Feauce |

If you were jumping UP 10 feet and OUT 20, I'd have you make two jump checks, one for height and one for distance. Fail either one and you've got a problem.

I would be more inclined to have the player make one check, and since the *out* is larger than the *up*, I'd have them roll for a long jump. Since they're not trying to land at the same elevation, I would add to the DC to account for the extra 10-15 feet they would have otherwise needed to go laterally to reach that height at the 20-foot mark.

If it's something the player likes to do a lot, or the setting lends itself to doing so often (such as an urban-based adventure/campaign), I would try to figure out more precise math ahead of time to have some quick figures to reference. I'd also have a reference for the arc formula for attempts that aren't on the cheat sheet.

Goes to show once again that, depending on your GM, YMMV quite a lot. :)

RedDogMT |

If you were jumping UP 10 feet and OUT 20, I'd have you make two jump checks, one for height and one for distance. Fail either one and you've got a problem.

Yeah, I can't support this ruling either. Making a jump is a single action. The rules do not talk about how high a jump can arc, but when characters get unusually high acrobatic bonuses like +20 or +30, I think a GM needs to work with the player a bit on what cool stuff he can do.

SlimGauge |

SlimGauge wrote:If you were jumping UP 10 feet and OUT 20, I'd have you make two jump checks, one for height and one for distance. Fail either one and you've got a problem.Yeah, I can't support this ruling either. Making a jump is a single action. The rules do not talk about how high a jump can arc, but when characters get unusually high acrobatic bonuses like +20 or +30, I think a GM needs to work with the player a bit on what cool stuff he can do.

Then simply take the harder of the two, but if that (single) check is failed, you no longer know if they didn't make the distance or didn't make the height.

Feauce |

SlimGauge wrote:

Indeed, the Pathfinder skill description apparently doesn't say. The more specific Jump skill from 3.0/3.5 stated the following:

At the midpoint of the jump, you attain a vertical height equal to one-quarter of the horizontal distance.

So you would need to be able to jump 40 feet to reach a 10-foot high point at the middle. Since the DC is equal to the distance needed, that would mean a DC of 40, which is a bit absurd under normal circumstances, but might be manageable depending on level and magic items.

Don't forget your 10-foot pole. :)

**EDIT:** I pulled a dumb and worked my math with the numbers in reverse. The above blunder has been fixed.

Feauce |

Isn't the DC for High Jump 4 for every foot, anyway? So, that would be a DC 40 for a 10' High Jump, same as a long jump that achieves 10' height in the middle.

Yes, but for a very different reason. If the jump was instead a 30-foot long jump with a 10-foot height difference at the end, a DC 40 wouldn't cut it. Not sure off the top of my head how the precise arc math works out, but it would probably be closer to a DC 50-55 or so for that scenario.

toastedamphibian |

Yes it would be. Though, 10ft out and 10ft up would be a 50 by my method, and I am not sure if that is fair or not... I need to rethink this a little.

Jumping Up 10ft and out 20ft or less should be 40, but out 20ft and up 1ft should be more than out 20ft and up zero. I don't see an immediately obvious and simple rule to make this happen yet.

Feauce |

Yes it would be. Though, 10ft out and 10ft up would be a 50 by my method, and I am not sure if that is fair or not... I need to rethink this a little.

Jumping Up 10ft and out 20ft or less should be 40, but out 20ft and up 1ft should be more than out 20ft and up zero. I don't see an immediately obvious and simple rule to make this happen yet.

A jump 20 feet out and zero up would be a DC 20. Adding a single foot would not add a DC of +4, because adding 10 feet would, by that logic, add +40 for the height. Only it doesn't. Half of that, as it turns out, because the distances involved make the math easier. So adding +2 to the DC for a 1-foot height could work. I don't think that's a workable solution for the generic case, though.

Adding the 10 feet up requires you to be capable of a longer jump, which makes it the equivalent of a 40-foot long jump, thus DC 40. A jump only 10 feet out and 10 feet high would be, for the most part, just as hard as a jump merely 10 feet high, adding maybe +2-5 to the DC of 40 for the base jump.

