Was the transatlantic crossing for Concorde too short to reach optimal cruising altitude?
$begingroup$
When listening to Gander/Shannon ATC on shortwave, you could hear Concorde communicate its planned flight levels at longitudes from 20 West till 50 West. What I remember is that it would continue climbing to, say, flight level 570 at 30 West and then descent. So, it seems that it never reached cruising altitude, it would climb until halfway on the Atlantic and then start to descent again.
altitude concorde
edited Nov 13 '18 at 21:44
Ari Brodsky
1093
asked Nov 13 '18 at 18:25
Count IblisCount Iblis
22626
$endgroup$
add a comment |
$begingroup$
When listening to Gander/Shannon ATC on shortwave, you could hear Concorde communicate its planned flight levels at longitudes from 20 West till 50 West. What I remember is that it would continue climbing to, say, flight level 570 at 30 West and then descent. So, it seems that it never reached cruising altitude, it would climb until halfway on the Atlantic and then start to descent again.
altitude concorde
edited Nov 13 '18 at 21:44
Ari Brodsky
1093
asked Nov 13 '18 at 18:25
Count IblisCount Iblis
22626
$endgroup$
add a comment |
$begingroup$
When listening to Gander/Shannon ATC on shortwave, you could hear Concorde communicate its planned flight levels at longitudes from 20 West till 50 West. What I remember is that it would continue climbing to, say, flight level 570 at 30 West and then descent. So, it seems that it never reached cruising altitude, it would climb until halfway on the Atlantic and then start to descent again.
altitude concorde
edited Nov 13 '18 at 21:44
Ari Brodsky
1093
asked Nov 13 '18 at 18:25
Count IblisCount Iblis
22626
$endgroup$
When listening to Gander/Shannon ATC on shortwave, you could hear Concorde communicate its planned flight levels at longitudes from 20 West till 50 West. What I remember is that it would continue climbing to, say, flight level 570 at 30 West and then descent. So, it seems that it never reached cruising altitude, it would climb until halfway on the Atlantic and then start to descent again.
altitude concorde
altitude concorde
edited Nov 13 '18 at 21:44
Ari Brodsky
1093
asked Nov 13 '18 at 18:25
Count IblisCount Iblis
22626
edited Nov 13 '18 at 21:44
Ari Brodsky
1093
asked Nov 13 '18 at 18:25
Count IblisCount Iblis
22626
edited Nov 13 '18 at 21:44
Ari Brodsky
1093
edited Nov 13 '18 at 21:44
Ari Brodsky
1093
edited Nov 13 '18 at 21:44
Ari Brodsky
1093
1093
asked Nov 13 '18 at 18:25
Count IblisCount Iblis
22626
asked Nov 13 '18 at 18:25
Count IblisCount Iblis
22626
asked Nov 13 '18 at 18:25
Count IblisCount Iblis
22626
22626
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The simple answer is that the Concorde had no single assigned altitude, it was allowed to climb freely above ~FL450; this is discussed in depth in episode 166 – Flying the Concorde (worth the listen as it answers just about every Concorde question!). As @pilothead alludes to in their answer it climbed as it burned fuel but the aircraft never actually initiated a climb, it simply drifted up as it burned fuel and became lighter.
Also discussed in the episode is the complex approach and departure procedure. Due to the fuel burn schedule the Concorde did not really have the ability to hold for more than a single lap in a holding pattern or do a stepped climb with some route adjustments like many airliners. They had a special departure procedure that was more or less runway to cruise with no interruptions and a similar descent option. So the flight was effectively a climb to cruise block then a glide down right to landing. Depending on the wind and conditions of any given day as well as the load on board, the cruise altitude could vary greatly.
edited Nov 13 '18 at 21:46
FreeMan
7,7451060125
answered Nov 13 '18 at 19:17
DaveDave
68.3k4130247
$endgroup$
24
$begingroup$
The Concorde cruise climb would actually be the most efficient cruise procedure for all aircraft, anyway, but no other aircraft are allowed this due to traffic density. Normal aircraft approximate this by step climbing a couple of thousand feet every few hours. Concorde was very much alone at her altitude, hence a gradual climb didn’t risk any loss of separation with other traffic.
