how would you explain ... a puzzle

T

The Gecko_imp

Been there
Thread starter
The last half mile towards my house is a gentle, flat, bump-less piece of tarmac. For years I am used to accelerating to 45 mph after the last intersection, switch to neutral and role right into my garage without power ... in my car. Speed “on final” has been adjusted to a “perfect” 36 mph passing a traffic sign, all I have to do is push the remote for the garage door. Sometimes I turn the ignition off right after the sign.

Now I tried this a few times on my CB. Guess what, she does not even get close to the garage. Not at 36, not at 40, barely at 45.

How do you explain this taken into account :
- my brake pads and disks show no unusual signs of wear
- gas mileage on the CB is an average 50 and seems good
- my gas consumption in the car is about 25% higher
- gross weight is 5 times the bike
- bike’s tires are good (not new or worn)

Where is this friction on the bike coming from ? Why does it not decrease gas mileage?
Thanks for helping with the puzzle.
 
1/2 a mile? When I'm coming home really late and don't want to wake the neighbors, I'll kill the engine and try to coast to my garage. That's proabably a bit less than 100 yards and I can't coast the entire way.
 
Thread starter
Gone in 60_imp said:
1/2 a mile? When I'm coming home really late and don't want to wake the neighbors, I'll kill the engine and try to coast to my garage. That's proabably a bit less than 100 yards and I can't coast the entire way.
Click to expand...
Admitted, my 1/2 mile is more like 1/4 mile ...
m in sc_imp said:
the car has a lot more mass therefore will coast longer @ same speed. just basic physics.
Click to expand...
okay, back to basics,
energy e=mv^2, so much more in the car,
but that ain’t what matters, remember two stones (big and small) falling from a tower arrive at the same time at the ground because the acceleration is the same. What matters here is the deceleration (negative acceleration) d equal some factor for the friction times the speed (not the mass or energy’), so it comes to d = f x v.
If the friction in the tires, wheels, bearings, drag .... etc was the same, car and bike would decelerate the same way.
Therefore the question ... can the slick bike, with two small tires, and light weight slow down faster ?
Nachodaddy_imp said:
You obviously need to gain some weight.
Click to expand...
Nacho, if you knew my “figure” you would understand that any extra pound at any part of my frame would be a sin !
 
I am not a physicist or engineer, but I think due to tires width, tread, size, construction and number of tires(4), a car has less rolling resistance, meaning less weight per sq inch.
A lot of cars are equipped with low rolling resistance tires to achieve a better fuel economy.
Never heard about a moto tire, maybe there are??
 
Only because I like pedantic arguments like this, I'll "weigh" in.

The simplest way to think about it is in terms of deceleration. Since both the motorcycle and the car start out at the same velocity, if the motorcycle travels less distance, it's because it decelerated to zero quicker. Why would that be? Since F=m*a, a=F/M where F is frictional and aerodynamic force in the rearward direction. If that force were identical between the car and the motorcycle, clearly the motorcycle would decelerate quicker since its mass is less.

The frictional and aerodynamic force for the car is certainly going to be more than for a motorcycle of course, but not enough in comparison with the mass difference. So the car takes longer to decelerate to zero. Same holds true for m-in-sc's train example.

There are certainly examples though of vehicles that weigh more, but travel shorter distances from speed than some other vehicle that weighs less. For most wheeled vehicles though, the heavier ones will decelerate more slowly due to frictional and drag forces.
 
People, already posted - inertial storage.

Think: Freight train stoppage.

There is a whole lotta friction during the best of times without asserting the brakes. Look how long they take to stop these not-so-light long bricks-in-the-wind even with ALL brakes asserted (a very long distance).
 
GoldOxide_imp said:
People, already posted - inertial storage.

Think: Freight train stoppage.

There is a whole lotta friction during the best of times without asserting the brakes. Look how long they take to stop these not-so-light long bricks-in-the-wind even with ALL brakes asserted (a very long distance).
Click to expand...
Well, yes in that case....but if you had a "bike" with skinny hard wheels and you disconnected the chain, and you towed it up to 40 mph, it would potentially go as far or farther than a car if you reduced friction enough. The main point is it's the relative ratio of friction and mass. A hockey puck on ice may travel farther than a car coasting on ice given the same starting speed (see...I brought it down to a level that you Canadians can grok :) )
 
tommymck_imp said:
GoldOxide_imp said:
People, already posted - inertial storage.

Think: Freight train stoppage.

There is a whole lotta friction during the best of times without asserting the brakes. Look how long they take to stop these not-so-light long bricks-in-the-wind even with ALL brakes asserted (a very long distance).
Click to expand...
Well, yes in that case....but if you had a "bike" with skinny hard wheels and you disconnected the chain, and you towed it up to 40 mph, it would potentially go as far or farther than a car if you reduced friction enough. The main point is it's the relative ratio of friction and mass. A hockey puck on ice may travel farther than a car coasting on ice given the same starting speed (see...I brought it down to a level that you Canadians can grok :) )
Click to expand...
Heh, heh, ... grok, ... I think I understand now. :)
 
My boat is '17...my friends is '20...same width. His out weighs mine by 2000lbs. Given the same speed, it takes a lot longer for his to come to a stop than it does mine. That's why he always gets upset with me when when I drive his boat and he wants to ski. He'll be in the water after a fall or get-off, waiting for me to come around. I keep nearly running him over when I drive his boat because it takes longer to stop. Usually I hit reverse in time.
 
m in sc_imp said:
having have slid a braked car along an iced road for an unusually long distance with the brakes locked i'd have to disagree.
I still think the car 'wins'.
mass will be a greater factor than coefficient of friction between the 2 examples.
5000 lbs car on 4 tires has less c.f. per "lbs" than a 1? lb puck with its full side dragging on the ice.
Click to expand...
Coast, not brake. Coefficient of static friction is less than coefficient of kinetic friction and if the tires are flatish and there is other losses typical of a coasting car, the puck will go farther. Regardless, the point is that if you lower the deceleration of the lighter object by enough to overcome the mass difference, it'll go farther. It's just a question of the ratio of F/M. In the extreme, add a passenger in a car, repeat the experiment using the same (now heavier) car with brakes applied and it'll come to a stop faster even though the car has less mass. It's a question of degree.
 
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