I was working with a sine graph and a 30-foot long jump with a 10-foot height at the end would be roughly a DC 45 for the added distance. The peak of the jump would be over the main 30-foot distance, approximately 22-23 feet into the jump. The remaining distance, the jumper would be falling to reach the desired height at the end. So it looks like I was over-estimating the difference in my last post.

toastedamphibian |

Yeah, it seems to assume a 45 degree jump, with equal vertical and horizontal forces. For long jumps. For vertical leaps... it assumes the same vertical velocity. Basically it pretends your being thrown at at 45 degree angle and just ignoring the forward travel. In otherwords, forward travel distance does not change the DC to attain a particular height. (If this is accurate, if humans are really that bad at converting forward thrust to vertical thrust, I have no idea)

Got a formula I'm happy with:

If the distance is less than or equal to twice the height, use the vertical jump DC, if not, use longjump DC +2 per foot of height.

Excel Up. =IF(2*H>=D,4H,D+2H)

Where H is height above or below origin and D is distance from origin.

Thoughts?

Edit: Down was displeasing, so removed.

bbangerter |

Use the greater of the two DC's for vertical vs horizontal difference.

1) It's simple.

2) If you take the 3.5 rule it fits with the existing math.

2a) Jumps are not calculated based on actual distance traveled through the arc. They are calculated either on height attained, or horizontal distance covered. This is why a 40' jump is DC 40 and a 10' height (at 4 DC per foot) is DC 40, and why the 1/4 height of horizontal distance matches.

2b) So if a character wants to jump 20' distance, and 10' height, take the greater DC of 40 for the 10' height. They could also make that same jump while traveling 40' distance, but may not want to or be able to due to a wall or other obstacle that exists 40' away but not 20' away.

3) Long jumps always contain a vertical element, that is already factored in, so adding additional DC for relatively small height differences should not be done.

3a) Given earth gravity of 32'/s^2, that is, in the first second you would naturally fall 16' if you had no initial upward velocity. Clearly a horizontal jump is not strictly horizontal, or you'd have to cross that distance in millisecond timing in order to not have fallen enough to just slam into the opposite wall.

And how does all of that factor in when the landing point is lower than the starting point instead of the other way around?

Finally, you can use the Acrobatics skill to make jumps or tosoften a fall.

At this point you are softening a fall, use the rules for that. A GM would certainly be within their right to give a +2 (or other) circumstance bonus for the horizontal distance covered. And maybe even a +2 per 10' height distance to a max of +5 or something.

Cattleman |

I think Bbangerter's post makes sense, though I'd also be fine with calcing the distance. It's a simple Pythagorean calc that takes only a couple seconds on google. I used that to allow someone a 30ft spell when they were 15ft away and the thing was 20ft higher than them, for example.

In some sense someone arguing that he's out of range makes some game sense, but the simulation says he's in range with room to spare.

EDIT: I'll stick with Bbang's post I think. I goofed a bit. I agree that taking the higher DC for height vs. length makes sense as well. hmm..

Feauce |

I think Bbangerter's post makes sense, though I'd also be fine with calcing the distance. It's a simple Pythagorean calc that takes only a couple seconds on google. I used that to allow someone a 30ft spell when they were 15ft away and the thing was 20ft higher than them, for example.

In some sense someone arguing that he's out of range makes some game sense, but the simulation says he's in range with room to spare.

EDIT: I'll stick with Bbang's post I think. I goofed a bit. I agree that taking the higher DC for height vs. length makes sense as well. hmm..

Okay, I've been doing some trig this morning and came up with the following equation, which appears to work for any arbitrary jump.

- DC = square(length) / (length - height)

Examples:

Original question, 20-foot wide gap, landing point 10 feet lower than starting position.

- DC = square(20) / (20 - -10)

- DC = 400 / 30

- DC = 13-1/3

Reverse the height difference, making the jump 20 feet wide and 10 feet *higher* instead.

- DC = square(20) / (20 - 10)

- DC = 400 / 10

- DC = 40

Make the jump longer from the last example, with a 30-foot wide gap and a landing 10 feet higher.