$endgroup$
– Cpt Reynolds
Nov 13 '18 at 19:45
4
$begingroup$
Link for what @CptReynolds said: Step Climb. As the aircraft changes weight the efficient altitude changes. Apparently conventional aircraft used to cruise climb, but since the skies are pretty busy now isn't no longer an option. Interesting read.
$endgroup$
– Nathan Cooper
Nov 14 '18 at 10:53
add a comment |
$begingroup$
Concorde had a 10,000fpm climb and a max altitude of 60,000ft, so time to climb was not a problem. It had an optimum cruise altitude that varied with weight, so as it burned fuel it climbed higher to stay on the optimum.
There were no other aircraft operating at those altitudes, so it would get clearances to climb 15,000ft at a time and would cruise climb throughout the trip until descent to destination was required.
answered Nov 13 '18 at 19:04
PilotheadPilothead
9,33722759
$endgroup$
$begingroup$
Is there a link to graph/formula of optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 14 '18 at 5:38
$begingroup$
@smci It is more complex than just weight. I found a repository of the flight manuals but it is thousands of pages. If you are interested avialogs.com/index.php/en/aircraft/europe-and-consortiums/…
$endgroup$
– Pilothead
Nov 14 '18 at 20:01
$begingroup$
can you just tell us the first-order approximation to the relationship between optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 15 '18 at 11:54
add a comment |
$begingroup$
Amazing how we humans skew altitude and distance. 60,000 feet up is 12 miles. Transatlantic 2400 miles.
The climb would be a gradient of 12 miles vertical/1200 miles horizontal x 100% = 1.0%
I daresay the Concorde could climb a bit faster.
A 1.0% gradient would be barely noticeable in an automobile.
Miles units cancel, answer expressed in %. Good job nitpickers!
answered Nov 13 '18 at 19:44
Robert DiGiovanniRobert DiGiovanni
2,6181316
$endgroup$
13
$begingroup$
This answer does not have any sources (and doesn't make any strong claims at all) and may be more fitting as a comment.
$endgroup$
– 0xdd
Nov 13 '18 at 20:40
1
$begingroup$
On the other hand, a freight train climbing that same grade (12 miles in 1200) could be handled by normal locomotive allocation if they didn't mind it going slower than normal. If speed is a factor, e.g. Fast container train, it will get helper units added mid-train simply to keep speed up.
$endgroup$
– Harper
Nov 13 '18 at 23:19
4
$begingroup$
Math nitpick: it should be "x 100%" (="x 1"), not just "x 100".
$endgroup$
– amI
Nov 14 '18 at 4:03
3
$begingroup$
@RobertDiGiovanni User aml is right though.12/1200 * 100 = 1200/1200 = 1and1% = 1/100, from which follows that LHS and RHS of your post's equation can't be equivalent.
$endgroup$
– Inarion
Nov 14 '18 at 12:38
8
$begingroup$
Better nitpick: There should be nox 100. 12/1200 is 1%. 12/1200 x 100 is 100%.
$endgroup$
– pipe
Nov 14 '18 at 12:39
|
show 5 more comments
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "528"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2faviation.stackexchange.com%2fquestions%2f57058%2fwas-the-transatlantic-crossing-for-concorde-too-short-to-reach-optimal-cruising%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The simple answer is that the Concorde had no single assigned altitude, it was allowed to climb freely above ~FL450; this is discussed in depth in episode 166 – Flying the Concorde (worth the listen as it answers just about every Concorde question!). As @pilothead alludes to in their answer it climbed as it burned fuel but the aircraft never actually initiated a climb, it simply drifted up as it burned fuel and became lighter.
