The technical answer is that displacement is the vector sum of the distances.
An example to illustrate the difference in less technical terms, distance travelled in one direction added to the same distance in the opposite direction will result in the total distance being twice the distance of each leg but the total displacement is 0.
After traversing 1/2 of a circular track with radius 'R', the body has effectively moved from one end of a diameter to the other end of the same diameter. The distance traveled is 1/2 the circumference = (pi)D/2 = (pi)R. The displacement is D = 2R. The ratio of displacement to distance = (2R)/(piR) = 2/pi= 0.63662 (rounded), independent of 'R'.
A degree is an angular measure and cannot be measured in millimetres. A 1 degree rise can be interpreted as a ratio of a rise (in millimetres) per a distance of horizontal displacement.
5 to 1 is greater
compression ratio = compressed size / uncompressed size the ratio should be between 1 and 0 (multiply with 100 to get the ratio in percent) a ratio greater than 1 means, the compressed size is actually greater than the uncompressed size a ratio just below 1 means bad compression the lower the ratio, the better the compression
Any point where x/y is greater than 1 has a ratio larger than one. For example, the point (2, 1) has a ratio of 2:1, or 2. (3, 1) has a ratio of 3, etc.
There's no firm relationship between the magnitudes of distance and displacement, except that displacement can never be greater than distance. So if you're looking for a ratio, I guess (distance)/(displacement) = or > 1
The ratio is 1.
There's no way to answer that, because it can be a different number in every situation. It can never be greater than ' 1 ', but the actual number depends on how squiggly the route is between the starting point and the ending point.
1 is to 1
Greater compression = greater fuel consumption = greater power
It's not always the same number. But whatever that ratio happens to be in a specific situation, it can never be less than ' 1 '.
After traversing 1/2 of a circular track with radius 'R', the body has effectively moved from one end of a diameter to the other end of the same diameter. The distance traveled is 1/2 the circumference = (pi)D/2 = (pi)R. The displacement is D = 2R. The ratio of displacement to distance = (2R)/(piR) = 2/pi= 0.63662 (rounded), independent of 'R'.
These have to do with the hydraulic density. In automobile, the density increases therefore making the hydraulic ratio to displace more than any hydraulic ratio in a displacement.
No, it's not possible to have an RF value greater than one. Rf values are a ratio of the distance a spot has traveled to the total distance of the solvent front. Since the distance of the front is always greater than the distance of a spot, the Rf value is always less than 1.
Distance input --------------- = Speed Ratio Distance output The distance input divided by the distance output equals the Speed Ratio.
There isn't a set displacement to horsepower ratio - hence, horsepower can't be determined by displacement alone.
It is ratio.