- DC = square(30) / (30 - 10)

- DC = 900 / 20

- DC = 45

The jumps that were posed by toastedamphibian: 20 feet wide and 1 foot up, and 20 feet wide and 5 feet up, respectively.

- DC = square(20) / (20 - 1)

- DC = 400 / 19

- DC = 21.0526315...

- DC = square(20) / (20 - 5)

- DC = 400 / 15

- DC = 26-2/3

If anyone wants, I can go through the math I used to come up with this formula.

Feauce |

5ft over and 10ft up

DC = 25/-5

DC = -5Bit of a problem there. Any situation where your height and length of your jump are equal, IE 10ft up and 10ft out, results in division by 0.

The formula is for adjusting long jumps. More specifically, where the height is less than the length. It is not intended for what is essentially a high jump.

Further, a 20-foot gap with a 15-foot high landing would be a DC of 80, double what a 10-foot high landing works out to be. The character would have one of two options: either start the jump before the actual gap, making it a longer jump (making the above a 30-foot wide jump reduces the DC to 60), or jump onto the surface and climb from there. To long-jump it directly without magical assistance would likely be impossible.

For jumps like the 10-foot out and up, I would take the DC 40 high jump and add between +5-10 to it for the additional distance.

Feauce |

Bit of a problem there. Any situation where your height and length of your jump are equal, IE 10ft up and 10ft out, results in division by 0.

To be more specific about the *why* of it...

The midpoint height being set to one-quarter of the total length places the initial angle of the jump at 45 degrees. Regardless of your initial velocity, gravity is not going to allow your jump's height to equal your distance beyond a couple of feet. Well, unless you're being shot from a cannon, but that's a different topic.

The reason you reach the division-by-zero point is because that's the point at which it becomes physically impossible for the long-jump form. A negative DC reflects this as well, for any point after that. Reaching either result indicates that you need to instead calculate it as a high jump, allow them to grab the side of whatever they're jumping at, or inform them that it's not possible.

In short, there isn't a problem with the formula. Rather, the formula is telling you something that isn't necessarily immediately obvious.

Feauce |

So, would you say that your assertion that it works for "any arbitrary jump" is verifiably false?

No. I would say that I missed a word. I should have specified that it works for "any arbitrary long jump" instead. Unfortunately, it's been too long to correct my previous post to reflect this.

toastedamphibian |

Your DCs compare poorly to what is in the book, and fail in a variety of situations that are poorly described ahead of time.

A jump check of 60 gets you 15 feet up, or 15 feet up and 30 feet out. Going less than 30 feet out should not be 33% harder.

Jumping 20ft up and 20ft over should not be 5x as difficult as jumping 20ft up.

Feauce |

Your DCs compare poorly to what is in the book, and fail in a variety of situations that are poorly described ahead of time.

Interesting, because for the scenarios listed previously in the thread, the numbers work out just fine. So what precisely are you going on about? Be more specific and I can address those issues. If all you've got is "your math sucks" then you have no argument (in the sense of having a discussion), and I'm done engaging with that.

A jump check of 60 gets you 15 feet up, or 15 feet up and 30 feet out.

The DCs for those two jumps will match for the same reason why a high jump of 10 feet will match a long jump 20 feet out and 10 feet up (DC 40). This was covered in an earlier post. (Post Link)

Going less than 30 feet out should not be 33% harder. Jumping 20ft up and 20ft over should not be 5x as difficult as jumping 20ft up.

Did you even read my follow-up post? I don't care much for excessively repeating myself, especially in the same thread, let alone on the same page. (Post Link)

Feauce |

Speaking of "more specific"...

Your DCs compare poorly to what is inthe book

*The* book, eh? With a connotation like that I probably shouldn't have to ask, but that's such a vague statement that I do. *Which* book are you talking about?

Show me which book and page(s) list DCs that "compare poorly" to what I've put forth, and I'll revisit the formula. I've already looked in the Pathfinder Core Rulebook and the Player's Handbook v3.5, the ones usually referred to as "*the* book" for their respective editions, and neither source mentions any DCs for attempting a jump in both directions at once.

If there is a *first-party* source with rules for doing precisely what this thread is about, I would be delighted to hear about it, especially for the chance to compare notes.

toastedamphibian |

You asked me my point, so I clarified. If it looks similiar to what I said before, that should not be surprising as my point did not change.