Also discussed in the episode is the complex approach and departure procedure. Due to the fuel burn schedule the Concorde did not really have the ability to hold for more than a single lap in a holding pattern or do a stepped climb with some route adjustments like many airliners. They had a special departure procedure that was more or less runway to cruise with no interruptions and a similar descent option. So the flight was effectively a climb to cruise block then a glide down right to landing. Depending on the wind and conditions of any given day as well as the load on board, the cruise altitude could vary greatly.
edited Nov 13 '18 at 21:46
FreeMan
7,7451060125
answered Nov 13 '18 at 19:17
DaveDave
68.3k4130247
$endgroup$
24
$begingroup$
The Concorde cruise climb would actually be the most efficient cruise procedure for all aircraft, anyway, but no other aircraft are allowed this due to traffic density. Normal aircraft approximate this by step climbing a couple of thousand feet every few hours. Concorde was very much alone at her altitude, hence a gradual climb didn’t risk any loss of separation with other traffic.
$endgroup$
– Cpt Reynolds
Nov 13 '18 at 19:45
4
$begingroup$
Link for what @CptReynolds said: Step Climb. As the aircraft changes weight the efficient altitude changes. Apparently conventional aircraft used to cruise climb, but since the skies are pretty busy now isn't no longer an option. Interesting read.
$endgroup$
– Nathan Cooper
Nov 14 '18 at 10:53
add a comment |
$begingroup$
The simple answer is that the Concorde had no single assigned altitude, it was allowed to climb freely above ~FL450; this is discussed in depth in episode 166 – Flying the Concorde (worth the listen as it answers just about every Concorde question!). As @pilothead alludes to in their answer it climbed as it burned fuel but the aircraft never actually initiated a climb, it simply drifted up as it burned fuel and became lighter.
Also discussed in the episode is the complex approach and departure procedure. Due to the fuel burn schedule the Concorde did not really have the ability to hold for more than a single lap in a holding pattern or do a stepped climb with some route adjustments like many airliners. They had a special departure procedure that was more or less runway to cruise with no interruptions and a similar descent option. So the flight was effectively a climb to cruise block then a glide down right to landing. Depending on the wind and conditions of any given day as well as the load on board, the cruise altitude could vary greatly.
edited Nov 13 '18 at 21:46
FreeMan
7,7451060125
answered Nov 13 '18 at 19:17
DaveDave
68.3k4130247
$endgroup$
24
$begingroup$
The Concorde cruise climb would actually be the most efficient cruise procedure for all aircraft, anyway, but no other aircraft are allowed this due to traffic density. Normal aircraft approximate this by step climbing a couple of thousand feet every few hours. Concorde was very much alone at her altitude, hence a gradual climb didn’t risk any loss of separation with other traffic.
$endgroup$
– Cpt Reynolds
Nov 13 '18 at 19:45
4
$begingroup$
Link for what @CptReynolds said: Step Climb. As the aircraft changes weight the efficient altitude changes. Apparently conventional aircraft used to cruise climb, but since the skies are pretty busy now isn't no longer an option. Interesting read.
$endgroup$
– Nathan Cooper
Nov 14 '18 at 10:53
add a comment |
$begingroup$
The simple answer is that the Concorde had no single assigned altitude, it was allowed to climb freely above ~FL450; this is discussed in depth in episode 166 – Flying the Concorde (worth the listen as it answers just about every Concorde question!). As @pilothead alludes to in their answer it climbed as it burned fuel but the aircraft never actually initiated a climb, it simply drifted up as it burned fuel and became lighter.
