 feet (rear drums)
2003 Wrangler: 167.4 feet (rear disc)
2007 Wrangler (4 door): 4592lbs 148 feet (rear disc, larger front disc)

So distance is all over the place, with little correlation to mass. I conclude then that this is all braking system and wheel/tire related, which further detracts from the claim by car ultra-lighters that they have an intrinsic advantage in stopping distance.
 
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  • #10
mgb_phys
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The UK driving test has a big list of stopping distances that you have to memorise. Of course you aren't asked what 73m looks like on the road - you just have to recite "18m reacting and 55m stopping at 60mph"
Many of the accidents on UK roads are presumably caused by drivers trying to use a theodolite to measure the distance to the car in front while driving.

A UK car show just found that the shortest stopping distance was for small sporty hatchbacks, typically < 25m from 70mph or a 1/3 the official distance.
This site lists the typical distances for lots of cars (100mkh = 62mph) http://www.movit.de/rahmen/stoptbl.htm [Broken]
 
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  • #11
mheslep
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thanks mgb_phys, very interesting as it lists the same vehicle 'empty' and 'fully loaded'. The fully loaded cases looks to be on average 3-4M longer and in some cases 15-20M longer! This then supports the case of the ultralight vehicle designers: they can stop shorter. I'm at a loss to explain why!
 
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  • #12
Mech_Engineer
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thanks mgb_phys, very interesting as it lists the same vehicle 'empty' and 'fully loaded'. The fully loaded cases looks to be on average 3-4M longer and in some cases 15-20M longer! This then supports the case of the ultralight vehicle designers: they can stop shorter. I'm at a loss to explain why!

No, what this proves is that increasing the weight while keeping the same brakes means the vehicle will take longer to stop. This is because brakes have an associated "power rating," which can be thought of in terms of horsepower or watts.

Since the brakes at maximum clamping force can only convert a specific amount of kinetic energy per second to heat, having more weight means more kinetic energy which in turn means it takes longer to convert all of the kinetic energy to heat.
 
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  • #13
mheslep
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thanks mgb_phys, very interesting as it lists the same vehicle 'empty' and 'fully loaded'. The fully loaded cases looks to be on average 3-4M longer and in some cases 15-20M longer! This then supports the case of the ultralight vehicle designers: they can stop shorter. I'm at a loss to explain why!

No, what this proves is that increasing the weight while keeping the same brakes means the vehicle will take longer to stop.
Isn't that what I asserted, expressed as a negative?

This is because brakes have an associated "power rating," which can in some cases be thought of in terms of horsepower or watts.

Since the brakes at maximum clamping force can only convert a specific amount of kinetic energy per second to heat, having more weight means more kinetic energy which in turn means it takes longer to convert all of the kinetic energy to heat.
Ah. Ok, I suppose I knew this but was skimming by it. You've expressed it clearly here and exposes my misconception: the original distance = v^2 / (Cf*g) equation, independent of mass, is not reflective of modern reality. That equation is derived assuming dynamic friction (locked, skidding tires) where the vehicle mass directly controls the stopping force due to friction. The modern reality is unlocked tires and the stopping force is due solely to the brake pressure, so that the stopping force is mostly independent of vehicle mass in the case of tire/surface static friction, rather is dependent on the brake pad pressure. In that case, for given brake horsepower and anti-lock braking, the lighter the vehicle the sooner it stops.

Edit: Mech_engineer - I see you had been saying essentially this above already; I missed the point because I was too focused on that equation. Thanks.
 
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  • #14
mheslep
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The point is that there isn't any fundamental reason an ultra-light car can stop faster than a heavy one, as long as the brakes and tires on each car are sized appropriately. Theoretically, if you are using the same size brakes and tires on two different weight cars, the lighter one will stop faster; but, this argument is not really applicable to a vehicle that is being designed from scratch and can have brakes designed accordingly.

The advertising claim that super-light cars stop faster than heavy ones is really just trying to sell them; it isn't necessarily based in fact. It could be argued that it is easier and cheaper to make a light car stop quickly, but that's about it (and it's easier and cheaper to do most anything performance-based in a lightweight car).
I think the point of the ultralight vehicles is that they can afford to put the same HP braking system in their 1600lb vehicle (planned) as is used in say a comparably sized 3000lb vehicle and thus they'll stop dramatically shorter. The ultralight designers referenced in the OP article are keenly aware of safety criticisms in their design so they are planning to take advantage of stopping distance.
 