You ask for examples of where I think the math works out poorly. All the scenarios I provided in that post and the preceding posts are those examples, that is why they are there.

As for the book:

At the midpoint of the jump, you attain a vertical height equal to one-quarter of the horizontal distance.

As I noted, a DC 60 check puts you 15ft up at the midpoint of a 60ft jump. 15ft up, 30 out. Your formula makes it 33% harder to do the same while traveling 2/3rd the distance.

I do not need there to be a direct rule to COMPARE. If the DC to jump 20ft out and 19ft up is 400, I can compare that to the official jump DCs. 19ft up and 20ft over being the same level of difficulty as jumping 100ft into the air or 400ft horizontally is a valid point for comparison. And I feel that they are a poor match.

Your math seems fine. Your results seem inconsistent with what is already avaliable.

Edit: It is entirely possible that your formula is a more accurate portrayal of reality than what is in the books.

Feauce |

You asked me my point, so I clarified. If it looks similiar to what I said before, that should not be surprising as my point did not change.

You ask for examples of where I think the math works out poorly. All the scenarios I provided in that post and the preceding posts are those examples, that is why they are there.

As for the book:

3.5 phb wrote:At the midpoint of the jump, you attain a vertical height equal to one-quarter of the horizontal distance.As I noted, a DC 60 check puts you 15ft up at the midpoint of a 60ft jump. 15ft up, 30 out. Your formula makes it 33% harder to do the same while traveling 2/3rd the distance.

I do not need there to be a direct rule to COMPARE. If the DC to jump 20ft out and 19ft up is 400, I can compare that to the official jump DCs. 19ft up and 20ft over being the same level of difficulty as jumping 100ft into the air or 400ft horizontally is a valid point for comparison. And I feel that they are a poor match.

Your math seems fine. Your results seem inconsistent with what is already avaliable.

It seems to me that you either still haven't read my earlier post where I went further into the *why* of it, or you don't understand it. Or maybe I could've written it better. I dunno.

Bear in mind, all of this is referring specifically to a *long jump* style of jumping. In this form, the distance covered is more important than the height attained. As the height increases, the difficulty of such a jump goes up by quite a lot, because the effective distance needed to attain said height gets much longer.

The way the formulas filtered down, gravity became irrelevant math-wise. That does not mean it's irrelevant logically.

The initial angle for a long jump is 45 degrees. This is known because of the maximum height during the jump. This also means that the closer your target point gets to this angle, the more difficult the jump will be to pull off. If your target point is higher than this angle, it's going into the territory of physically impossible. Muscle power can only do so much.

This is why a longer jump to the same height can be easier, because it puts the target point at a lower angle instead of keeping it close to the 45 degree maximum.

Once the height approaches the jump distance (within 5 feet seems good as a Rule of Thumb), and certainly once the height exceeds it, setting the DC according to the height is more appropriate. How to figure that DC is outside the scope of the formula I have. Using the height for a high jump DC and adding half of the "out" distance would probably be a good place to start, though.

toastedamphibian |

Hmm, internet ate my post. Going to try something shorter...

I think you are realistically correct, but your results fit poorly in the existing system.

Jump skill seems to make no allowance for angel. Jumping at a 90 degree angle (a High Jump) gives the same vertical distance as a long jump with the same DC.

They did not use parabolic equations to set their DCs, the math they used was very simplified and linear. Yours is very not linear.

Additionally, if you wish other people to be able to use it, you should probably repost it with explicit instructions for when to use it and when not to.

Feauce |

Hmm, internet ate my post. Going to try something shorter...

I think you are realistically correct, but your results fit poorly in the existing system.

My sympathies for the consumed post. Though I still disagree in regards to the formula not cooperating with the system.

Jumping at a 90 degree angle (a High Jump) gives the same vertical distance as a long jump with the same DC.

This is on purpose. As stated in my last post, a long jump cares more about distance, while the high jump is exclusively concerned about height. The difficulty to reach the same height (albeit using different jumping methods), however, is the same. This also shows that the developers were thinking about arcing travel paths at some level. The scenario that comes to mind for me is a cavern or dungeon hallway with a ceiling, and if you attempt to jump too far, you will hit said ceiling and probably fall.