Also discussed in the episode is the complex approach and departure procedure. Due to the fuel burn schedule the Concorde did not really have the ability to hold for more than a single lap in a holding pattern or do a stepped climb with some route adjustments like many airliners. They had a special departure procedure that was more or less runway to cruise with no interruptions and a similar descent option. So the flight was effectively a climb to cruise block then a glide down right to landing. Depending on the wind and conditions of any given day as well as the load on board, the cruise altitude could vary greatly.
edited Nov 13 '18 at 21:46
FreeMan
7,7451060125
answered Nov 13 '18 at 19:17
DaveDave
68.3k4130247
$endgroup$
The simple answer is that the Concorde had no single assigned altitude, it was allowed to climb freely above ~FL450; this is discussed in depth in episode 166 – Flying the Concorde (worth the listen as it answers just about every Concorde question!). As @pilothead alludes to in their answer it climbed as it burned fuel but the aircraft never actually initiated a climb, it simply drifted up as it burned fuel and became lighter.
Also discussed in the episode is the complex approach and departure procedure. Due to the fuel burn schedule the Concorde did not really have the ability to hold for more than a single lap in a holding pattern or do a stepped climb with some route adjustments like many airliners. They had a special departure procedure that was more or less runway to cruise with no interruptions and a similar descent option. So the flight was effectively a climb to cruise block then a glide down right to landing. Depending on the wind and conditions of any given day as well as the load on board, the cruise altitude could vary greatly.
edited Nov 13 '18 at 21:46
FreeMan
7,7451060125
answered Nov 13 '18 at 19:17
DaveDave
68.3k4130247
edited Nov 13 '18 at 21:46
FreeMan
7,7451060125
edited Nov 13 '18 at 21:46
FreeMan
7,7451060125
edited Nov 13 '18 at 21:46
FreeMan
7,7451060125
7,7451060125
answered Nov 13 '18 at 19:17
DaveDave
68.3k4130247
answered Nov 13 '18 at 19:17
DaveDave
68.3k4130247
answered Nov 13 '18 at 19:17
DaveDave
68.3k4130247
68.3k4130247
24
$begingroup$
The Concorde cruise climb would actually be the most efficient cruise procedure for all aircraft, anyway, but no other aircraft are allowed this due to traffic density. Normal aircraft approximate this by step climbing a couple of thousand feet every few hours. Concorde was very much alone at her altitude, hence a gradual climb didn’t risk any loss of separation with other traffic.
$endgroup$
– Cpt Reynolds
Nov 13 '18 at 19:45
4
$begingroup$
Link for what @CptReynolds said: Step Climb. As the aircraft changes weight the efficient altitude changes. Apparently conventional aircraft used to cruise climb, but since the skies are pretty busy now isn't no longer an option. Interesting read.
$endgroup$
– Nathan Cooper
Nov 14 '18 at 10:53
add a comment |
24
$begingroup$
The Concorde cruise climb would actually be the most efficient cruise procedure for all aircraft, anyway, but no other aircraft are allowed this due to traffic density. Normal aircraft approximate this by step climbing a couple of thousand feet every few hours. Concorde was very much alone at her altitude, hence a gradual climb didn’t risk any loss of separation with other traffic.
$endgroup$
– Cpt Reynolds
Nov 13 '18 at 19:45
4
$begingroup$
Link for what @CptReynolds said: Step Climb. As the aircraft changes weight the efficient altitude changes. Apparently conventional aircraft used to cruise climb, but since the skies are pretty busy now isn't no longer an option. Interesting read.
$endgroup$
– Nathan Cooper
Nov 14 '18 at 10:53
24
24
$begingroup$
The Concorde cruise climb would actually be the most efficient cruise procedure for all aircraft, anyway, but no other aircraft are allowed this due to traffic density. Normal aircraft approximate this by step climbing a couple of thousand feet every few hours. Concorde was very much alone at her altitude, hence a gradual climb didn’t risk any loss of separation with other traffic.
$endgroup$
– Cpt Reynolds
Nov 13 '18 at 19:45
$begingroup$
The Concorde cruise climb would actually be the most efficient cruise procedure for all aircraft, anyway, but no other aircraft are allowed this due to traffic density. Normal aircraft approximate this by step climbing a couple of thousand feet every few hours. Concorde was very much alone at her altitude, hence a gradual climb didn’t risk any loss of separation with other traffic.
