Any idea 'roughly' how much generic cromoly forks weigh?

Just trying to work out how much weight I've taken off the bike by swapping out the old forks. The old forks were generic cromoly, think basic Columbus, probably slightly heavier than 531. I'm only after an approximate. Cheers :-)

...weigh it?

A full carbon fork replacing a steel fork can take off a little less than a pound

In google we trust ;)
http://www.smartcycles.com/bike_weight.htm

Weigh the bike with the new forks, then weigh the bike with the old forks, subtract the latter from the former to see how much weight saved.

Not difficult is it ;)

maybe 8

maybe 8

Yahoo answers comes to the rescue.
http://answers.yahoo.com/question/index?qid=20080413212642AAXxz2C
1 lB = 2 - 3 medium sized potatoes.

Carbon is aproxametley 3.5 rads lighter than cromoly.
i think it would be somewhere around rad to the power of max +1 mark.

Yahoo answers comes to the rescue.
http://answers.yahoo.com/question/index?qid=20080413212642AAXxz2C
1 lB = 2 - 3 medium sized potatoes.

those were small/ish ones though

Carbon is aproxametley 3.5 rads lighter than cromoly.
i think it would be somewhere around rad to the power of max +1 mark.

Yeah, but how many potatoes is that?

those were small/ish ones though

New potatoes?

rad= radish's, he's talking shite again

i disagree.

about a can of coke off a plane

1 lB = 2/3 potatoes

  = 3.5 cups of sugar
  = 3.75 cups of flour
  = 453.6 grams
  = this many dried mushrooms

i'll go along with that*

*medium sized??

Train A, traveling 70 miles per hour (mph), leaves Westford heading toward Eastford, 260 miles away. At the same time Train B, traveling 60 mph, leaves Eastford heading toward Westford. When do the two trains meet? How far from each city do they meet?

                               ![](http://mathforum.org/dr.math/faq/faq.two.trains.gif)

Train A, traveling 70 miles per hour (mph), leaves Westford heading toward Eastford, 260 miles away. At the same time Train B, traveling 60 mph, leaves Eastford heading toward Westford. When do the two trains meet? How far from each city do they meet?

                               ![](http://mathforum.org/dr.math/faq/faq.two.trains.gif)

Who cares?! There's only one track! They're all going to die!

that, may be true, but it is not the correct answer.

To solve this problem, we'll use the distance formula:

Distance = Rate x Time

Since an equation remains true as long as we perform the same operation on both sides, we can divide both sides by rate:

Distance
---------- = Time
Rate

or by time:

Distance
---------- = Rate
Time

So rate is defined as distance divided by time, which is a ratio.
Speed is another word that is used for rate. When a problem says that a train is moving at a speed of 40 mph, you can understand this to mean that the train's rate is 40 mph, which means it will travel 40 miles in one hour.

Let's start by listing the information given:

Speed of Train A: 70 mph
Speed of Train B: 60 mph
Distance between Westford and Eastford: 260 miles

Method I: We'll use the notion of relative speed 1 (or relative rate) in order to express the rates of the two trains in one number that can then be used in the distance formula. Imagine you're on Train A. You're going 70 mph, so your speed relative to the trees, houses, and other non-moving things outside the train is 70 mph. (All of those objects look as if they're going by at 70 mph.) Now imagine you're the engineer and you can see Train B coming toward you - not on the same track, of course! Since Train B is moving 60 mph, it will look as if it's approaching faster than if it were sitting still in the station - a lot faster than the trees and houses appear to be moving.
The relative speed of the two trains is the sum of the speeds they are traveling. (If you're on either of the trains, this is the speed you appear to be moving when you see the other train.) In our problem, the relative speed of the two trains is 70 mph + 60 mph = 130 mph. What if the trains were traveling in the same direction? Then we'd need to subtract the speed of the slower train from the speed of the faster train, and their relative speed would be 10 mph.
At this point we know two of the three unknowns: rate and distance, so we can solve the problem for time. Remember that time = distance/rate, the distance traveled is 260 miles, and the relative speed is 130 mph:

t = 260 miles/130 mph

t = 2 hrs.

[INDENT] We find that the trains meet two hours after leaving their respective cities. [/INDENT]

Cool, I always wondered how to calculate that.
Never had it in school.

Weigh the bike with the new forks, then weigh the bike with the old forks, subtract the latter from the former to see how much weight saved.

Not difficult is it ;)

Correct. If the old forks hadn't have gone to the great God egay, many moons ago ;)

I saw the aswer here:
http://www.londonfgss.com/post717922-3.html

[*]t = 260 miles/130 mph
t = 2 hrs.

[INDENT] We find that the trains meet two hours after leaving their respective cities. [/INDENT]

it's true look

IIIIIIIIIIIII0IIIIIIIIIIII

I saw the aswer here:
http://www.londonfgss.com/post717922-3.html

Cool, cool, thanks Smallfurry that's eggzackly the answer I was looking for :-)

that, may be true, but it is not the correct answer.

To solve this problem, we'll use the distance formula:

[*]Distance = Rate x Time

Since an equation remains true as long as we perform the same operation on both sides, we can divide both sides by rate:

[*]

[*]Distance
---------- = Time
Rate

or by time:

[*]

[*]Distance
---------- = Rate
Time

So rate is defined as distance divided by time, which is a ratio.
Speed is another word that is used for rate. When a problem says that a train is moving at a speed of 40 mph, you can understand this to mean that the train's rate is 40 mph, which means it will travel 40 miles in one hour.

Let's start by listing the information given:

[*]Speed of Train A: 70 mph
Speed of Train B: 60 mph
Distance between Westford and Eastford: 260 miles

Method I: We'll use the notion of relative speed 1 (or relative rate) in order to express the rates of the two trains in one number that can then be used in the distance formula. Imagine you're on Train A. You're going 70 mph, so your speed relative to the trees, houses, and other non-moving things outside the train is 70 mph. (All of those objects look as if they're going by at 70 mph.) Now imagine you're the engineer and you can see Train B coming toward you - not on the same track, of course! Since Train B is moving 60 mph, it will look as if it's approaching faster than if it were sitting still in the station - a lot faster than the trees and houses appear to be moving.
The relative speed of the two trains is the sum of the speeds they are traveling. (If you're on either of the trains, this is the speed you appear to be moving when you see the other train.) In our problem, the relative speed of the two trains is 70 mph + 60 mph = 130 mph. What if the trains were traveling in the same direction? Then we'd need to subtract the speed of the slower train from the speed of the faster train, and their relative speed would be 10 mph.
At this point we know two of the three unknowns: rate and distance, so we can solve the problem for time. Remember that time = distance/rate, the distance traveled is 260 miles, and the relative speed is 130 mph:

[*]t = 260 miles/130 mph
t = 2 hrs.

[INDENT] We find that the trains meet two hours after leaving their respective cities. [/INDENT]

WRONG.
Dave was in Milton Keynes, and Robina was is Cardiff, so the question is irrelevant