  • #15
Mech_Engineer
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One of my favorite articles Road and Track has ever published is the August 2003 "http://www.roadandtrack.com/article.asp?section_id=3&article_id=663&page_number=1" [Broken]" where they have a 0-100-0 deathmatch. Basically they haul butt to 100 mph as fast as they possibly can, and then slam on the brakes to get back to zero. Shortest time in each class wins.

Not only is it incredible the amount of time some of the vehicles take to do it, but lots of useful data was recorded about each vehicle, that can be used to compare them in a sort of an apples to apples test.

Here is an interesting graph from that article:

http://www.roadandtrack.com/assets/image/7162003125156.gif [Broken]
http://www.roadandtrack.com/article.asp?section_id=3&article_id=663&page_number=5 [Broken]

The Viper puts down an average of 237 hp getting from 0-100 mph, but has an average braking power from 100-0 mph of 547 hp. Looking at the graph we can see the braking curve is very linear, so we probably have a good estimate of the braking system's maximum power dissipation (taking into account traction available from the tires as well)...

But look at this next graph:

http://www.roadandtrack.com/assets/image/7162003125016.gif [Broken]
http://www.roadandtrack.com/article.asp?section_id=3&article_id=663&page_number=9 [Broken]

In the "exotic" class, the Saleen S7 is pitted against the Lamborghini Murcielago. The S7 weighs in at 3050 lb, a full 1140 lbs lighter than the Lamborghini. Yet, the Lamborghini stops 70 feet shorter and 0.8 seconds faster from 100mph than the S7. Why?

Both cars have the exact same tires fitted (Pirelli P Zero Rosso's, 245/ 35ZR-18 front and 335/ 30ZR-18 rear), so the answer has to be a combination of more traction available to the Lamborghini because it weighs more, and the fact that the Saleen does not have ABS. The Saleen should have more braking power available, since it has 1" larger discs in the front and 0.8" larger dics in the rear, but its traction is limited by its lighter weight, and its lack of ABS causes the tires to lock up easily...

The effects of no ABS can be seen in the graph, where the Lamborghini's braking curve is completely linear all the way to from 100 to 0 mph, while the Saleen's fluctuates wildly since the driver has to modulate the pedal to try and make up for the lack of ABS. Even though the Saleen was much faster to 100 mph, it ironically loses the 0-100-0 because the Lamborghini is HEAVIER (more traction available from the ame set of tires) and has ABS. The Lamborghini puts down an average of 606 braking hp, versus the Saleen's "paltry" 370 braking hp.

So there you have it, a case where being heavier means a shorter stopping distance... :wink:
 
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  • #16
mheslep
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It may be important to point out that the ultralight concept car proposed in the Moore - Lovins paper not some kind of tiny toy car. It is a five-six seater roomy design and comparable in passenger room to the Ford Taurus, and thus Moore-Levins has room for Taurus sized brakes. However the Moore-Lovins design is 854kg and the Taurus is 1423kg.
 
  • #17
mheslep
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One of my favorite articles Road and Track has ever published is the 2003 "Power Trip" where they have a 0-100-0 deathmatch. Basically they haul butt to 100 mph as fast as they possibly can, and then slam on the brakes to get back to zero. Shortest time in each class wins.

Not only is it incredible the amount of time some of the vehicles take to do it, but lots of useful data was recorded about each vehicle, that can be used to compare them in a sort of an apples to apples test.

Here is an interesting graph from that article:

http://www.roadandtrack.com/assets/image/7162003125156.gif [Broken]
http://www.roadandtrack.com/article.asp?section_id=3&article_id=663&page_number=5 [Broken]

The Viper puts down an average of 237 hp getting from 0-100 mph, but has an average braking power from 100-0 mph of 547 hp. Looking at the graph we can see the braking curve is very linear, so we probably have a good estimate of the braking system's maximum power dissipation (taking into account traction available from the tires as well)...
Interesting, note the stopping g's: I'll call the 911 stopping time from 100 mph ~4.4 secs so that is just over one G. I wonder if there is an upper G limit beyond which it doesn't improve safety on average to stop any faster. That is, lots of minor injuries in numerous high G stops - no impact vs severe injuries in the relatively rare impact.
 