$endgroup$
– Cpt Reynolds
Nov 13 '18 at 19:45
4
4
$begingroup$
Link for what @CptReynolds said: Step Climb. As the aircraft changes weight the efficient altitude changes. Apparently conventional aircraft used to cruise climb, but since the skies are pretty busy now isn't no longer an option. Interesting read.
$endgroup$
– Nathan Cooper
Nov 14 '18 at 10:53
$begingroup$
Link for what @CptReynolds said: Step Climb. As the aircraft changes weight the efficient altitude changes. Apparently conventional aircraft used to cruise climb, but since the skies are pretty busy now isn't no longer an option. Interesting read.
$endgroup$
– Nathan Cooper
Nov 14 '18 at 10:53
add a comment |
$begingroup$
Concorde had a 10,000fpm climb and a max altitude of 60,000ft, so time to climb was not a problem. It had an optimum cruise altitude that varied with weight, so as it burned fuel it climbed higher to stay on the optimum.
There were no other aircraft operating at those altitudes, so it would get clearances to climb 15,000ft at a time and would cruise climb throughout the trip until descent to destination was required.
answered Nov 13 '18 at 19:04
PilotheadPilothead
9,33722759
$endgroup$
$begingroup$
Is there a link to graph/formula of optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 14 '18 at 5:38
$begingroup$
@smci It is more complex than just weight. I found a repository of the flight manuals but it is thousands of pages. If you are interested avialogs.com/index.php/en/aircraft/europe-and-consortiums/…
$endgroup$
– Pilothead
Nov 14 '18 at 20:01
$begingroup$
can you just tell us the first-order approximation to the relationship between optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 15 '18 at 11:54
add a comment |
$begingroup$
Concorde had a 10,000fpm climb and a max altitude of 60,000ft, so time to climb was not a problem. It had an optimum cruise altitude that varied with weight, so as it burned fuel it climbed higher to stay on the optimum.
There were no other aircraft operating at those altitudes, so it would get clearances to climb 15,000ft at a time and would cruise climb throughout the trip until descent to destination was required.
answered Nov 13 '18 at 19:04
PilotheadPilothead
9,33722759
$endgroup$
$begingroup$
Is there a link to graph/formula of optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 14 '18 at 5:38
$begingroup$
@smci It is more complex than just weight. I found a repository of the flight manuals but it is thousands of pages. If you are interested avialogs.com/index.php/en/aircraft/europe-and-consortiums/…
$endgroup$
– Pilothead
Nov 14 '18 at 20:01
$begingroup$
can you just tell us the first-order approximation to the relationship between optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 15 '18 at 11:54
add a comment |
$begingroup$
Concorde had a 10,000fpm climb and a max altitude of 60,000ft, so time to climb was not a problem. It had an optimum cruise altitude that varied with weight, so as it burned fuel it climbed higher to stay on the optimum.
There were no other aircraft operating at those altitudes, so it would get clearances to climb 15,000ft at a time and would cruise climb throughout the trip until descent to destination was required.
answered Nov 13 '18 at 19:04
PilotheadPilothead
9,33722759
$endgroup$
Concorde had a 10,000fpm climb and a max altitude of 60,000ft, so time to climb was not a problem. It had an optimum cruise altitude that varied with weight, so as it burned fuel it climbed higher to stay on the optimum.