Further, as the rules reckon distance, succeeding at a 5-foot high jump does not mean that a 4-foot dwarf jumps and grabs a ledge just above his head and pulls himself up. What it does mean is that said dwarf ends up with his boots at the 5-foot mark, standing on top of the ledge. In fact, a Climb skill check would be more appropriate for the former scenario, since very little jumping was involved.

Jump skill seems to make no allowance for angle. ... They did not use parabolic equations to set their DCs, the math they used was very simplified and linear. Yours is very not linear.

Certainly it's non-linear, because jumping is non-linear by its very nature. The arcing path of a creature jumping is parabolic in shape, and while the books don't explicitly say so, they do in fact take angle into account.

I see two main reasons why the DCs for a high jump are +4/foot instead of +1/foot for a long jump. What entered the minds of the developers that came up with these numbers, I have no idea. All I know is that this is what makes logical sense to me.

Obviously, moving 10 feet vertically off of the ground is much more difficult than moving 10 feet horizontally (at the local scale) across the surface. This, of course, is mostly due to the first reason: gravity. The second reason is that our bodies are not built for jumping. Other creatures with more specialized or differently developed body structures would have a racial bonus to the Jump/Acrobatics skill.

Additionally, if you wish other people to be able to use it, you should probably repost it with explicit instructions for when to use it and when not to.

The topic at hand was about long jumps, and I feel like the posts I made directly after stated quite specifically what the limitations were and when a different calculation would be appropriate. At the end of the day, a GM can use whatever they like to set the DCs at their table.

That being said, this is a good suggestion. I would like to eventually post this formula, along with the math that led to it, where it could be referenced more easily. That's a project for another day, unfortunately. Hopefully soon.

toastedamphibian |

There was in fact a rule restricting long jump distance by vertical clearance in some edition... 3.0 I think, removed in 3.5 for simplicity.

If we ignore forward momentum, which affects the DCs equally, the math they used to determine long and high jumps is rather clear I think: Result of X can move X feat horizontally and X/4 vertically.

Reaching height X at 45 degree angle should be harder than reaching X at 90 degree angle realistically . Again, from standing. But it is not. High and long jump DCs being different seems to be a myth.

If you get around to posting this again some day, shoot me a PM if you don't mind. I look forward to seeing it.

Any chance you would like to offer a critique of my solution? Ultimately much more simplistic, but gives results I feel are more in line with the realities of the game world.

If the distance is less than or equal to twice the height, use the vertical jump DC, if not, use longjump DC +2 per foot of height.

Excel Up. =IF(2*H>=D,4H,D+2H)

Where H is height above or below origin and D is distance from origin.

Feauce |

Reaching height X at 45 degree angle should be harder than reaching X at 90 degree angle realistically.

Reaching height X *at* a 45 degree angle without mechanical, alchemical, or magical assistance is not possible, for reasons I've already covered. Leaving that aside, the difficulty of reaching height X with a jump that *starts* from a 45 degree angle versus reaching the same height from one that starts at a 90 degree angle are indeed the same, in game terms.

Again, this is both not an accident and also largely irrelevant, but is something you've been stuck on almost this entire time. The mechanics of these two jumping styles have very little to do with each other. Additionally, as I've stated before, the solution I've presented has absolutely *nothing* to do with high jumps.

The fact that by either coincidence or design some of my solution's DCs match some high jump DCs means as much as the fact that the Craft check DCs for making a bastard sword and a suit of half-plate are the same. It doesn't mean anything.

Again, from standing. But it is not.

It would be more difficult from both a standing jump and from a running jump. High jumps are not exempt from needing a running start.

High and long jump DCs being different seems to be a myth.

You only reach that conclusion because you pick the handful of scenarios that the DCs either work out to be the same or show that such a jump isn't possible using the method you've chosen. You keep harping on the same "issues" that aren't really issues, and absolutely refuse to address anything that I say when I explain why it works the way it does.