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  • #18
Mech_Engineer
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So I tried charting the results of the Road & Track tests to see if there were any correlations...

It appears that available braking power linearly correlates to vehicle weight (regardless of brake size, and ignoring the S7 due to lack of ABS), perhaps due to additional friction available with the ground? It looks like this is because all of the braking systems are powerful enough to lock up the tires given the chance, and so instead braking force depends on tire compound, F/R weight distribution, and vehicle weight. The Mercedes-Benz SL55 AMG is the heaviest vehicle in the article at 4520lbs, but also puts down the most braking power at 638 hp, and stopping from 100-0 in 312 feet.

Also, Braking time/distance very slightly correlate to vehicle weight. It is my contention that this is because this is a very narrow sample of vehicles on the market, and they are all performance-oriented vehicles with powerful multi-piston 4-wheel disc brakes. If a large sample was taken including trucks, SUVs, economy cars, and super-lightweights, I suspect the correlation could possibly disappear or at least become less available.
 

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  • #19
Mech_Engineer
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Interesting, note the stopping g's: I'll call the 911 stopping time from 100 mph ~4.4 secs so that is just over one G. I wonder if there is an upper G limit beyond which it doesn't improve safety on average to stop any faster. That is, lots of minor injuries in numerous high G stops - no impact vs severe injuries in the relatively rare impact.

I don't think it has to do with injuries so much as available coefficient of friction between the tires and the road. I suspect passengers could go through a 2g stop without suffering major injury, but the friction required for that kind of force would require very sticky tires that would be expensive and have horrible wear characteristics.
 
  • #20
mheslep
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So I tried charting the results of the Road & Track tests to see if there were any correlations...
nice plots, thanks for posting.

It appears that available braking power linearly correlates to vehicle weight (regardless of brake size, and ignoring the S7 due to lack of ABS), perhaps due to additional friction available with the ground? It looks like this is because all of the braking systems are powerful enough to lock up the tires given the chance, and so instead braking force depends on tire compound, F/R weight distribution, and vehicle weight. The Mercedes-Benz SL55 AMG is the heaviest vehicle in the article at 4520lbs, but also puts down the most braking power at 638 hp, and stopping from 100-0 in 312 feet.
I assume you derive braking HP as the kinetic energy at 100mph / time to stop. That power would be provided by either the work of the tires against the road, or the brake pads if the tires are not locked, and we don't necessarily know which is the case. Hopefully with ABS or a very good R&T driver most of the work is done by the brakes, plus what Ill call a small 'braking rolling resistance?' work from the increased tire / road surface under braking conditions.

Also, Braking time/distance very slightly correlate to vehicle weight. It is my contention that this is because this is a very narrow sample of vehicles on the market, and they are all performance-oriented vehicles with powerful multi-piston 4-wheel disc brakes. If a large sample was taken including trucks, SUVs, economy cars, and super-lightweights, I suspect the correlation could possibly disappear or at least become less available.
There is only one stock super light made that I know of, the Mercedes carbon fiber McLaren. I'll take a look at its numbers time permitting. The rest is all still concept.
 
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  • #22
shamrock5585
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duh!

Ok everyone... i understand you are all hardcore mathematicians and love to show off your enormous brain muscles but can we keep things simple sometimes... very simple stuff here! It was said right in the opening statement... he answered his own question... kinetic energy is dependant of mass at a given velocity... therefore, more mass = more energy... if you have more energy then it will require more energy to stop... very easy to see... have you ever wondered why it takes trains a long time to stop?
 
  • #23
Mech_Engineer
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Ok everyone... i understand you are all hardcore mathematicians and love to show off your enormous brain muscles but can we keep things simple sometimes... very simple stuff here! It was said right in the opening statement... he answered his own question... kinetic energy is dependant of mass at a given velocity... therefore, more mass = more energy... if you have more energy then it will require more energy to stop... very easy to see... have you ever wondered why it takes trains a long time to stop?

Flexing your brain muscles is healthy.

We all obviously know that heavier vehicles have more kinetic energy. The point of this thread is that some manufacturers are advertising superlight vehicles as safer because they stop faster, which isn't necessarily true since it takes more than mass to determine how fast a car can stop.
 