There were no other aircraft operating at those altitudes, so it would get clearances to climb 15,000ft at a time and would cruise climb throughout the trip until descent to destination was required.
answered Nov 13 '18 at 19:04
PilotheadPilothead
9,33722759
answered Nov 13 '18 at 19:04
PilotheadPilothead
9,33722759
answered Nov 13 '18 at 19:04
PilotheadPilothead
9,33722759
answered Nov 13 '18 at 19:04
PilotheadPilothead
9,33722759
9,33722759
$begingroup$
Is there a link to graph/formula of optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 14 '18 at 5:38
$begingroup$
@smci It is more complex than just weight. I found a repository of the flight manuals but it is thousands of pages. If you are interested avialogs.com/index.php/en/aircraft/europe-and-consortiums/…
$endgroup$
– Pilothead
Nov 14 '18 at 20:01
$begingroup$
can you just tell us the first-order approximation to the relationship between optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 15 '18 at 11:54
add a comment |
$begingroup$
Is there a link to graph/formula of optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 14 '18 at 5:38
$begingroup$
@smci It is more complex than just weight. I found a repository of the flight manuals but it is thousands of pages. If you are interested avialogs.com/index.php/en/aircraft/europe-and-consortiums/…
$endgroup$
– Pilothead
Nov 14 '18 at 20:01
$begingroup$
can you just tell us the first-order approximation to the relationship between optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 15 '18 at 11:54
$begingroup$
Is there a link to graph/formula of optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 14 '18 at 5:38
$begingroup$
Is there a link to graph/formula of optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 14 '18 at 5:38
$begingroup$
@smci It is more complex than just weight. I found a repository of the flight manuals but it is thousands of pages. If you are interested avialogs.com/index.php/en/aircraft/europe-and-consortiums/…
$endgroup$
– Pilothead
Nov 14 '18 at 20:01
$begingroup$
@smci It is more complex than just weight. I found a repository of the flight manuals but it is thousands of pages. If you are interested avialogs.com/index.php/en/aircraft/europe-and-consortiums/…
$endgroup$
– Pilothead
Nov 14 '18 at 20:01
$begingroup$
can you just tell us the first-order approximation to the relationship between optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 15 '18 at 11:54
$begingroup$
can you just tell us the first-order approximation to the relationship between optimum cruise altitude vs weight?
$endgroup$
– smci
Nov 15 '18 at 11:54
add a comment |
$begingroup$
Amazing how we humans skew altitude and distance. 60,000 feet up is 12 miles. Transatlantic 2400 miles.
The climb would be a gradient of 12 miles vertical/1200 miles horizontal x 100% = 1.0%
I daresay the Concorde could climb a bit faster.
A 1.0% gradient would be barely noticeable in an automobile.
Miles units cancel, answer expressed in %. Good job nitpickers!
answered Nov 13 '18 at 19:44
Robert DiGiovanniRobert DiGiovanni
2,6181316
$endgroup$
13
$begingroup$
This answer does not have any sources (and doesn't make any strong claims at all) and may be more fitting as a comment.
$endgroup$
– 0xdd
Nov 13 '18 at 20:40
1
$begingroup$
On the other hand, a freight train climbing that same grade (12 miles in 1200) could be handled by normal locomotive allocation if they didn't mind it going slower than normal. If speed is a factor, e.g. Fast container train, it will get helper units added mid-train simply to keep speed up.
$endgroup$
– Harper
Nov 13 '18 at 23:19
4
$begingroup$
Math nitpick: it should be "x 100%" (="x 1"), not just "x 100".
$endgroup$
– amI
Nov 14 '18 at 4:03
3
$begingroup$
@RobertDiGiovanni User aml is right though.12/1200 * 100 = 1200/1200 = 1and1% = 1/100, from which follows that LHS and RHS of your post's equation can't be equivalent.
$endgroup$
– Inarion
Nov 14 '18 at 12:38
8
$begingroup$
Better nitpick: There should be nox 100. 12/1200 is 1%. 12/1200 x 100 is 100%.
$endgroup$
– pipe
Nov 14 '18 at 12:39
|
show 5 more comments
$begingroup$
Amazing how we humans skew altitude and distance. 60,000 feet up is 12 miles. Transatlantic 2400 miles.
The climb would be a gradient of 12 miles vertical/1200 miles horizontal x 100% = 1.0%
I daresay the Concorde could climb a bit faster.