At this point, based on this behavior, I can only assume that you don't understand the bulk of what I've been saying, have no interest in understanding, and so have no recourse except to behave as though it doesn't matter.

Any chance you would like to offer a critique of my solution? Ultimately much more simplistic, but gives results I feel are more in line with the realities of the game world.

me wrote:If the distance is less than or equal to twice the height, use the vertical jump DC, if not, use longjump DC +2 per foot of height.

Excel Up. =IF(2*H>=D,4H,D+2H)

Where H is height above or below origin and D is distance from origin.

You haven't given my solution a fair assessment, so why should I? In any case, whatever you want to do as a GM at your table makes little difference.

At this point, this entire exchange is fruitless. At every turn, you seek to just dismiss what I've shown to be a working solution completely out of hand. You have nothing of value to offer in the discussion of the solution I've put forth, and I have no reason whatsoever to continue trying to reason with someone that has no argument. So, as stated before, I'm done with you.

Feauce |

In case anyone wants to see what I can reasonably post of the proof (in the mathematical sense), or wants to work the math out for themselves, here is the way I reached the above solution.

The first bit came from the formulas for projectile motion.

The maximum height for any given jump is given, so we don't need to calculate that. This is especially good, because we don't have precise assumptions about the values for gravity or the initial velocity.

Since the maximum height at the peak of motion is a known ratio, I started with the formula under the heading: Relation between horizontal range and maximum height (Math Render). This allows one to solve for *theta*, giving us the initial angle of the jump. Solving for *theta*, this gives us the following: *theta* = arctan(4*h* / *R*)

Within this context, *h* is our maximum jumping height and *R* is the full long jump distance. Since *h* is one-quarter of *R*, the fraction simplifies to 1 (one), giving us an angle of *pi*/4 radians or 45 degrees.

This is all well and good, but we will still need a way to figure out the initial velocity, to substitute into the last equation. Thankfully, we have one. The formula for Displacement under the heading: Kinematic quantities of projectile motion (Math Render). We know the initial angle, and we have a point along the curve (the midpoint) to use as a known position. Therefore, we can solve this and figure out what the initial velocity is. In this context, *g* is gravity, and *x* and *y* are the coordinates for the midpoint of the jump. To allow simplification (mathematically speaking), these two coordinates need to be in a common form. Therefore, *x* gets substituted with *R*/2 and *y* with *R*/4. This gives us the following: *v* = sqrt(*R**g*)

The last task is to figure out what the *R* value is for any curve given these qualities. For that, a formula from the trajectory of a projectile does the trick nicely.

More specifically, the Height at *x* formula under this heading: Conditions at an arbitrary distance *x* (Math Render). We know the initial angle and the starting point (0, 0), so before doing any other substitutions, this simplifies to the following: *y* = *x* - (*g*/square(*v*))square(*x*)

We know the substitution value for *v*, which also allows us to remove *g* from the equation altogether. This simplifies to the following: *y* = *x* - (1/*R*)square(*x*)

Solving for *R* gives us the previously presented equation: *R* = square(*x*) / (*x* - *y*)

In this context, *R* is once again the full long jump distance, but it is no longer the actual distance jumped. Instead, it becomes the equivalent distance for a jump given the point (*x*, *y*) on the curve. For this solution, *x* is the actual length of the jump being attempted, and *y* is the height difference at the end of the jump. Be aware that this height difference can be negative if jumping to a lower elevation, and subtracting a negative is the same as adding the positive value.

And so, we are left with the version I posted previously: DC = square(length) / (length - height)

As described before, and as toastedamphibian has been pointing out repeatedly, the *height* value must be less than the *length*. When equal, this causes division by zero, and when greater, results in a negative total distance, which is impossible in this context. Either of these results means that a skill-based *long jump* is not appropriate for the attempt, and the high jump rules or some other solution needs to be used instead.

**P.S.**: In case anyone is curious, I've used GraphSketch.com to plot the arcs of long jumps from 10 to 60 feet in increments of ten. They're plotted out on an 80x80 grid so the curves don't get distorted. Hopefully this is helpful for visualization. (Graph Link)

toastedamphibian |

Leaving that aside, the difficulty of reaching height X with a jump thatstartsfrom a 45 degree angle versus reaching the same height from one that starts at a 90 degree angle are indeed the same, in game terms.