  • #24
shamrock5585
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true it takes more than mass to determine how fast the car can stop... yes you will have less friction, but with some good brakes you can stop faster than a heavy car with the same breaks... also advertisement is retarded... they advertise that the car can stop faster... stop what? stop driving under 60 miles per hour?
 
  • #25
Mech_Engineer
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yes you will have less friction, but with some good brakes you can stop faster than a heavy car with the same breaks...

Unless, the brakes being used are strong enough to lock up the tires on both cars.

The braking time/distance of a vehicle is directly tied to how much decelerating force the vehicle's tires can impart on the road. If the brakes are so strong that full force causes the tires to lock up, the brakes' power has to be modulated through an ABS system. A car stops the fastest when its tires are imparting the maximum force possible without locking, so just taking the brakes off a heavier car and putting them on a lighter one won't necessarily make it stop faster; the lighter car can still only stop as quickly as is allowed by not locking up the tires.
 
  • #26
shamrock5585
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fair enough... seeing as almost all cars these days have abs, they're advertisement would be true. that is a pretty interesting concept tho that by locking the tires the stopping distance could be independant of mass. it is kind of a hard concept to grasp. i don't know if i buy it. if a really light vehicle and a tractor trailor are driving the same speed and they lock up the brakes, i would think that the tractor trailor would skid further even though it has more tires, hence more contact to the road resulting in more friction... but then again I've never seen this happen... sounds like something to send to mythbusters!
 
  • #27
Mech_Engineer
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fair enough... seeing as almost all cars these days have abs, they're advertisement would be true.

The thing I've been trying to argue this whole time is that there is not fundamental physical property that dictates a significantly lighter car will stop significantly faster than a short one. You can stop any vehicle as fast as any other vehicle, given a properly designed braking system and similar tire compounds.

You have two trade-offs; on the one hand your kinetic energy is proportional to the vehicle's mass, on the other hand the vehicle's available frictional force is proportional to its mass. These two properties tend to balance each other such that when a vehicle's mass increases, its kinetic energy increases proportionally AND the fricitonal force available to decelerate it increases proportionally as well. So as long as the brakes on the vehicle are strong enough to be able to lock up the wheels (e.g. they have been "properly sized"), the largest factor dictating stopping distance is available tire friction, not mass.

that is a pretty interesting concept tho that by locking the tires the stopping distance could be independant of mass. it is kind of a hard concept to grasp. i don't know if i buy it. if a really light vehicle and a tractor trailor are driving the same speed and they lock up the brakes, i would think that the tractor trailor would skid further even though it has more tires, hence more contact to the road resulting in more friction... but then again I've never seen this happen... sounds like something to send to mythbusters!

The two vehicles would stop in the same distance as long as the coefficient of friction between the road and the tires was the same for the semi-truck and the car. In practical applications this wouldn't necessarily be true, since there tend to be large variations in friction depending on what brand/type of tires are being used.

Stopping distance being independent of mass is similar to the fact that friction is independent of area. If you derived the simplified equation for stopping distance using the kinematic equations, you would see that the mass term in the calculation for friction, and the mass term in the calculation of kinetic energy cabcel out, leaving only g, Cf, and v as variables.
 
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  • #28
shamrock5585
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so basically your saying they could make mac trucks that can stop real easy like a vw bug but they choose to use crappy tires and breaks so that the truck takes a while to stop... I am sure companies don't want to protect theyre precious cargo or anything...
 
  • #29
Mech_Engineer
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shamrock5585 said:
so basically your saying they could make mac trucks that can stop real easy like a vw bug but they choose to use crappy tires and breaks so that the truck takes a while to stop... I am sure companies don't want to protect theyre precious cargo or anything...

I didn't say anything of the sort. Please read my previous post:

... It could be argued that it is easier and cheaper to make a light car stop quickly, but that's about it (and it's easier and cheaper to do most anything performance-based in a lightweight car).

It is easier to make a lighter car stop faster, but "heavy" cars can as well given the right brakes.

I never said it was EASY to stop heavy vehicles. You are of course using the most extreme example on the road, but even semi-trucks post some impressive braking numbers considering their weight (semis might weigh 20-30 times as much as many cars on the road, yet their braking distances are not 20-30times as large). Additionally, semi-trucks use low rolling resistance high-mileage tires to lower maintinence and fuel costs, but these tires also have harder rubber compounds and thus lower frictional coefficients.
 