A 1.0% gradient would be barely noticeable in an automobile.
Miles units cancel, answer expressed in %. Good job nitpickers!
answered Nov 13 '18 at 19:44
Robert DiGiovanniRobert DiGiovanni
2,6181316
$endgroup$
13
$begingroup$
This answer does not have any sources (and doesn't make any strong claims at all) and may be more fitting as a comment.
$endgroup$
– 0xdd
Nov 13 '18 at 20:40
1
$begingroup$
On the other hand, a freight train climbing that same grade (12 miles in 1200) could be handled by normal locomotive allocation if they didn't mind it going slower than normal. If speed is a factor, e.g. Fast container train, it will get helper units added mid-train simply to keep speed up.
$endgroup$
– Harper
Nov 13 '18 at 23:19
4
$begingroup$
Math nitpick: it should be "x 100%" (="x 1"), not just "x 100".
$endgroup$
– amI
Nov 14 '18 at 4:03
3
$begingroup$
@RobertDiGiovanni User aml is right though.12/1200 * 100 = 1200/1200 = 1and1% = 1/100, from which follows that LHS and RHS of your post's equation can't be equivalent.
$endgroup$
– Inarion
Nov 14 '18 at 12:38
8
$begingroup$
Better nitpick: There should be nox 100. 12/1200 is 1%. 12/1200 x 100 is 100%.
$endgroup$
– pipe
Nov 14 '18 at 12:39
|
show 5 more comments
$begingroup$
Amazing how we humans skew altitude and distance. 60,000 feet up is 12 miles. Transatlantic 2400 miles.
The climb would be a gradient of 12 miles vertical/1200 miles horizontal x 100% = 1.0%
I daresay the Concorde could climb a bit faster.
A 1.0% gradient would be barely noticeable in an automobile.
Miles units cancel, answer expressed in %. Good job nitpickers!
answered Nov 13 '18 at 19:44
Robert DiGiovanniRobert DiGiovanni
2,6181316
$endgroup$
Amazing how we humans skew altitude and distance. 60,000 feet up is 12 miles. Transatlantic 2400 miles.
The climb would be a gradient of 12 miles vertical/1200 miles horizontal x 100% = 1.0%
I daresay the Concorde could climb a bit faster.
A 1.0% gradient would be barely noticeable in an automobile.
Miles units cancel, answer expressed in %. Good job nitpickers!
answered Nov 13 '18 at 19:44
Robert DiGiovanniRobert DiGiovanni
2,6181316
edited Nov 15 '18 at 7:59
answered Nov 13 '18 at 19:44
Robert DiGiovanniRobert DiGiovanni
2,6181316
answered Nov 13 '18 at 19:44
Robert DiGiovanniRobert DiGiovanni
2,6181316
answered Nov 13 '18 at 19:44
Robert DiGiovanniRobert DiGiovanni
2,6181316
2,6181316
13
$begingroup$
This answer does not have any sources (and doesn't make any strong claims at all) and may be more fitting as a comment.
$endgroup$
– 0xdd
Nov 13 '18 at 20:40
1
$begingroup$
On the other hand, a freight train climbing that same grade (12 miles in 1200) could be handled by normal locomotive allocation if they didn't mind it going slower than normal. If speed is a factor, e.g. Fast container train, it will get helper units added mid-train simply to keep speed up.
$endgroup$
– Harper
Nov 13 '18 at 23:19
4
$begingroup$
Math nitpick: it should be "x 100%" (="x 1"), not just "x 100".
$endgroup$
– amI
Nov 14 '18 at 4:03
3
$begingroup$
@RobertDiGiovanni User aml is right though.12/1200 * 100 = 1200/1200 = 1and1% = 1/100, from which follows that LHS and RHS of your post's equation can't be equivalent.
$endgroup$
– Inarion
Nov 14 '18 at 12:38
8
$begingroup$
Better nitpick: There should be nox 100. 12/1200 is 1%. 12/1200 x 100 is 100%.