Yes.

The mechanics of these two jumping styles have very little to do with each other. Additionally, as I've stated before, the solution I've presented has absolutelynothingto do with high jumps.

As you noted, they are identical in game terms.

The fact that by either coincidence or design some of my solution's DCs match some high jump DCs means as much as the fact that the Craft check DCs for making a bastard sword and a suit of half-plate are the same. It doesn't mean anything.

It should. If two uses of the same skill have the same DC, they should be similarly difficult. Because they share the same Difficulty Class.

It would be more difficult from both a standing jump and from a running jump. High jumps are not exempt from needing a running start.

They are not. That was not my point. As noted, they affect the DC identically. The point is to remove the variable of forward momentum from the conceptualization of the physics.

High and long jump DCs being different seems to be a myth.

You only reach that conclusion because you pick the handful of scenarios that the DCs either work out to be the same or show that such a jump isn't possible using the method you've chosen.

This sentence is referring to the core rules. Not your formula. In the core rules, High Jumps are Long Jumps that don't go any where.

At this point, this entire exchange is fruitless. At every turn, you seek to just dismiss what I've shown to be a working solution completely out of hand. You have nothing of value to offer in the discussion of the solution I've put forth, and I have no reason whatsoever to continue trying to reason with someone that has no argument. So, as stated before, I'm done with you.

As you wish. Your numbers widely differ from the the long jump DC's. You should consider reworking those as well, for consistency.

Cevah |

SlimGauge wrote:

Actually...

PRDa horizontal jump's peak height is one-fourth of the horizontal distance

/cevah

RedDogMT |

RedDogMT wrote:SlimGauge wrote:Actually...

PRD

Above and Beneath the Streets wrote:a horizontal jump's peak height is one-fourth of the horizontal distance/cevah

Actually...your reference has very little to do with this conversation. It is not a chat about long jump arcs. I said the rules to not talk about how high a jump arc CAN be. So, tell me how it would work to get an arc 1/2 of the horizontal distance? How about 3/4?

Cevah |

Cevah wrote:Actually...your reference has very little to do with this conversation. It is not a chat about long jump arcs. I said the rules to not talk about how high a jump arc CAN be. So, tell me how it would work to get an arc 1/2 of the horizontal distance? How about 3/4?RedDogMT wrote:SlimGauge wrote:Actually...

PRD

Above and Beneath the Streets wrote:a horizontal jump's peak height is one-fourth of the horizontal distance

To jump 10' up, you need to make the 4x DC (=40). You also happen to get that high when you jump horizontally 40', which has the exact same DC. In that long jump, you achieve that height at the halfway point. So at 20' feet from the start, you are 10' up.

For the posited jump of *jumping UP 10 feet and OUT 20*, that exactly matches what happens when you broad jump 40'. Therefore the DC is 40.

For the OP, jumping from a 40' building to a 30' building is essentially jumping down rather than falling down. The rules there indicate DC 15 to ignore damage from the first 10' fallen. [See acrobatics rules.]

/cevah

toastedamphibian |

How about if the distance is zero. Pick a height, is the distance less than half of that DC? High jump DC. 10ft up and zero over is the same as 40 over. That puts you 10ft up and 20 out. The listed math sets all jumps at a virtual 45 degrees. Changing that requires the whole thing to be redone.

Half: so, say you want to get on a pillar 10ft up and 20ft away: DC 40

3/4: 10ft up and 15ft out. I would say DC 40. The DC to jump 10ft x 0 ft is 40, the DC to go 10ft by 20ft is 40. The DC to go 10ft x 15ft should fall between those numbers.

Edit:

Or 40 and 45 by Feauce's method.

toastedamphibian |

For the OP, jumping from a 40' building to a 30' building is essentially jumping down rather than falling down. The rules there indicate DC 15 to ignore damage from the first 10' fallen. [See acrobatics rules.]/cevah

It is not necessarily that simple. That 30ft building could be 10ft away, 100ft away, or a mile. Still need a way to calculate a jump DC.