  • #30
Mech_Engineer
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Some more on semi truck braking-

According to the National Safety Council's Defensive Driving Course for Professional Truck Drivers, the stopping distance for an 80,000 pound tractor trailer, traveling at 60 mph, is 426 feet.
http://www.truck-accident-lawfirm.com/CM/Resources/Trucking-Accidents-Collision.asp [Broken]

A 5000lb Chevy Silverado takes 128 feet to stop from 60 mph. So a semi tractor trailer weighing 16 times that of a Silverado requires only 3.3 times the stopping distance (conservatively). That sounds like pretty serious braking performance to me...

Additionally, Federal Motor Vehicle Safety Standard No. 121, Air brake systems states that a truck tractor loaded with an unbraked control trailer (trailer that loads the tractor to its GVWR) shall stop from 60mph in less than 355 feet. So that same truck, assuming its GVWR is indeed 80,000lbs and the truck is not overloaded, should be able to stop from 60 mph in 355 ft without the aid of its trailer's brakes, a mere 2.6 times that of the Silverado.

http://www.access.gpo.gov/nara/cfr/waisidx_01/49cfr571_01.html
 
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  • #31
shamrock5585
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ok fools here's the math to prove it... you proved that your work from the friction and the momentum of the vehicle add up to be your total force and that will be independant of mass... but then we got good old Newton on my side... F=m*a now the force could end up being the same if your math is correct... but if you divide the equal force by a larger mass then your acceleration (negative so its deceleration) will be smaller therefore the care with less mass will have a faster deceleration!
 
  • #32
Mech_Engineer
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I don't know what it is you're not getting, but your "math" was ineffective to say the least. You lost me at adding work and momentum to get force...
 
  • #33
mheslep
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fair enough... seeing as almost all cars these days have abs, they're advertisement would be true. that is a pretty interesting concept tho that by locking the tires the stopping distance could be independant of mass. it is kind of a hard concept to grasp. i don't know if i buy it. ...
Yes, IIRC, when I learned to drive back sometime in Model T days and everyone just skidded on locked tires to a stop, they used to just declare in driving courses that all 4 wheel cars (regardless of weight) stopped from 60mph on average in something like 160ft (i think) on dry payment; some multiple of that on wet pavement, etc. You had to know it for the test.
 
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  • #34
Stingray
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ok fools here's the math to prove it... you proved that your work from the friction and the momentum of the vehicle add up to be your total force and that will be independant of mass... but then we got good old Newton on my side... F=m*a now the force could end up being the same if your math is correct... but if you divide the equal force by a larger mass then your acceleration (negative so its deceleration) will be smaller therefore the care with less mass will have a faster deceleration!

Lose the attitude. You don't know what you're talking about.

To a first approximation, every car can brake at the capacity of its tires (at least if you don't keep doing it so much that everything overheats). The braking force those tires generate is roughly proportional to the vertical load on them. The total force is therefore proportional to the car's mass. Newton's second law then says that the maximum deceleration is independent of mass.

This is not the greatest approximation in real life, but it is pretty much correct that properly designed vehicles all stop in similar distances on similar tires. One problem in real life is distributing the braking force optimally to all tires. No fixed ratio between the front and rear axles will always be best. This is largely fixed nowadays with electronic brake distribution, but is not a trivial problem. There are also tradeoffs between stopping distance and stability. Locking up the rear tires and not the fronts will cause a car to spin, for example.

It is also not quite true that the maximum braking force a tire can produce is directly proportional to load. The effective friction coefficient actually decreases a little at higher loads. This is a relatively small effect, but very important in racing.
 
  • #35
shamrock5585
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The braking force those tires generate is roughly proportional to the vertical load on them. The total force is therefore proportional to the car's mass. Newton's second law then says that the maximum deceleration is independent of mass.

ok the original argument says that, from the math... your friction force working against you and your force due to momentum will end up being the same for small and large vehicles because in the equations mass cancels out... but that's just your force... so your force, which is the same for a mac truck and a small car, can be plugged into F=m*a right? so for a small car your force is divided by a smaller number... rendering your deceleration to be larger which means you will stop faster... i don't see a flaw in that logic but if I am wrong I am wrong
 

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