$endgroup$
– pipe
Nov 14 '18 at 12:39
|
show 5 more comments
13
$begingroup$
This answer does not have any sources (and doesn't make any strong claims at all) and may be more fitting as a comment.
$endgroup$
– 0xdd
Nov 13 '18 at 20:40
1
$begingroup$
On the other hand, a freight train climbing that same grade (12 miles in 1200) could be handled by normal locomotive allocation if they didn't mind it going slower than normal. If speed is a factor, e.g. Fast container train, it will get helper units added mid-train simply to keep speed up.
$endgroup$
– Harper
Nov 13 '18 at 23:19
4
$begingroup$
Math nitpick: it should be "x 100%" (="x 1"), not just "x 100".
$endgroup$
– amI
Nov 14 '18 at 4:03
3
$begingroup$
@RobertDiGiovanni User aml is right though.12/1200 * 100 = 1200/1200 = 1and1% = 1/100, from which follows that LHS and RHS of your post's equation can't be equivalent.
$endgroup$
– Inarion
Nov 14 '18 at 12:38
8
$begingroup$
Better nitpick: There should be nox 100. 12/1200 is 1%. 12/1200 x 100 is 100%.
$endgroup$
– pipe
Nov 14 '18 at 12:39
13
13
$begingroup$
This answer does not have any sources (and doesn't make any strong claims at all) and may be more fitting as a comment.
$endgroup$
– 0xdd
Nov 13 '18 at 20:40
$begingroup$
This answer does not have any sources (and doesn't make any strong claims at all) and may be more fitting as a comment.
$endgroup$
– 0xdd
Nov 13 '18 at 20:40
1
1
$begingroup$
On the other hand, a freight train climbing that same grade (12 miles in 1200) could be handled by normal locomotive allocation if they didn't mind it going slower than normal. If speed is a factor, e.g. Fast container train, it will get helper units added mid-train simply to keep speed up.
$endgroup$
– Harper
Nov 13 '18 at 23:19
$begingroup$
On the other hand, a freight train climbing that same grade (12 miles in 1200) could be handled by normal locomotive allocation if they didn't mind it going slower than normal. If speed is a factor, e.g. Fast container train, it will get helper units added mid-train simply to keep speed up.
$endgroup$
– Harper
Nov 13 '18 at 23:19
4
4
$begingroup$
Math nitpick: it should be "x 100%" (="x 1"), not just "x 100".
$endgroup$
– amI
Nov 14 '18 at 4:03
$begingroup$
Math nitpick: it should be "x 100%" (="x 1"), not just "x 100".
$endgroup$
– amI
Nov 14 '18 at 4:03
3
3
$begingroup$
@RobertDiGiovanni User aml is right though.
12/1200 * 100 = 1200/1200 = 1 and 1% = 1/100, from which follows that LHS and RHS of your post's equation can't be equivalent.$endgroup$
– Inarion
Nov 14 '18 at 12:38
$begingroup$
@RobertDiGiovanni User aml is right though.
12/1200 * 100 = 1200/1200 = 1 and 1% = 1/100, from which follows that LHS and RHS of your post's equation can't be equivalent.$endgroup$
– Inarion
Nov 14 '18 at 12:38
8
8
$begingroup$
Better nitpick: There should be no
x 100. 12/1200 is 1%. 12/1200 x 100 is 100%.$endgroup$
– pipe
Nov 14 '18 at 12:39
$begingroup$
Better nitpick: There should be no
x 100. 12/1200 is 1%. 12/1200 x 100 is 100%.$endgroup$
– pipe
Nov 14 '18 at 12:39
|
show 5 more comments
Thanks for contributing an answer to Aviation Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2faviation.stackexchange.com%2fquestions%2f57058%2fwas-the-transatlantic-crossing-for-concorde-too-short-to-reach-optimal-cruising%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
